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Fake proof that $1$ is the solution of $x^2+x+1=0$
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So I have this false proof and I am honestly confused why this is happening.
Consider $x^2+x+1=0$, then $x+1=-x^2$.
Now by simply dividing the equation by $x$ we get $x+1+1/x=0$. Substituting $x+1=-x^2$ we get $-x^2+1/x=0$ we get $x=1$ as a solution. Why is this substitution introducing a new solution, 1? Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution? Why?
I think this might be a really silly question but it is stomping me :P
fake-proofs
asked Jul 22 at 17:08
Sorfosh
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up vote
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So I have this false proof and I am honestly confused why this is happening.
Consider $x^2+x+1=0$, then $x+1=-x^2$.
Now by simply dividing the equation by $x$ we get $x+1+1/x=0$. Substituting $x+1=-x^2$ we get $-x^2+1/x=0$ we get $x=1$ as a solution. Why is this substitution introducing a new solution, 1? Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution? Why?
I think this might be a really silly question but it is stomping me :P
fake-proofs
asked Jul 22 at 17:08
Sorfosh
910616
5
$x^3-1=(x-1)(x^2+x+1)$
– Lord Shark the Unknown
Jul 22 at 17:09
I guess the problem is somewhere there that you rewrite an equation and then substitute something you have made out of the equation into the original.
– mrtaurho
Jul 22 at 17:11
@LordSharktheUnknown I am not sure how that answers my question, I am asking why is this happening.
– Sorfosh
Jul 22 at 17:12
4
@Sorfosh Actually you multiplied by $1-1/x$, but the principle is the same.
– Lord Shark the Unknown
Jul 22 at 17:13
1
You could make even worse examples. $x=-(x^2+1)implies x^2=x^4+2x^2+1$ which we could use to rewrite the original as $x^4+2x^2+x+2=0$. That one has four separate solutions, including $frac 12times (1pm sqrt -7)$. But all these rewrites do is to find other equations that our original roots must satisfy.
– lulu
Jul 22 at 17:21
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up vote
10
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up vote
10
down vote
favorite
So I have this false proof and I am honestly confused why this is happening.
Consider $x^2+x+1=0$, then $x+1=-x^2$.
Now by simply dividing the equation by $x$ we get $x+1+1/x=0$. Substituting $x+1=-x^2$ we get $-x^2+1/x=0$ we get $x=1$ as a solution. Why is this substitution introducing a new solution, 1? Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution? Why?
I think this might be a really silly question but it is stomping me :P
fake-proofs
asked Jul 22 at 17:08
Sorfosh
910616
So I have this false proof and I am honestly confused why this is happening.
Consider $x^2+x+1=0$, then $x+1=-x^2$.
Now by simply dividing the equation by $x$ we get $x+1+1/x=0$. Substituting $x+1=-x^2$ we get $-x^2+1/x=0$ we get $x=1$ as a solution. Why is this substitution introducing a new solution, 1? Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution? Why?
I think this might be a really silly question but it is stomping me :P
fake-proofs
asked Jul 22 at 17:08
Sorfosh
910616
asked Jul 22 at 17:08
Sorfosh
910616
asked Jul 22 at 17:08
Sorfosh
910616
asked Jul 22 at 17:08
Sorfosh
910616
910616
5
$x^3-1=(x-1)(x^2+x+1)$
– Lord Shark the Unknown
Jul 22 at 17:09
I guess the problem is somewhere there that you rewrite an equation and then substitute something you have made out of the equation into the original.
– mrtaurho
Jul 22 at 17:11
@LordSharktheUnknown I am not sure how that answers my question, I am asking why is this happening.
– Sorfosh
Jul 22 at 17:12
4
@Sorfosh Actually you multiplied by $1-1/x$, but the principle is the same.
– Lord Shark the Unknown
Jul 22 at 17:13
1
You could make even worse examples. $x=-(x^2+1)implies x^2=x^4+2x^2+1$ which we could use to rewrite the original as $x^4+2x^2+x+2=0$. That one has four separate solutions, including $frac 12times (1pm sqrt -7)$. But all these rewrites do is to find other equations that our original roots must satisfy.
– lulu
Jul 22 at 17:21
 |Â
show 2 more comments
5
$x^3-1=(x-1)(x^2+x+1)$
– Lord Shark the Unknown
Jul 22 at 17:09
I guess the problem is somewhere there that you rewrite an equation and then substitute something you have made out of the equation into the original.
– mrtaurho
Jul 22 at 17:11
@LordSharktheUnknown I am not sure how that answers my question, I am asking why is this happening.
– Sorfosh
Jul 22 at 17:12
4
@Sorfosh Actually you multiplied by $1-1/x$, but the principle is the same.
– Lord Shark the Unknown
Jul 22 at 17:13
1
You could make even worse examples. $x=-(x^2+1)implies x^2=x^4+2x^2+1$ which we could use to rewrite the original as $x^4+2x^2+x+2=0$. That one has four separate solutions, including $frac 12times (1pm sqrt -7)$. But all these rewrites do is to find other equations that our original roots must satisfy.
– lulu
Jul 22 at 17:21
5
5
$x^3-1=(x-1)(x^2+x+1)$
– Lord Shark the Unknown
Jul 22 at 17:09
$x^3-1=(x-1)(x^2+x+1)$
– Lord Shark the Unknown
Jul 22 at 17:09
I guess the problem is somewhere there that you rewrite an equation and then substitute something you have made out of the equation into the original.
– mrtaurho
Jul 22 at 17:11
I guess the problem is somewhere there that you rewrite an equation and then substitute something you have made out of the equation into the original.
– mrtaurho
Jul 22 at 17:11
@LordSharktheUnknown I am not sure how that answers my question, I am asking why is this happening.
– Sorfosh
Jul 22 at 17:12
@LordSharktheUnknown I am not sure how that answers my question, I am asking why is this happening.
– Sorfosh
Jul 22 at 17:12
4
4
@Sorfosh Actually you multiplied by $1-1/x$, but the principle is the same.
– Lord Shark the Unknown
Jul 22 at 17:13
@Sorfosh Actually you multiplied by $1-1/x$, but the principle is the same.
– Lord Shark the Unknown
Jul 22 at 17:13
1
1
You could make even worse examples. $x=-(x^2+1)implies x^2=x^4+2x^2+1$ which we could use to rewrite the original as $x^4+2x^2+x+2=0$. That one has four separate solutions, including $frac 12times (1pm sqrt -7)$. But all these rewrites do is to find other equations that our original roots must satisfy.
– lulu
Jul 22 at 17:21
You could make even worse examples. $x=-(x^2+1)implies x^2=x^4+2x^2+1$ which we could use to rewrite the original as $x^4+2x^2+x+2=0$. That one has four separate solutions, including $frac 12times (1pm sqrt -7)$. But all these rewrites do is to find other equations that our original roots must satisfy.
– lulu
Jul 22 at 17:21
 |Â
show 2 more comments
9 Answers
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Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes, exactly right. Substituting one equation in another has the potential to introduce more solutions, because you've thrown away information. Specifically, you threw away the original equations you were given. A solution to the original equation will be a solution to your new one, but the new one may have others.
For a silly example, consider
$$
x + x = x + 4.
$$
There is only one solution to the equation: $x = 4$. But suppose we substitute $x = 4$ on the left side for one of the $x$'s; then we get $x + 4 = x + 4$, or $4 = 4$, which has infinitely many solutions (every real number $x$).
