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Prove that if $|x|<1$ then $x^6<1$
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I am trying to prove that if $|x|<1$ then $x^6<1$ and that if $x^6<1$ then $|x|<1$.
For the first part, I thought of first considering $0<x<1$. Multiplying by $x$ (which is positive) I then get $0<x^2<x<1$. And repeating the multiplication I would end up with $x^6<1$.
Then I would consider $-1<x<0$. But now multiplying by $x$ (which is negative) gives $-x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$.
There is probably a more elegant way of going about this?
For the second part I thought of using the contrapositive and prove $xleq -1$ or $xgeq 1$ then $x^6geq 1$, and then I would try to use the previous proof to break this into two cases.
$xleq -1$ then $x^6geq 1$ and $xgeq 1$ then $x^6geq 1$.
proof-explanation
edited Jul 18 at 3:57
Xander Henderson
13.1k83150
asked Jul 18 at 1:13
JennyToy
20818
add a comment |Â
up vote
2
down vote
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I am trying to prove that if $|x|<1$ then $x^6<1$ and that if $x^6<1$ then $|x|<1$.
For the first part, I thought of first considering $0<x<1$. Multiplying by $x$ (which is positive) I then get $0<x^2<x<1$. And repeating the multiplication I would end up with $x^6<1$.
Then I would consider $-1<x<0$. But now multiplying by $x$ (which is negative) gives $-x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$.
There is probably a more elegant way of going about this?
For the second part I thought of using the contrapositive and prove $xleq -1$ or $xgeq 1$ then $x^6geq 1$, and then I would try to use the previous proof to break this into two cases.
$xleq -1$ then $x^6geq 1$ and $xgeq 1$ then $x^6geq 1$.
proof-explanation
edited Jul 18 at 3:57
Xander Henderson
13.1k83150
asked Jul 18 at 1:13
JennyToy
20818
4
Use that $x^6=(-x)^6$
– Mateus Rocha
Jul 18 at 1:16
2
Hint: $;dfracxx=|x|^5+|x|^4+ldots+1 gt 0,$, so $x^6-1$ and $|x|-1$ have the same sign.
– dxiv
Jul 18 at 1:18
2
In general, if $|x| < 1$, then $|x|^n < 1$, as your logic in the $0 < x < 1$ case proves. Also, $x le |x|$ and $|x^n| = |x|^n$. Putting it together gives $$x^n le |x^n| = |x|^n < 1.$$
– Theo Bendit
Jul 18 at 1:23
Does anyone else want to write up an intermediate value theorem answer? I can't right now, but if I don't see one soon I'll put one up.
– rnrstopstraffic
Jul 23 at 1:55
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to prove that if $|x|<1$ then $x^6<1$ and that if $x^6<1$ then $|x|<1$.
For the first part, I thought of first considering $0<x<1$. Multiplying by $x$ (which is positive) I then get $0<x^2<x<1$. And repeating the multiplication I would end up with $x^6<1$.
Then I would consider $-1<x<0$. But now multiplying by $x$ (which is negative) gives $-x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$.
There is probably a more elegant way of going about this?
For the second part I thought of using the contrapositive and prove $xleq -1$ or $xgeq 1$ then $x^6geq 1$, and then I would try to use the previous proof to break this into two cases.
$xleq -1$ then $x^6geq 1$ and $xgeq 1$ then $x^6geq 1$.
proof-explanation
edited Jul 18 at 3:57
Xander Henderson
13.1k83150
asked Jul 18 at 1:13
JennyToy
20818
I am trying to prove that if $|x|<1$ then $x^6<1$ and that if $x^6<1$ then $|x|<1$.
For the first part, I thought of first considering $0<x<1$. Multiplying by $x$ (which is positive) I then get $0<x^2<x<1$. And repeating the multiplication I would end up with $x^6<1$.
Then I would consider $-1<x<0$. But now multiplying by $x$ (which is negative) gives $-x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$.