In your example, you ended up with the equation $−x^2+frac1x=0$, which is equivalent to the equation $-x^3 + 1 = 0$ (assuming $x ne 0$), or $x^3 - 1 = 0$. This factors as $x^3 - 1 = (x - 1)(x^2 + x + 1)$ (as was pointed out in a comment), so you have introduced exactly 1 new solution.
answered Jul 22 at 17:17
6005
34.7k750122
1
@6004 I see, so how do we know which manipulations make us lose information and which do not? Clearly simple adding subtracting, or even multiplying by constants (But not by $x$ as that would introduce $x=0$ as a solution) leave the solutions alone. Is there a name for operations that conserve solutions?
– Sorfosh
Jul 22 at 17:23
2
@Sorfosh I think it's a good question. A good way to tell is, "can I undo this operation and go back to the original equation(s) from before?" For adding and subtracting from both sides, or multiplying by a nonzero constant, you can see that you can undo the operation. For multiplying by x, you cannot undo the operation, because if x is 0 you can't divide by it. For substitution, in general it cannot be undone. But if you substitute and keep the original equation, that is OK.
– 6005
Jul 22 at 17:29
Also, some substitution can be undone. It is common if you have two equations and two variables to substitute one equation in the other. For example if we have $x + y = 3$ and $y = 4$ we substitute $y = 4$ into the first equation. This can be undone even if we throw away the old first equation, but we need to keep around the second equation $y = 4$.
– 6005
Jul 22 at 17:30
@Sorfosh One important partial answer is: if $f$ is an injective operation, then doing $f$ to both sides of an equation gives an equivalent equation. For example, we can always add $x$ to both sides of an equation because, no matter what number $c$ you pick, the function $f(z)=z+c$ is injective, so - no matter what $x$ - is the "add $x$ to both sides" operation is undoable. By contrast, the function $f(z)=xcdot z$ isn't injective if $x=0$, so multiplying both sides of an equation by $x$ loses information unless we already know $xnot=0$.
– Noah Schweber
Jul 22 at 17:38
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The equation you have ended up with is not the same as the original. In fact, multiplying each side by $-x$, you get:
$$
x^3 - 1 = 0
$$
And it is clear that $1$ is a solution of this equation. The way you ended up with this fact is that, as Lord Shark pointed out, we can factor $x^3 - 1 = (x-1)(x^2+x+1)$, and you essentially showed that:
$$
x^2 + x+1 = 0
Rightarrow
(x-1)(x^2+x+1)=0
$$
Which is pretty clear when you restate it like that.
answered Jul 22 at 17:16
Sambo
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Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes. One important fact about mathematics is that when we manipulate an equation (or proposition, or ...), we can lose information along the way. E.g. from $$x^2=4$$ we can deduce $$0cdot x^2=0cdot 4,$$ and from that deduce $$0cdot x^2=0.$$ $x=17$ is certainly a solution to this new equation, but not the original equation. Basically, we haven't done anything wrong - if $x^2=4$ then $0cdot x^2=0$ - but we also haven't gotten an equivalent statement.
As a slightly less trivial example, from $$-y=4$$ we can deduce $$vert -yvert=vert 4vert,$$ or equivalently $$vert yvert=4;$$ $y=4$ satisfies this new equation, but not the original one. Again, we haven't made any mistakes, but we've lost information: the new equation $vert yvert=4$ tells us less than the old equation $-y=4$.
answered Jul 22 at 17:20
Noah Schweber
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You cheated yourself in a very interesting way! This is really illuminating!
When we solve equations, we usually just open a new line and write down some consequence of the condition. For example:
$$2x-3=2$$
$$2x=5$$
$$x=5/2$$
And then we make a note that the solution is $5/2$, without really thinking about what was happening. In fact, opening a new line and write an equation below the previous one has a very definite meaning: it is a consequence of the previous line. So usually, fake roots can sneak in, as we do not necessarily want to apply equivalent transformations, we only want to collect logical consequences of the equation. The strategy is that if we find a logical consequence that we can solve, then it simplyfies our task, as the solutions of the original equation must be among the solutions of the consequence. Throwing away extraneous roots are not so hard, after all...
Now back to your problem. You correctly deduced that if $x$ is a root, then it must also satisfy
$$E_1: x+1=-x^2$$
$$E_2: x+1+1/x=0$$
And then in particular, it also has to satisfy
$$E_3: -x^2+1/x=0$$
This is all true. But not in the other way around. Note that your original equation is equaivalent to $E_1$ which is also equivalent to $E_2$. You can also take the conjunction of two equivalent statements, to obtain an equivalent statement again. So $E_1wedge E_2$ is also equivalent to your original condition. However, $E_3$ is just a consequence of $E_1wedge E_2$, but from $E_3$, you cannot deduce $E_1$ or $E_2$ separately.
So long story short: not all your transformations were equivalent ones, hence extraneous roots can (and in this case did) occur.
answered Jul 22 at 17:23
A. Pongrácz
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The problem lies within an implicit assumption, i.e. that your equation holds.
First, we need to specify the domain of $x$. So let $xinmathbbR$.
We're now searching for all solutions to the equation $x^2+x+1=0$, i.e. the equations solution space. By substituting in a few random values for $x$, we can easily deduce that this solution space, let's call it $L$, is a strict subset of the real numbers $mathbbR$, i.e. $Lsubsetneq mathbbR$.
So while both $x+1+1/x =0$ and $x+1 = -x^2$ still have the same solution space $L$
(all actions taken to this point are invertible, as $x=0$ is no solution)
the key point is that $Lsubsetneq mathbbR$ means the equations don't hold for all values.
This means, as long as $xin L$, the equation $x+1 = -x^2$ holds, and we can substitute $x+1$ by $-x^2$.
However, for all $xnotin L$, the equation $x+1 = -x^2$ does not hold, and therefore, your substitution may create new, erroneous solutions.
answered Jul 22 at 18:31
Sudix
7911316
Thank you, Sudix, this is the only answer which really explains why substitution can introduce extraneous solutions.
– Mike Earnest
Jul 23 at 15:49
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LOGIC. Or maybe just GRAMMAR. $ ;x^2+x+1=0iff$ $ (xne 1land x^2+x+1=0)iff$ $ (xne 1land -x^2+1/x=0)iff$ $ (xne 1 land x^3=1).$
What you have shown is that $x^2+x+1=0implies x^3=1$ but you have NOT shown
that $ x^2+x+1=0iff x^3=1.$
SO $x^3=1$ does NOT "give a solution $x=1$" to the original equation. $x^3=1$ is only a consequence of $x^2+x+1=0$..
If you omit the "$implies$" or "$iff$", either in symbols or in words, in your writing, then you will lose track of whether you have $Aimplies B$ or $Bimplies A$ or $Aiff B.$
answered Jul 23 at 5:41
DanielWainfleet
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Substituting a function of $x$ in an equation obviously gives results other than desired.
For instance:
$$x-1=4$$
we have $$x=5$$ $implies$
$$x^2=25$$ $implies$
$$4=frac100x^2 tag1$$
Putting $(1)$ in original equation we get
$$x-1=frac100x^2$$ which leads to cubic giving you other solutions including your $5$
answered Jul 22 at 17:43
Ekaveera Kumar Sharma
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You need always to have in mind that "if there is a solution" then it should be 1 as you guessed. But this poly does not have real roots....
answered Jul 22 at 18:45
dmtri
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A tricky problem:
Let $x not =0$, real, henceforth :
1) Originally : $f(x) = x^2+x+1;$
2) $g(x) = (1/x)f(x) = x+1+1/x.$
Note:
a) $f(x)=0$, $g(x)=0$ have no real zeroes.
b) $f(x) not = g(x)$, if $x not =1$.
Consider the difference :
$d(x):=f(x)-g(x) =$
$ xg(x)-g(x)=(x-1)g(x)=$
$x^2-1/x=0.$
1) $x not =1$:
$d(x): =f(x)-g(x)=$
$ (x-1)g(x)=0$ , has no real zeroes.
2) $x=1$ :
$d(1)= f(1)-g(1)=0=$
$ 0g(0)=0$; where $g(0) not =0$.
Combining :
$d(x) = 0$ is equivalent to $g(x) = 0$, if $xnot =1$.
By considering $d(x)$ you have introduced an additional zero at $x=1$.(Lord Shark's comment)
answered Jul 22 at 20:45
Peter Szilas
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9 Answers
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Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes, exactly right. Substituting one equation in another has the potential to introduce more solutions, because you've thrown away information. Specifically, you threw away the original equations you were given. A solution to the original equation will be a solution to your new one, but the new one may have others.
For a silly example, consider
$$
x + x = x + 4.
$$
There is only one solution to the equation: $x = 4$. But suppose we substitute $x = 4$ on the left side for one of the $x$'s; then we get $x + 4 = x + 4$, or $4 = 4$, which has infinitely many solutions (every real number $x$).
In your example, you ended up with the equation $−x^2+frac1x=0$, which is equivalent to the equation $-x^3 + 1 = 0$ (assuming $x ne 0$), or $x^3 - 1 = 0$. This factors as $x^3 - 1 = (x - 1)(x^2 + x + 1)$ (as was pointed out in a comment), so you have introduced exactly 1 new solution.
answered Jul 22 at 17:17
6005
34.7k750122
1
@6004 I see, so how do we know which manipulations make us lose information and which do not? Clearly simple adding subtracting, or even multiplying by constants (But not by $x$ as that would introduce $x=0$ as a solution) leave the solutions alone. Is there a name for operations that conserve solutions?
– Sorfosh
Jul 22 at 17:23
2
@Sorfosh I think it's a good question. A good way to tell is, "can I undo this operation and go back to the original equation(s) from before?" For adding and subtracting from both sides, or multiplying by a nonzero constant, you can see that you can undo the operation. For multiplying by x, you cannot undo the operation, because if x is 0 you can't divide by it. For substitution, in general it cannot be undone. But if you substitute and keep the original equation, that is OK.
– 6005
Jul 22 at 17:29
Also, some substitution can be undone. It is common if you have two equations and two variables to substitute one equation in the other. For example if we have $x + y = 3$ and $y = 4$ we substitute $y = 4$ into the first equation. This can be undone even if we throw away the old first equation, but we need to keep around the second equation $y = 4$.
– 6005
Jul 22 at 17:30
@Sorfosh One important partial answer is: if $f$ is an injective operation, then doing $f$ to both sides of an equation gives an equivalent equation. For example, we can always add $x$ to both sides of an equation because, no matter what number $c$ you pick, the function $f(z)=z+c$ is injective, so - no matter what $x$ - is the "add $x$ to both sides" operation is undoable. By contrast, the function $f(z)=xcdot z$ isn't injective if $x=0$, so multiplying both sides of an equation by $x$ loses information unless we already know $xnot=0$.
– Noah Schweber
Jul 22 at 17:38
add a comment |Â
up vote
6
down vote
accepted
Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes, exactly right. Substituting one equation in another has the potential to introduce more solutions, because you've thrown away information. Specifically, you threw away the original equations you were given. A solution to the original equation will be a solution to your new one, but the new one may have others.
For a silly example, consider
$$
x + x = x + 4.
$$
There is only one solution to the equation: $x = 4$. But suppose we substitute $x = 4$ on the left side for one of the $x$'s; then we get $x + 4 = x + 4$, or $4 = 4$, which has infinitely many solutions (every real number $x$).
In your example, you ended up with the equation $−x^2+frac1x=0$, which is equivalent to the equation $-x^3 + 1 = 0$ (assuming $x ne 0$), or $x^3 - 1 = 0$. This factors as $x^3 - 1 = (x - 1)(x^2 + x + 1)$ (as was pointed out in a comment), so you have introduced exactly 1 new solution.
answered Jul 22 at 17:17
6005
34.7k750122
1
@6004 I see, so how do we know which manipulations make us lose information and which do not? Clearly simple adding subtracting, or even multiplying by constants (But not by $x$ as that would introduce $x=0$ as a solution) leave the solutions alone. Is there a name for operations that conserve solutions?
– Sorfosh
Jul 22 at 17:23
2
@Sorfosh I think it's a good question. A good way to tell is, "can I undo this operation and go back to the original equation(s) from before?" For adding and subtracting from both sides, or multiplying by a nonzero constant, you can see that you can undo the operation. For multiplying by x, you cannot undo the operation, because if x is 0 you can't divide by it. For substitution, in general it cannot be undone. But if you substitute and keep the original equation, that is OK.
– 6005
Jul 22 at 17:29
Also, some substitution can be undone. It is common if you have two equations and two variables to substitute one equation in the other. For example if we have $x + y = 3$ and $y = 4$ we substitute $y = 4$ into the first equation. This can be undone even if we throw away the old first equation, but we need to keep around the second equation $y = 4$.
– 6005
Jul 22 at 17:30
@Sorfosh One important partial answer is: if $f$ is an injective operation, then doing $f$ to both sides of an equation gives an equivalent equation. For example, we can always add $x$ to both sides of an equation because, no matter what number $c$ you pick, the function $f(z)=z+c$ is injective, so - no matter what $x$ - is the "add $x$ to both sides" operation is undoable. By contrast, the function $f(z)=xcdot z$ isn't injective if $x=0$, so multiplying both sides of an equation by $x$ loses information unless we already know $xnot=0$.
– Noah Schweber
Jul 22 at 17:38
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes, exactly right. Substituting one equation in another has the potential to introduce more solutions, because you've thrown away information. Specifically, you threw away the original equations you were given. A solution to the original equation will be a solution to your new one, but the new one may have others.
For a silly example, consider
$$
x + x = x + 4.
$$
There is only one solution to the equation: $x = 4$. But suppose we substitute $x = 4$ on the left side for one of the $x$'s; then we get $x + 4 = x + 4$, or $4 = 4$, which has infinitely many solutions (every real number $x$).
In your example, you ended up with the equation $−x^2+frac1x=0$, which is equivalent to the equation $-x^3 + 1 = 0$ (assuming $x ne 0$), or $x^3 - 1 = 0$. This factors as $x^3 - 1 = (x - 1)(x^2 + x + 1)$ (as was pointed out in a comment), so you have introduced exactly 1 new solution.
answered Jul 22 at 17:17
6005
34.7k750122
Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes, exactly right. Substituting one equation in another has the potential to introduce more solutions, because you've thrown away information. Specifically, you threw away the original equations you were given. A solution to the original equation will be a solution to your new one, but the new one may have others.
For a silly example, consider
$$
x + x = x + 4.
$$
There is only one solution to the equation: $x = 4$. But suppose we substitute $x = 4$ on the left side for one of the $x$'s; then we get $x + 4 = x + 4$, or $4 = 4$, which has infinitely many solutions (every real number $x$).