There is probably a more elegant way of going about this?
For the second part I thought of using the contrapositive and prove $xleq -1$ or $xgeq 1$ then $x^6geq 1$, and then I would try to use the previous proof to break this into two cases.
$xleq -1$ then $x^6geq 1$ and $xgeq 1$ then $x^6geq 1$.
proof-explanation
edited Jul 18 at 3:57
Xander Henderson
13.1k83150
asked Jul 18 at 1:13
JennyToy
20818
edited Jul 18 at 3:57
Xander Henderson
13.1k83150
edited Jul 18 at 3:57
Xander Henderson
13.1k83150
edited Jul 18 at 3:57
Xander Henderson
13.1k83150
13.1k83150
asked Jul 18 at 1:13
JennyToy
20818
asked Jul 18 at 1:13
JennyToy
20818
asked Jul 18 at 1:13
JennyToy
20818
20818
4
Use that $x^6=(-x)^6$
– Mateus Rocha
Jul 18 at 1:16
2
Hint: $;dfracxx=|x|^5+|x|^4+ldots+1 gt 0,$, so $x^6-1$ and $|x|-1$ have the same sign.
– dxiv
Jul 18 at 1:18
2
In general, if $|x| < 1$, then $|x|^n < 1$, as your logic in the $0 < x < 1$ case proves. Also, $x le |x|$ and $|x^n| = |x|^n$. Putting it together gives $$x^n le |x^n| = |x|^n < 1.$$
– Theo Bendit
Jul 18 at 1:23
Does anyone else want to write up an intermediate value theorem answer? I can't right now, but if I don't see one soon I'll put one up.
– rnrstopstraffic
Jul 23 at 1:55
add a comment |Â
4
Use that $x^6=(-x)^6$
– Mateus Rocha
Jul 18 at 1:16
2
Hint: $;dfracxx=|x|^5+|x|^4+ldots+1 gt 0,$, so $x^6-1$ and $|x|-1$ have the same sign.
– dxiv
Jul 18 at 1:18
2
In general, if $|x| < 1$, then $|x|^n < 1$, as your logic in the $0 < x < 1$ case proves. Also, $x le |x|$ and $|x^n| = |x|^n$. Putting it together gives $$x^n le |x^n| = |x|^n < 1.$$
– Theo Bendit
Jul 18 at 1:23
Does anyone else want to write up an intermediate value theorem answer? I can't right now, but if I don't see one soon I'll put one up.
– rnrstopstraffic
Jul 23 at 1:55
4
4
Use that $x^6=(-x)^6$
– Mateus Rocha
Jul 18 at 1:16
Use that $x^6=(-x)^6$
– Mateus Rocha
Jul 18 at 1:16
2
2
Hint: $;dfracxx=|x|^5+|x|^4+ldots+1 gt 0,$, so $x^6-1$ and $|x|-1$ have the same sign.
– dxiv
Jul 18 at 1:18
Hint: $;dfracxx=|x|^5+|x|^4+ldots+1 gt 0,$, so $x^6-1$ and $|x|-1$ have the same sign.
– dxiv
Jul 18 at 1:18
2
2
In general, if $|x| < 1$, then $|x|^n < 1$, as your logic in the $0 < x < 1$ case proves. Also, $x le |x|$ and $|x^n| = |x|^n$. Putting it together gives $$x^n le |x^n| = |x|^n < 1.$$
– Theo Bendit
Jul 18 at 1:23
In general, if $|x| < 1$, then $|x|^n < 1$, as your logic in the $0 < x < 1$ case proves. Also, $x le |x|$ and $|x^n| = |x|^n$. Putting it together gives $$x^n le |x^n| = |x|^n < 1.$$
– Theo Bendit
Jul 18 at 1:23
Does anyone else want to write up an intermediate value theorem answer? I can't right now, but if I don't see one soon I'll put one up.