In your example, you ended up with the equation $−x^2+frac1x=0$, which is equivalent to the equation $-x^3 + 1 = 0$ (assuming $x ne 0$), or $x^3 - 1 = 0$. This factors as $x^3 - 1 = (x - 1)(x^2 + x + 1)$ (as was pointed out in a comment), so you have introduced exactly 1 new solution.
answered Jul 22 at 17:17
6005
34.7k750122
answered Jul 22 at 17:17
6005
34.7k750122
answered Jul 22 at 17:17
6005
34.7k750122
answered Jul 22 at 17:17
6005
34.7k750122
34.7k750122
1
@6004 I see, so how do we know which manipulations make us lose information and which do not? Clearly simple adding subtracting, or even multiplying by constants (But not by $x$ as that would introduce $x=0$ as a solution) leave the solutions alone. Is there a name for operations that conserve solutions?
– Sorfosh
Jul 22 at 17:23
2
@Sorfosh I think it's a good question. A good way to tell is, "can I undo this operation and go back to the original equation(s) from before?" For adding and subtracting from both sides, or multiplying by a nonzero constant, you can see that you can undo the operation. For multiplying by x, you cannot undo the operation, because if x is 0 you can't divide by it. For substitution, in general it cannot be undone. But if you substitute and keep the original equation, that is OK.
– 6005
Jul 22 at 17:29
Also, some substitution can be undone. It is common if you have two equations and two variables to substitute one equation in the other. For example if we have $x + y = 3$ and $y = 4$ we substitute $y = 4$ into the first equation. This can be undone even if we throw away the old first equation, but we need to keep around the second equation $y = 4$.
– 6005
Jul 22 at 17:30
@Sorfosh One important partial answer is: if $f$ is an injective operation, then doing $f$ to both sides of an equation gives an equivalent equation. For example, we can always add $x$ to both sides of an equation because, no matter what number $c$ you pick, the function $f(z)=z+c$ is injective, so - no matter what $x$ - is the "add $x$ to both sides" operation is undoable. By contrast, the function $f(z)=xcdot z$ isn't injective if $x=0$, so multiplying both sides of an equation by $x$ loses information unless we already know $xnot=0$.
– Noah Schweber
Jul 22 at 17:38
add a comment |Â
1
@6004 I see, so how do we know which manipulations make us lose information and which do not? Clearly simple adding subtracting, or even multiplying by constants (But not by $x$ as that would introduce $x=0$ as a solution) leave the solutions alone. Is there a name for operations that conserve solutions?
– Sorfosh
Jul 22 at 17:23
2
@Sorfosh I think it's a good question. A good way to tell is, "can I undo this operation and go back to the original equation(s) from before?" For adding and subtracting from both sides, or multiplying by a nonzero constant, you can see that you can undo the operation. For multiplying by x, you cannot undo the operation, because if x is 0 you can't divide by it. For substitution, in general it cannot be undone. But if you substitute and keep the original equation, that is OK.
– 6005
Jul 22 at 17:29
Also, some substitution can be undone. It is common if you have two equations and two variables to substitute one equation in the other. For example if we have $x + y = 3$ and $y = 4$ we substitute $y = 4$ into the first equation. This can be undone even if we throw away the old first equation, but we need to keep around the second equation $y = 4$.
– 6005
Jul 22 at 17:30
@Sorfosh One important partial answer is: if $f$ is an injective operation, then doing $f$ to both sides of an equation gives an equivalent equation. For example, we can always add $x$ to both sides of an equation because, no matter what number $c$ you pick, the function $f(z)=z+c$ is injective, so - no matter what $x$ - is the "add $x$ to both sides" operation is undoable. By contrast, the function $f(z)=xcdot z$ isn't injective if $x=0$, so multiplying both sides of an equation by $x$ loses information unless we already know $xnot=0$.
– Noah Schweber
Jul 22 at 17:38
1
1
@6004 I see, so how do we know which manipulations make us lose information and which do not? Clearly simple adding subtracting, or even multiplying by constants (But not by $x$ as that would introduce $x=0$ as a solution) leave the solutions alone. Is there a name for operations that conserve solutions?
– Sorfosh
Jul 22 at 17:23
@6004 I see, so how do we know which manipulations make us lose information and which do not? Clearly simple adding subtracting, or even multiplying by constants (But not by $x$ as that would introduce $x=0$ as a solution) leave the solutions alone. Is there a name for operations that conserve solutions?
– Sorfosh
Jul 22 at 17:23
2
2
@Sorfosh I think it's a good question. A good way to tell is, "can I undo this operation and go back to the original equation(s) from before?" For adding and subtracting from both sides, or multiplying by a nonzero constant, you can see that you can undo the operation. For multiplying by x, you cannot undo the operation, because if x is 0 you can't divide by it. For substitution, in general it cannot be undone. But if you substitute and keep the original equation, that is OK.
– 6005
Jul 22 at 17:29
@Sorfosh I think it's a good question. A good way to tell is, "can I undo this operation and go back to the original equation(s) from before?" For adding and subtracting from both sides, or multiplying by a nonzero constant, you can see that you can undo the operation. For multiplying by x, you cannot undo the operation, because if x is 0 you can't divide by it. For substitution, in general it cannot be undone. But if you substitute and keep the original equation, that is OK.
– 6005
Jul 22 at 17:29
Also, some substitution can be undone. It is common if you have two equations and two variables to substitute one equation in the other. For example if we have $x + y = 3$ and $y = 4$ we substitute $y = 4$ into the first equation. This can be undone even if we throw away the old first equation, but we need to keep around the second equation $y = 4$.
– 6005
Jul 22 at 17:30
Also, some substitution can be undone. It is common if you have two equations and two variables to substitute one equation in the other. For example if we have $x + y = 3$ and $y = 4$ we substitute $y = 4$ into the first equation. This can be undone even if we throw away the old first equation, but we need to keep around the second equation $y = 4$.
– 6005
Jul 22 at 17:30
@Sorfosh One important partial answer is: if $f$ is an injective operation, then doing $f$ to both sides of an equation gives an equivalent equation. For example, we can always add $x$ to both sides of an equation because, no matter what number $c$ you pick, the function $f(z)=z+c$ is injective, so - no matter what $x$ - is the "add $x$ to both sides" operation is undoable. By contrast, the function $f(z)=xcdot z$ isn't injective if $x=0$, so multiplying both sides of an equation by $x$ loses information unless we already know $xnot=0$.
– Noah Schweber
Jul 22 at 17:38
@Sorfosh One important partial answer is: if $f$ is an injective operation, then doing $f$ to both sides of an equation gives an equivalent equation. For example, we can always add $x$ to both sides of an equation because, no matter what number $c$ you pick, the function $f(z)=z+c$ is injective, so - no matter what $x$ - is the "add $x$ to both sides" operation is undoable. By contrast, the function $f(z)=xcdot z$ isn't injective if $x=0$, so multiplying both sides of an equation by $x$ loses information unless we already know $xnot=0$.
– Noah Schweber
Jul 22 at 17:38
add a comment |Â
up vote
4
down vote
The equation you have ended up with is not the same as the original. In fact, multiplying each side by $-x$, you get:
$$
x^3 - 1 = 0
$$
And it is clear that $1$ is a solution of this equation. The way you ended up with this fact is that, as Lord Shark pointed out, we can factor $x^3 - 1 = (x-1)(x^2+x+1)$, and you essentially showed that:
$$
x^2 + x+1 = 0
Rightarrow
(x-1)(x^2+x+1)=0
$$
Which is pretty clear when you restate it like that.
answered Jul 22 at 17:16
Sambo
1,2561427
add a comment |Â
up vote
4
down vote
The equation you have ended up with is not the same as the original. In fact, multiplying each side by $-x$, you get:
$$
x^3 - 1 = 0
$$
And it is clear that $1$ is a solution of this equation. The way you ended up with this fact is that, as Lord Shark pointed out, we can factor $x^3 - 1 = (x-1)(x^2+x+1)$, and you essentially showed that:
$$
x^2 + x+1 = 0
Rightarrow
(x-1)(x^2+x+1)=0
$$
Which is pretty clear when you restate it like that.
answered Jul 22 at 17:16
Sambo
1,2561427
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The equation you have ended up with is not the same as the original. In fact, multiplying each side by $-x$, you get:
$$
x^3 - 1 = 0
$$
And it is clear that $1$ is a solution of this equation. The way you ended up with this fact is that, as Lord Shark pointed out, we can factor $x^3 - 1 = (x-1)(x^2+x+1)$, and you essentially showed that:
$$
x^2 + x+1 = 0
Rightarrow
(x-1)(x^2+x+1)=0
$$
Which is pretty clear when you restate it like that.
answered Jul 22 at 17:16
Sambo
1,2561427
The equation you have ended up with is not the same as the original. In fact, multiplying each side by $-x$, you get:
$$
x^3 - 1 = 0
$$
And it is clear that $1$ is a solution of this equation. The way you ended up with this fact is that, as Lord Shark pointed out, we can factor $x^3 - 1 = (x-1)(x^2+x+1)$, and you essentially showed that:
$$
x^2 + x+1 = 0
Rightarrow
(x-1)(x^2+x+1)=0
$$
Which is pretty clear when you restate it like that.
answered Jul 22 at 17:16
Sambo
1,2561427
answered Jul 22 at 17:16
Sambo
1,2561427
answered Jul 22 at 17:16
Sambo
1,2561427
answered Jul 22 at 17:16
Sambo
1,2561427
1,2561427
add a comment |Â
add a comment |Â
up vote
3
down vote
Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes. One important fact about mathematics is that when we manipulate an equation (or proposition, or ...), we can lose information along the way. E.g. from $$x^2=4$$ we can deduce $$0cdot x^2=0cdot 4,$$ and from that deduce $$0cdot x^2=0.$$ $x=17$ is certainly a solution to this new equation, but not the original equation. Basically, we haven't done anything wrong - if $x^2=4$ then $0cdot x^2=0$ - but we also haven't gotten an equivalent statement.
As a slightly less trivial example, from $$-y=4$$ we can deduce $$vert -yvert=vert 4vert,$$ or equivalently $$vert yvert=4;$$ $y=4$ satisfies this new equation, but not the original one. Again, we haven't made any mistakes, but we've lost information: the new equation $vert yvert=4$ tells us less than the old equation $-y=4$.
answered Jul 22 at 17:20
Noah Schweber
111k9140261
add a comment |Â
up vote
3
down vote
Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes. One important fact about mathematics is that when we manipulate an equation (or proposition, or ...), we can lose information along the way. E.g. from $$x^2=4$$ we can deduce $$0cdot x^2=0cdot 4,$$ and from that deduce $$0cdot x^2=0.$$ $x=17$ is certainly a solution to this new equation, but not the original equation. Basically, we haven't done anything wrong - if $x^2=4$ then $0cdot x^2=0$ - but we also haven't gotten an equivalent statement.
As a slightly less trivial example, from $$-y=4$$ we can deduce $$vert -yvert=vert 4vert,$$ or equivalently $$vert yvert=4;$$ $y=4$ satisfies this new equation, but not the original one. Again, we haven't made any mistakes, but we've lost information: the new equation $vert yvert=4$ tells us less than the old equation $-y=4$.
answered Jul 22 at 17:20
Noah Schweber
111k9140261
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes. One important fact about mathematics is that when we manipulate an equation (or proposition, or ...), we can lose information along the way. E.g. from $$x^2=4$$ we can deduce $$0cdot x^2=0cdot 4,$$ and from that deduce $$0cdot x^2=0.$$ $x=17$ is certainly a solution to this new equation, but not the original equation. Basically, we haven't done anything wrong - if $x^2=4$ then $0cdot x^2=0$ - but we also haven't gotten an equivalent statement.
As a slightly less trivial example, from $$-y=4$$ we can deduce $$vert -yvert=vert 4vert,$$ or equivalently $$vert yvert=4;$$ $y=4$ satisfies this new equation, but not the original one. Again, we haven't made any mistakes, but we've lost information: the new equation $vert yvert=4$ tells us less than the old equation $-y=4$.
answered Jul 22 at 17:20
Noah Schweber
111k9140261
Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?
Yes. One important fact about mathematics is that when we manipulate an equation (or proposition, or ...), we can lose information along the way. E.g. from $$x^2=4$$ we can deduce $$0cdot x^2=0cdot 4,$$ and from that deduce $$0cdot x^2=0.$$ $x=17$ is certainly a solution to this new equation, but not the original equation. Basically, we haven't done anything wrong - if $x^2=4$ then $0cdot x^2=0$ - but we also haven't gotten an equivalent statement.
As a slightly less trivial example, from $$-y=4$$ we can deduce $$vert -yvert=vert 4vert,$$ or equivalently $$vert yvert=4;$$ $y=4$ satisfies this new equation, but not the original one. Again, we haven't made any mistakes, but we've lost information: the new equation $vert yvert=4$ tells us less than the old equation $-y=4$.
answered Jul 22 at 17:20
Noah Schweber
111k9140261
answered Jul 22 at 17:20
Noah Schweber
111k9140261
answered Jul 22 at 17:20
Noah Schweber
111k9140261
answered Jul 22 at 17:20
Noah Schweber
111k9140261
111k9140261
add a comment |Â
add a comment |Â
up vote
2
down vote
You cheated yourself in a very interesting way! This is really illuminating!
When we solve equations, we usually just open a new line and write down some consequence of the condition. For example:
$$2x-3=2$$
$$2x=5$$
$$x=5/2$$
And then we make a note that the solution is $5/2$, without really thinking about what was happening. In fact, opening a new line and write an equation below the previous one has a very definite meaning: it is a consequence of the previous line. So usually, fake roots can sneak in, as we do not necessarily want to apply equivalent transformations, we only want to collect logical consequences of the equation. The strategy is that if we find a logical consequence that we can solve, then it simplyfies our task, as the solutions of the original equation must be among the solutions of the consequence. Throwing away extraneous roots are not so hard, after all...
Now back to your problem. You correctly deduced that if $x$ is a root, then it must also satisfy
$$E_1: x+1=-x^2$$
$$E_2: x+1+1/x=0$$
And then in particular, it also has to satisfy
$$E_3: -x^2+1/x=0$$
This is all true. But not in the other way around. Note that your original equation is equaivalent to $E_1$ which is also equivalent to $E_2$. You can also take the conjunction of two equivalent statements, to obtain an equivalent statement again. So $E_1wedge E_2$ is also equivalent to your original condition. However, $E_3$ is just a consequence of $E_1wedge E_2$, but from $E_3$, you cannot deduce $E_1$ or $E_2$ separately.