– rnrstopstraffic
Jul 23 at 1:55
Does anyone else want to write up an intermediate value theorem answer? I can't right now, but if I don't see one soon I'll put one up.
– rnrstopstraffic
Jul 23 at 1:55
add a comment |Â
6 Answers
6
active
oldest
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up vote
3
down vote
accepted
The argument that dxiv made in a comment is, I think, the most elegant argument. As comments are ephemeral, I'll reiterate it here (with some additional details): note that
$$ fracxx = |x|^5 + |x|^4 + |x|^3 + |x|^2 + |x| + 1. tag1$$
This is actually a specific case of a much more general result: for any natural number $n$,
$$ fract^n-1t-1 = sum_j=0^n-1 t^j. $$
This can be proved by an induction argument after noticing that
$$t^n+1 - 1 = t^n+1 - t^n + t^n -1 = t^n(t-1) + t^n - 1.$$ In this particular case, we take $t = |x|$. In any event, since all of the terms on the right-hand side of (1) are nonnegative (and $1>0$, i.e. $1$ is strictly positive), it follows that
$$ fracxx > 0. tag2$$
Note that
$$ fracab > 0 iff (a>0 land b > 0) lor (a< 0land b < 0). $$
That is, a fraction is positive if and only if both the numerator and denominator have the same sign. Applying this to (2), either
$$ |x|^6 - 1 < 0 qquadtextandqquad |x| - 1 < 0, tag3 $$
or
$$ |x|^6 - 1 > 0 qquadtextandqquad |x| - 1 > 0. $$
Therefore if we assume that $|x|<1$ it immediately follows from (3) that $|x|^6<1$, and vice versa.
Alternatively, we can make your approach work. The first part of your argument is basically an induction argument. We can actually strengthen that result a bit and get the following:
Proposition: If $|x| < 1$ then $|x|^n < 1$ for any natural number $n$.
Proof: The proof is by induction. Assume that $|x|< 1$, and note that $|x|^1 < 1$, providing a base for the induction. Now suppose that $|x|^k < 1$. Then
$$ |x|^k < 1 implies |x|^k cdot |x| < 1 cdot |x| implies |x|^k+1 < |x|. $$
But $|x|<1$, which proves the result.$tag*$blacksquare$$
Taking $n = 6$ gives gives the result in the original question. Then, as you note, the converse requires an additional argument. Following your instinct of attempting contraposition, you might consider the following proposition:
Proposition: Let $n$ be a natural number.
- If $|x| < 1$ then $|x|^n < 1$,
- if $|x| = 1$ then $|x|^n = 1$, and
- if $|x| > 1$ then $|x|^n > 1$.
There may be more elegant arguments, but the induction proof above can be used, mutatis mutandis, to prove the two new statements. This gives a kind of trichotomy, from which you can conclude that if $|x|^n < 1$, then $|x| < 1$.
answered Jul 18 at 1:40
Xander Henderson
13.1k83150
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The case $x=0$ is clear. So we can assume that $x ne 0$. Then we have that $ frac1x>1$. Hence there ist $t>0$ such that $ frac1x=1+t.$
Then we get, by Bernoulli:
$frac1x^6=frac1=(1+t)^6 ge 1+6t >1$ and the result follows.
answered Jul 18 at 5:27
Fred
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First, note that $f(x)=x^6-1$ only has real roots at $-1$ and $1$. If it would be necessary to prove this, you could factor into $(x^3-1)(x^3+1)$, apply the sum/difference of cubes formulas, show that the quadratic factors yield non-real roots, and then appeal to the Fundamental Theorem of Algebra to say that $-1$ and $1$ are the unique real roots.
Also note that $x^6ge 0$, so $x^6-1ge -1$.
Now note that $f(-2)=64>0$, $f(2)=64>0$, and $f(0)=-1<0$. Then by the Intermediate Value Theorem, $f(x)>0$ on $(-infty,-1)cup (1,infty)$ and $f(x)<0$ on $(-1,1)$.