So long story short: not all your transformations were equivalent ones, hence extraneous roots can (and in this case did) occur.
answered Jul 22 at 17:23
A. Pongrácz
2,079120
add a comment |Â
up vote
2
down vote
You cheated yourself in a very interesting way! This is really illuminating!
When we solve equations, we usually just open a new line and write down some consequence of the condition. For example:
$$2x-3=2$$
$$2x=5$$
$$x=5/2$$
And then we make a note that the solution is $5/2$, without really thinking about what was happening. In fact, opening a new line and write an equation below the previous one has a very definite meaning: it is a consequence of the previous line. So usually, fake roots can sneak in, as we do not necessarily want to apply equivalent transformations, we only want to collect logical consequences of the equation. The strategy is that if we find a logical consequence that we can solve, then it simplyfies our task, as the solutions of the original equation must be among the solutions of the consequence. Throwing away extraneous roots are not so hard, after all...
Now back to your problem. You correctly deduced that if $x$ is a root, then it must also satisfy
$$E_1: x+1=-x^2$$
$$E_2: x+1+1/x=0$$
And then in particular, it also has to satisfy
$$E_3: -x^2+1/x=0$$
This is all true. But not in the other way around. Note that your original equation is equaivalent to $E_1$ which is also equivalent to $E_2$. You can also take the conjunction of two equivalent statements, to obtain an equivalent statement again. So $E_1wedge E_2$ is also equivalent to your original condition. However, $E_3$ is just a consequence of $E_1wedge E_2$, but from $E_3$, you cannot deduce $E_1$ or $E_2$ separately.
So long story short: not all your transformations were equivalent ones, hence extraneous roots can (and in this case did) occur.
answered Jul 22 at 17:23
A. Pongrácz
2,079120
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You cheated yourself in a very interesting way! This is really illuminating!
When we solve equations, we usually just open a new line and write down some consequence of the condition. For example:
$$2x-3=2$$
$$2x=5$$
$$x=5/2$$
And then we make a note that the solution is $5/2$, without really thinking about what was happening. In fact, opening a new line and write an equation below the previous one has a very definite meaning: it is a consequence of the previous line. So usually, fake roots can sneak in, as we do not necessarily want to apply equivalent transformations, we only want to collect logical consequences of the equation. The strategy is that if we find a logical consequence that we can solve, then it simplyfies our task, as the solutions of the original equation must be among the solutions of the consequence. Throwing away extraneous roots are not so hard, after all...
Now back to your problem. You correctly deduced that if $x$ is a root, then it must also satisfy
$$E_1: x+1=-x^2$$
$$E_2: x+1+1/x=0$$
And then in particular, it also has to satisfy
$$E_3: -x^2+1/x=0$$
This is all true. But not in the other way around. Note that your original equation is equaivalent to $E_1$ which is also equivalent to $E_2$. You can also take the conjunction of two equivalent statements, to obtain an equivalent statement again. So $E_1wedge E_2$ is also equivalent to your original condition. However, $E_3$ is just a consequence of $E_1wedge E_2$, but from $E_3$, you cannot deduce $E_1$ or $E_2$ separately.
So long story short: not all your transformations were equivalent ones, hence extraneous roots can (and in this case did) occur.
answered Jul 22 at 17:23
A. Pongrácz
2,079120
You cheated yourself in a very interesting way! This is really illuminating!
When we solve equations, we usually just open a new line and write down some consequence of the condition. For example:
$$2x-3=2$$
$$2x=5$$
$$x=5/2$$
And then we make a note that the solution is $5/2$, without really thinking about what was happening. In fact, opening a new line and write an equation below the previous one has a very definite meaning: it is a consequence of the previous line. So usually, fake roots can sneak in, as we do not necessarily want to apply equivalent transformations, we only want to collect logical consequences of the equation. The strategy is that if we find a logical consequence that we can solve, then it simplyfies our task, as the solutions of the original equation must be among the solutions of the consequence. Throwing away extraneous roots are not so hard, after all...
Now back to your problem. You correctly deduced that if $x$ is a root, then it must also satisfy
$$E_1: x+1=-x^2$$
$$E_2: x+1+1/x=0$$
And then in particular, it also has to satisfy
$$E_3: -x^2+1/x=0$$
This is all true. But not in the other way around. Note that your original equation is equaivalent to $E_1$ which is also equivalent to $E_2$. You can also take the conjunction of two equivalent statements, to obtain an equivalent statement again. So $E_1wedge E_2$ is also equivalent to your original condition. However, $E_3$ is just a consequence of $E_1wedge E_2$, but from $E_3$, you cannot deduce $E_1$ or $E_2$ separately.
So long story short: not all your transformations were equivalent ones, hence extraneous roots can (and in this case did) occur.
answered Jul 22 at 17:23
A. Pongrácz
2,079120
answered Jul 22 at 17:23
A. Pongrácz
2,079120
answered Jul 22 at 17:23
A. Pongrácz
2,079120
answered Jul 22 at 17:23
A. Pongrácz
2,079120
2,079120
add a comment |Â
add a comment |Â
up vote
1
down vote
The problem lies within an implicit assumption, i.e. that your equation holds.
First, we need to specify the domain of $x$. So let $xinmathbbR$.
We're now searching for all solutions to the equation $x^2+x+1=0$, i.e. the equations solution space. By substituting in a few random values for $x$, we can easily deduce that this solution space, let's call it $L$, is a strict subset of the real numbers $mathbbR$, i.e. $Lsubsetneq mathbbR$.
So while both $x+1+1/x =0$ and $x+1 = -x^2$ still have the same solution space $L$
(all actions taken to this point are invertible, as $x=0$ is no solution)
the key point is that $Lsubsetneq mathbbR$ means the equations don't hold for all values.
This means, as long as $xin L$, the equation $x+1 = -x^2$ holds, and we can substitute $x+1$ by $-x^2$.
However, for all $xnotin L$, the equation $x+1 = -x^2$ does not hold, and therefore, your substitution may create new, erroneous solutions.
answered Jul 22 at 18:31
Sudix
7911316
Thank you, Sudix, this is the only answer which really explains why substitution can introduce extraneous solutions.
– Mike Earnest
Jul 23 at 15:49
add a comment |Â
up vote
1
down vote
The problem lies within an implicit assumption, i.e. that your equation holds.
First, we need to specify the domain of $x$. So let $xinmathbbR$.
We're now searching for all solutions to the equation $x^2+x+1=0$, i.e. the equations solution space. By substituting in a few random values for $x$, we can easily deduce that this solution space, let's call it $L$, is a strict subset of the real numbers $mathbbR$, i.e. $Lsubsetneq mathbbR$.
So while both $x+1+1/x =0$ and $x+1 = -x^2$ still have the same solution space $L$
(all actions taken to this point are invertible, as $x=0$ is no solution)
the key point is that $Lsubsetneq mathbbR$ means the equations don't hold for all values.
This means, as long as $xin L$, the equation $x+1 = -x^2$ holds, and we can substitute $x+1$ by $-x^2$.
However, for all $xnotin L$, the equation $x+1 = -x^2$ does not hold, and therefore, your substitution may create new, erroneous solutions.
answered Jul 22 at 18:31
Sudix
7911316
Thank you, Sudix, this is the only answer which really explains why substitution can introduce extraneous solutions.
– Mike Earnest
Jul 23 at 15:49
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The problem lies within an implicit assumption, i.e. that your equation holds.
First, we need to specify the domain of $x$. So let $xinmathbbR$.
We're now searching for all solutions to the equation $x^2+x+1=0$, i.e. the equations solution space. By substituting in a few random values for $x$, we can easily deduce that this solution space, let's call it $L$, is a strict subset of the real numbers $mathbbR$, i.e. $Lsubsetneq mathbbR$.
So while both $x+1+1/x =0$ and $x+1 = -x^2$ still have the same solution space $L$
(all actions taken to this point are invertible, as $x=0$ is no solution)
the key point is that $Lsubsetneq mathbbR$ means the equations don't hold for all values.
This means, as long as $xin L$, the equation $x+1 = -x^2$ holds, and we can substitute $x+1$ by $-x^2$.
However, for all $xnotin L$, the equation $x+1 = -x^2$ does not hold, and therefore, your substitution may create new, erroneous solutions.
answered Jul 22 at 18:31
Sudix
7911316
The problem lies within an implicit assumption, i.e. that your equation holds.
First, we need to specify the domain of $x$. So let $xinmathbbR$.
We're now searching for all solutions to the equation $x^2+x+1=0$, i.e. the equations solution space. By substituting in a few random values for $x$, we can easily deduce that this solution space, let's call it $L$, is a strict subset of the real numbers $mathbbR$, i.e. $Lsubsetneq mathbbR$.
So while both $x+1+1/x =0$ and $x+1 = -x^2$ still have the same solution space $L$
(all actions taken to this point are invertible, as $x=0$ is no solution)
the key point is that $Lsubsetneq mathbbR$ means the equations don't hold for all values.
This means, as long as $xin L$, the equation $x+1 = -x^2$ holds, and we can substitute $x+1$ by $-x^2$.
However, for all $xnotin L$, the equation $x+1 = -x^2$ does not hold, and therefore, your substitution may create new, erroneous solutions.
answered Jul 22 at 18:31
Sudix
7911316
edited Jul 22 at 18:56
answered Jul 22 at 18:31
Sudix
7911316
answered Jul 22 at 18:31
Sudix
7911316
answered Jul 22 at 18:31
Sudix
7911316
7911316
Thank you, Sudix, this is the only answer which really explains why substitution can introduce extraneous solutions.
– Mike Earnest
Jul 23 at 15:49
add a comment |Â
Thank you, Sudix, this is the only answer which really explains why substitution can introduce extraneous solutions.
– Mike Earnest
Jul 23 at 15:49
Thank you, Sudix, this is the only answer which really explains why substitution can introduce extraneous solutions.
– Mike Earnest
Jul 23 at 15:49
Thank you, Sudix, this is the only answer which really explains why substitution can introduce extraneous solutions.
– Mike Earnest
Jul 23 at 15:49
add a comment |Â
up vote
1
down vote
LOGIC. Or maybe just GRAMMAR. $ ;x^2+x+1=0iff$ $ (xne 1land x^2+x+1=0)iff$ $ (xne 1land -x^2+1/x=0)iff$ $ (xne 1 land x^3=1).$
What you have shown is that $x^2+x+1=0implies x^3=1$ but you have NOT shown
that $ x^2+x+1=0iff x^3=1.$
SO $x^3=1$ does NOT "give a solution $x=1$" to the original equation. $x^3=1$ is only a consequence of $x^2+x+1=0$..
If you omit the "$implies$" or "$iff$", either in symbols or in words, in your writing, then you will lose track of whether you have $Aimplies B$ or $Bimplies A$ or $Aiff B.$
answered Jul 23 at 5:41
DanielWainfleet
31.6k31543
add a comment |Â
up vote
1
down vote
LOGIC. Or maybe just GRAMMAR. $ ;x^2+x+1=0iff$ $ (xne 1land x^2+x+1=0)iff$ $ (xne 1land -x^2+1/x=0)iff$ $ (xne 1 land x^3=1).$
What you have shown is that $x^2+x+1=0implies x^3=1$ but you have NOT shown
that $ x^2+x+1=0iff x^3=1.$
SO $x^3=1$ does NOT "give a solution $x=1$" to the original equation. $x^3=1$ is only a consequence of $x^2+x+1=0$..
If you omit the "$implies$" or "$iff$", either in symbols or in words, in your writing, then you will lose track of whether you have $Aimplies B$ or $Bimplies A$ or $Aiff B.$
answered Jul 23 at 5:41
DanielWainfleet
31.6k31543
add a comment |Â
up vote
1
down vote
up vote
1
down vote
LOGIC. Or maybe just GRAMMAR. $ ;x^2+x+1=0iff$ $ (xne 1land x^2+x+1=0)iff$ $ (xne 1land -x^2+1/x=0)iff$ $ (xne 1 land x^3=1).$
What you have shown is that $x^2+x+1=0implies x^3=1$ but you have NOT shown
that $ x^2+x+1=0iff x^3=1.$
SO $x^3=1$ does NOT "give a solution $x=1$" to the original equation. $x^3=1$ is only a consequence of $x^2+x+1=0$..
If you omit the "$implies$" or "$iff$", either in symbols or in words, in your writing, then you will lose track of whether you have $Aimplies B$ or $Bimplies A$ or $Aiff B.$
answered Jul 23 at 5:41
DanielWainfleet
31.6k31543
LOGIC. Or maybe just GRAMMAR. $ ;x^2+x+1=0iff$ $ (xne 1land x^2+x+1=0)iff$ $ (xne 1land -x^2+1/x=0)iff$ $ (xne 1 land x^3=1).$
What you have shown is that $x^2+x+1=0implies x^3=1$ but you have NOT shown
that $ x^2+x+1=0iff x^3=1.$
SO $x^3=1$ does NOT "give a solution $x=1$" to the original equation. $x^3=1$ is only a consequence of $x^2+x+1=0$..
If you omit the "$implies$" or "$iff$", either in symbols or in words, in your writing, then you will lose track of whether you have $Aimplies B$ or $Bimplies A$ or $Aiff B.$
answered Jul 23 at 5:41
DanielWainfleet
31.6k31543
answered Jul 23 at 5:41
DanielWainfleet
31.6k31543
answered Jul 23 at 5:41
DanielWainfleet
31.6k31543
answered Jul 23 at 5:41
DanielWainfleet
31.6k31543
31.6k31543
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Substituting a function of $x$ in an equation obviously gives results other than desired.
For instance:
$$x-1=4$$
we have $$x=5$$ $implies$
$$x^2=25$$ $implies$
$$4=frac100x^2 tag1$$
Putting $(1)$ in original equation we get
$$x-1=frac100x^2$$ which leads to cubic giving you other solutions including your $5$
answered Jul 22 at 17:43
Ekaveera Kumar Sharma
5,15611122
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up vote
0
down vote
Substituting a function of $x$ in an equation obviously gives results other than desired.
For instance:
$$x-1=4$$
we have $$x=5$$ $implies$
$$x^2=25$$ $implies$
$$4=frac100x^2 tag1$$
Putting $(1)$ in original equation we get
$$x-1=frac100x^2$$ which leads to cubic giving you other solutions including your $5$
answered Jul 22 at 17:43
Ekaveera Kumar Sharma
5,15611122
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Substituting a function of $x$ in an equation obviously gives results other than desired.