Then $$lvert xrvert<1 iff -1<x<1\ iff -1<x^6-1<0\ iff 0<x^6<1 $$
answered Jul 23 at 5:04
rnrstopstraffic
80747
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The first part for $0<x<1$ is OK.
Then I would consider $−1<x<0$. But now multiplying by $x$ (which is negative) gives $−x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$. There is probably a more elegant way of going about this?
Note that $-1<x<0$ multiplied by $x$ (which is negative) results in $-xcolorred>x^2$ (because an inequality multiplied by a negative number will have reversed inequality sign).
Now multiply $-1<x<0$ by $-1$ to get: $1>-x$ (again note the inequality sign reversal).
Can you merge the two points and finish?
Also note, the case $x=0$ is trivial, which could be included in the first part as $0le x<1$.
answered Jul 23 at 8:06
farruhota
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Since $6$ is even $x^6=|x|^6$ and since $|x|<1$ we have $ x^6=|x|^6<1$
Btw multiplying by something negative changes the order ($<$ becomes $>$)
answered Jul 18 at 1:18
H. Walter
536
(1) This only addresses the forward implication (i.e. $|x|<1$ implies that $|x|^6 < 1$), and (2) it seems to assume the result, as if you assume that $|x|<1$ implies that $x^6 < 1$, it kind of renders the exercise moot, no?
– Xander Henderson
Jul 18 at 3:46
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Note that $$|x|^6=x^6 $$
If you raise a positive number which is less than one to any positive power, the result is less than one.
Similarly if you raise a positive number which is greater than one to any positive power, the result is greater than one.
Thus $$x^6<1 iff |x|^6 <1 iff |x|<1$$
answered Jul 18 at 1:23
Mohammad Riazi-Kermani
27.5k41852
2
You assume that "if you raise a positive number which is less than one to any positive power, the result is less than one." It seems to me that if one can make that assumption, then the exercise is moot.
– Xander Henderson
Jul 18 at 1:51
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The argument that dxiv made in a comment is, I think, the most elegant argument. As comments are ephemeral, I'll reiterate it here (with some additional details): note that
$$ fracxx = |x|^5 + |x|^4 + |x|^3 + |x|^2 + |x| + 1. tag1$$
This is actually a specific case of a much more general result: for any natural number $n$,
$$ fract^n-1t-1 = sum_j=0^n-1 t^j. $$
This can be proved by an induction argument after noticing that
$$t^n+1 - 1 = t^n+1 - t^n + t^n -1 = t^n(t-1) + t^n - 1.$$ In this particular case, we take $t = |x|$. In any event, since all of the terms on the right-hand side of (1) are nonnegative (and $1>0$, i.e. $1$ is strictly positive), it follows that
$$ fracxx > 0. tag2$$
Note that
$$ fracab > 0 iff (a>0 land b > 0) lor (a< 0land b < 0). $$
That is, a fraction is positive if and only if both the numerator and denominator have the same sign. Applying this to (2), either
$$ |x|^6 - 1 < 0 qquadtextandqquad |x| - 1 < 0, tag3 $$
or
$$ |x|^6 - 1 > 0 qquadtextandqquad |x| - 1 > 0. $$
Therefore if we assume that $|x|<1$ it immediately follows from (3) that $|x|^6<1$, and vice versa.
Alternatively, we can make your approach work. The first part of your argument is basically an induction argument. We can actually strengthen that result a bit and get the following:
Proposition: If $|x| < 1$ then $|x|^n < 1$ for any natural number $n$.
Proof: The proof is by induction. Assume that $|x|< 1$, and note that $|x|^1 < 1$, providing a base for the induction. Now suppose that $|x|^k < 1$. Then
$$ |x|^k < 1 implies |x|^k cdot |x| < 1 cdot |x| implies |x|^k+1 < |x|. $$
But $|x|<1$, which proves the result.$tag*$blacksquare$$
Taking $n = 6$ gives gives the result in the original question. Then, as you note, the converse requires an additional argument. Following your instinct of attempting contraposition, you might consider the following proposition:
Proposition: Let $n$ be a natural number.
- If $|x| < 1$ then $|x|^n < 1$,
- if $|x| = 1$ then $|x|^n = 1$, and
- if $|x| > 1$ then $|x|^n > 1$.
There may be more elegant arguments, but the induction proof above can be used, mutatis mutandis, to prove the two new statements. This gives a kind of trichotomy, from which you can conclude that if $|x|^n < 1$, then $|x| < 1$.
answered Jul 18 at 1:40
Xander Henderson
13.1k83150
add a comment |Â
up vote
3
down vote
accepted
The argument that dxiv made in a comment is, I think, the most elegant argument. As comments are ephemeral, I'll reiterate it here (with some additional details): note that
$$ fracxx = |x|^5 + |x|^4 + |x|^3 + |x|^2 + |x| + 1. tag1$$
This is actually a specific case of a much more general result: for any natural number $n$,
$$ fract^n-1t-1 = sum_j=0^n-1 t^j. $$
This can be proved by an induction argument after noticing that
$$t^n+1 - 1 = t^n+1 - t^n + t^n -1 = t^n(t-1) + t^n - 1.$$ In this particular case, we take $t = |x|$. In any event, since all of the terms on the right-hand side of (1) are nonnegative (and $1>0$, i.e. $1$ is strictly positive), it follows that
$$ fracxx > 0. tag2$$
Note that
$$ fracab > 0 iff (a>0 land b > 0) lor (a< 0land b < 0). $$
That is, a fraction is positive if and only if both the numerator and denominator have the same sign. Applying this to (2), either
$$ |x|^6 - 1 < 0 qquadtextandqquad |x| - 1 < 0, tag3 $$
or
$$ |x|^6 - 1 > 0 qquadtextandqquad |x| - 1 > 0. $$
Therefore if we assume that $|x|<1$ it immediately follows from (3) that $|x|^6<1$, and vice versa.
Alternatively, we can make your approach work. The first part of your argument is basically an induction argument. We can actually strengthen that result a bit and get the following:
Proposition: If $|x| < 1$ then $|x|^n < 1$ for any natural number $n$.
Proof: The proof is by induction. Assume that $|x|< 1$, and note that $|x|^1 < 1$, providing a base for the induction. Now suppose that $|x|^k < 1$. Then
$$ |x|^k < 1 implies |x|^k cdot |x| < 1 cdot |x| implies |x|^k+1 < |x|. $$
But $|x|<1$, which proves the result.$tag*$blacksquare$$
Taking $n = 6$ gives gives the result in the original question. Then, as you note, the converse requires an additional argument. Following your instinct of attempting contraposition, you might consider the following proposition:
Proposition: Let $n$ be a natural number.
- If $|x| < 1$ then $|x|^n < 1$,
- if $|x| = 1$ then $|x|^n = 1$, and
- if $|x| > 1$ then $|x|^n > 1$.
There may be more elegant arguments, but the induction proof above can be used, mutatis mutandis, to prove the two new statements. This gives a kind of trichotomy, from which you can conclude that if $|x|^n < 1$, then $|x| < 1$.
answered Jul 18 at 1:40
Xander Henderson
13.1k83150
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The argument that dxiv made in a comment is, I think, the most elegant argument. As comments are ephemeral, I'll reiterate it here (with some additional details): note that
$$ fracxx = |x|^5 + |x|^4 + |x|^3 + |x|^2 + |x| + 1. tag1$$
This is actually a specific case of a much more general result: for any natural number $n$,
$$ fract^n-1t-1 = sum_j=0^n-1 t^j. $$
This can be proved by an induction argument after noticing that
$$t^n+1 - 1 = t^n+1 - t^n + t^n -1 = t^n(t-1) + t^n - 1.$$ In this particular case, we take $t = |x|$. In any event, since all of the terms on the right-hand side of (1) are nonnegative (and $1>0$, i.e. $1$ is strictly positive), it follows that
$$ fracxx > 0. tag2$$
Note that
$$ fracab > 0 iff (a>0 land b > 0) lor (a< 0land b < 0). $$
That is, a fraction is positive if and only if both the numerator and denominator have the same sign. Applying this to (2), either
$$ |x|^6 - 1 < 0 qquadtextandqquad |x| - 1 < 0, tag3 $$
or
$$ |x|^6 - 1 > 0 qquadtextandqquad |x| - 1 > 0. $$
Therefore if we assume that $|x|<1$ it immediately follows from (3) that $|x|^6<1$, and vice versa.