For instance:
$$x-1=4$$
we have $$x=5$$ $implies$
$$x^2=25$$ $implies$
$$4=frac100x^2 tag1$$
Putting $(1)$ in original equation we get
$$x-1=frac100x^2$$ which leads to cubic giving you other solutions including your $5$
answered Jul 22 at 17:43
Ekaveera Kumar Sharma
5,15611122
Substituting a function of $x$ in an equation obviously gives results other than desired.
For instance:
$$x-1=4$$
we have $$x=5$$ $implies$
$$x^2=25$$ $implies$
$$4=frac100x^2 tag1$$
Putting $(1)$ in original equation we get
$$x-1=frac100x^2$$ which leads to cubic giving you other solutions including your $5$
answered Jul 22 at 17:43
Ekaveera Kumar Sharma
5,15611122
answered Jul 22 at 17:43
Ekaveera Kumar Sharma
5,15611122
answered Jul 22 at 17:43
Ekaveera Kumar Sharma
5,15611122
answered Jul 22 at 17:43
Ekaveera Kumar Sharma
5,15611122
5,15611122
add a comment |Â
add a comment |Â
up vote
0
down vote
You need always to have in mind that "if there is a solution" then it should be 1 as you guessed. But this poly does not have real roots....
answered Jul 22 at 18:45
dmtri
315213
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0
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You need always to have in mind that "if there is a solution" then it should be 1 as you guessed. But this poly does not have real roots....
answered Jul 22 at 18:45
dmtri
315213
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up vote
0
down vote
up vote
0
down vote
You need always to have in mind that "if there is a solution" then it should be 1 as you guessed. But this poly does not have real roots....
answered Jul 22 at 18:45
dmtri
315213
You need always to have in mind that "if there is a solution" then it should be 1 as you guessed. But this poly does not have real roots....
answered Jul 22 at 18:45
dmtri
315213
answered Jul 22 at 18:45
dmtri
315213
answered Jul 22 at 18:45
dmtri
315213
answered Jul 22 at 18:45
dmtri
315213
315213
add a comment |Â
add a comment |Â
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0
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A tricky problem:
Let $x not =0$, real, henceforth :
1) Originally : $f(x) = x^2+x+1;$
2) $g(x) = (1/x)f(x) = x+1+1/x.$
Note:
a) $f(x)=0$, $g(x)=0$ have no real zeroes.
b) $f(x) not = g(x)$, if $x not =1$.
Consider the difference :
$d(x):=f(x)-g(x) =$
$ xg(x)-g(x)=(x-1)g(x)=$
$x^2-1/x=0.$
1) $x not =1$:
$d(x): =f(x)-g(x)=$
$ (x-1)g(x)=0$ , has no real zeroes.
2) $x=1$ :
$d(1)= f(1)-g(1)=0=$
$ 0g(0)=0$; where $g(0) not =0$.
Combining :
$d(x) = 0$ is equivalent to $g(x) = 0$, if $xnot =1$.
By considering $d(x)$ you have introduced an additional zero at $x=1$.(Lord Shark's comment)
answered Jul 22 at 20:45
Peter Szilas
7,9252617
add a comment |Â
up vote
0
down vote
A tricky problem:
Let $x not =0$, real, henceforth :
1) Originally : $f(x) = x^2+x+1;$
2) $g(x) = (1/x)f(x) = x+1+1/x.$
Note:
a) $f(x)=0$, $g(x)=0$ have no real zeroes.
b) $f(x) not = g(x)$, if $x not =1$.
Consider the difference :
$d(x):=f(x)-g(x) =$
$ xg(x)-g(x)=(x-1)g(x)=$
$x^2-1/x=0.$
1) $x not =1$:
$d(x): =f(x)-g(x)=$
$ (x-1)g(x)=0$ , has no real zeroes.
2) $x=1$ :
$d(1)= f(1)-g(1)=0=$
$ 0g(0)=0$; where $g(0) not =0$.
Combining :
$d(x) = 0$ is equivalent to $g(x) = 0$, if $xnot =1$.
By considering $d(x)$ you have introduced an additional zero at $x=1$.(Lord Shark's comment)
answered Jul 22 at 20:45
Peter Szilas
7,9252617
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A tricky problem:
Let $x not =0$, real, henceforth :
1) Originally : $f(x) = x^2+x+1;$
2) $g(x) = (1/x)f(x) = x+1+1/x.$
Note:
a) $f(x)=0$, $g(x)=0$ have no real zeroes.
b) $f(x) not = g(x)$, if $x not =1$.
Consider the difference :
$d(x):=f(x)-g(x) =$
$ xg(x)-g(x)=(x-1)g(x)=$
$x^2-1/x=0.$
1) $x not =1$:
$d(x): =f(x)-g(x)=$
$ (x-1)g(x)=0$ , has no real zeroes.
2) $x=1$ :
$d(1)= f(1)-g(1)=0=$
$ 0g(0)=0$; where $g(0) not =0$.
Combining :
$d(x) = 0$ is equivalent to $g(x) = 0$, if $xnot =1$.
By considering $d(x)$ you have introduced an additional zero at $x=1$.(Lord Shark's comment)
answered Jul 22 at 20:45
Peter Szilas
7,9252617
A tricky problem:
Let $x not =0$, real, henceforth :
1) Originally : $f(x) = x^2+x+1;$
2) $g(x) = (1/x)f(x) = x+1+1/x.$
Note:
a) $f(x)=0$, $g(x)=0$ have no real zeroes.
b) $f(x) not = g(x)$, if $x not =1$.
Consider the difference :
$d(x):=f(x)-g(x) =$
$ xg(x)-g(x)=(x-1)g(x)=$
$x^2-1/x=0.$
1) $x not =1$:
$d(x): =f(x)-g(x)=$
$ (x-1)g(x)=0$ , has no real zeroes.
2) $x=1$ :
$d(1)= f(1)-g(1)=0=$
$ 0g(0)=0$; where $g(0) not =0$.
Combining :
$d(x) = 0$ is equivalent to $g(x) = 0$, if $xnot =1$.
By considering $d(x)$ you have introduced an additional zero at $x=1$.(Lord Shark's comment)
answered Jul 22 at 20:45
Peter Szilas
7,9252617
edited Jul 22 at 20:50
answered Jul 22 at 20:45
Peter Szilas
7,9252617
answered Jul 22 at 20:45
Peter Szilas
7,9252617
answered Jul 22 at 20:45
Peter Szilas
7,9252617
7,9252617
add a comment |Â
add a comment |Â
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5
$x^3-1=(x-1)(x^2+x+1)$
– Lord Shark the Unknown
Jul 22 at 17:09
I guess the problem is somewhere there that you rewrite an equation and then substitute something you have made out of the equation into the original.
– mrtaurho
Jul 22 at 17:11
@LordSharktheUnknown I am not sure how that answers my question, I am asking why is this happening.
– Sorfosh
Jul 22 at 17:12
4
@Sorfosh Actually you multiplied by $1-1/x$, but the principle is the same.
– Lord Shark the Unknown
Jul 22 at 17:13
1
You could make even worse examples. $x=-(x^2+1)implies x^2=x^4+2x^2+1$ which we could use to rewrite the original as $x^4+2x^2+x+2=0$. That one has four separate solutions, including $frac 12times (1pm sqrt -7)$. But all these rewrites do is to find other equations that our original roots must satisfy.
– lulu
Jul 22 at 17:21