Alternatively, we can make your approach work. The first part of your argument is basically an induction argument. We can actually strengthen that result a bit and get the following:
Proposition: If $|x| < 1$ then $|x|^n < 1$ for any natural number $n$.
Proof: The proof is by induction. Assume that $|x|< 1$, and note that $|x|^1 < 1$, providing a base for the induction. Now suppose that $|x|^k < 1$. Then
$$ |x|^k < 1 implies |x|^k cdot |x| < 1 cdot |x| implies |x|^k+1 < |x|. $$
But $|x|<1$, which proves the result.$tag*$blacksquare$$
Taking $n = 6$ gives gives the result in the original question. Then, as you note, the converse requires an additional argument. Following your instinct of attempting contraposition, you might consider the following proposition:
Proposition: Let $n$ be a natural number.
- If $|x| < 1$ then $|x|^n < 1$,
- if $|x| = 1$ then $|x|^n = 1$, and
- if $|x| > 1$ then $|x|^n > 1$.
There may be more elegant arguments, but the induction proof above can be used, mutatis mutandis, to prove the two new statements. This gives a kind of trichotomy, from which you can conclude that if $|x|^n < 1$, then $|x| < 1$.
answered Jul 18 at 1:40
Xander Henderson
13.1k83150
The argument that dxiv made in a comment is, I think, the most elegant argument. As comments are ephemeral, I'll reiterate it here (with some additional details): note that
$$ fracxx = |x|^5 + |x|^4 + |x|^3 + |x|^2 + |x| + 1. tag1$$
This is actually a specific case of a much more general result: for any natural number $n$,
$$ fract^n-1t-1 = sum_j=0^n-1 t^j. $$
This can be proved by an induction argument after noticing that
$$t^n+1 - 1 = t^n+1 - t^n + t^n -1 = t^n(t-1) + t^n - 1.$$ In this particular case, we take $t = |x|$. In any event, since all of the terms on the right-hand side of (1) are nonnegative (and $1>0$, i.e. $1$ is strictly positive), it follows that
$$ fracxx > 0. tag2$$
Note that
$$ fracab > 0 iff (a>0 land b > 0) lor (a< 0land b < 0). $$
That is, a fraction is positive if and only if both the numerator and denominator have the same sign. Applying this to (2), either
$$ |x|^6 - 1 < 0 qquadtextandqquad |x| - 1 < 0, tag3 $$
or
$$ |x|^6 - 1 > 0 qquadtextandqquad |x| - 1 > 0. $$
Therefore if we assume that $|x|<1$ it immediately follows from (3) that $|x|^6<1$, and vice versa.
Alternatively, we can make your approach work. The first part of your argument is basically an induction argument. We can actually strengthen that result a bit and get the following:
Proposition: If $|x| < 1$ then $|x|^n < 1$ for any natural number $n$.
Proof: The proof is by induction. Assume that $|x|< 1$, and note that $|x|^1 < 1$, providing a base for the induction. Now suppose that $|x|^k < 1$. Then
$$ |x|^k < 1 implies |x|^k cdot |x| < 1 cdot |x| implies |x|^k+1 < |x|. $$
But $|x|<1$, which proves the result.$tag*$blacksquare$$
Taking $n = 6$ gives gives the result in the original question. Then, as you note, the converse requires an additional argument. Following your instinct of attempting contraposition, you might consider the following proposition:
Proposition: Let $n$ be a natural number.
- If $|x| < 1$ then $|x|^n < 1$,
- if $|x| = 1$ then $|x|^n = 1$, and
- if $|x| > 1$ then $|x|^n > 1$.
There may be more elegant arguments, but the induction proof above can be used, mutatis mutandis, to prove the two new statements. This gives a kind of trichotomy, from which you can conclude that if $|x|^n < 1$, then $|x| < 1$.
answered Jul 18 at 1:40
Xander Henderson
13.1k83150
edited Jul 18 at 13:01
answered Jul 18 at 1:40
Xander Henderson
13.1k83150
answered Jul 18 at 1:40
Xander Henderson
13.1k83150
answered Jul 18 at 1:40
Xander Henderson
13.1k83150
13.1k83150
add a comment |Â
add a comment |Â
up vote
1
down vote
The case $x=0$ is clear. So we can assume that $x ne 0$. Then we have that $ frac1x>1$. Hence there ist $t>0$ such that $ frac1x=1+t.$
Then we get, by Bernoulli:
$frac1x^6=frac1=(1+t)^6 ge 1+6t >1$ and the result follows.
answered Jul 18 at 5:27
Fred
37.6k1237
add a comment |Â
up vote
1
down vote
The case $x=0$ is clear. So we can assume that $x ne 0$. Then we have that $ frac1x>1$. Hence there ist $t>0$ such that $ frac1x=1+t.$
Then we get, by Bernoulli:
$frac1x^6=frac1=(1+t)^6 ge 1+6t >1$ and the result follows.
answered Jul 18 at 5:27
Fred
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The case $x=0$ is clear. So we can assume that $x ne 0$. Then we have that $ frac1x>1$. Hence there ist $t>0$ such that $ frac1x=1+t.$
Then we get, by Bernoulli:
$frac1x^6=frac1=(1+t)^6 ge 1+6t >1$ and the result follows.
answered Jul 18 at 5:27
Fred
37.6k1237
The case $x=0$ is clear. So we can assume that $x ne 0$. Then we have that $ frac1x>1$. Hence there ist $t>0$ such that $ frac1x=1+t.$
Then we get, by Bernoulli:
$frac1x^6=frac1=(1+t)^6 ge 1+6t >1$ and the result follows.
answered Jul 18 at 5:27
Fred
37.6k1237
answered Jul 18 at 5:27
Fred
37.6k1237
answered Jul 18 at 5:27
Fred
37.6k1237
answered Jul 18 at 5:27
Fred
37.6k1237
37.6k1237
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1
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First, note that $f(x)=x^6-1$ only has real roots at $-1$ and $1$. If it would be necessary to prove this, you could factor into $(x^3-1)(x^3+1)$, apply the sum/difference of cubes formulas, show that the quadratic factors yield non-real roots, and then appeal to the Fundamental Theorem of Algebra to say that $-1$ and $1$ are the unique real roots.
Also note that $x^6ge 0$, so $x^6-1ge -1$.
Now note that $f(-2)=64>0$, $f(2)=64>0$, and $f(0)=-1<0$. Then by the Intermediate Value Theorem, $f(x)>0$ on $(-infty,-1)cup (1,infty)$ and $f(x)<0$ on $(-1,1)$.
Then $$lvert xrvert<1 iff -1<x<1\ iff -1<x^6-1<0\ iff 0<x^6<1 $$
answered Jul 23 at 5:04
rnrstopstraffic
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First, note that $f(x)=x^6-1$ only has real roots at $-1$ and $1$. If it would be necessary to prove this, you could factor into $(x^3-1)(x^3+1)$, apply the sum/difference of cubes formulas, show that the quadratic factors yield non-real roots, and then appeal to the Fundamental Theorem of Algebra to say that $-1$ and $1$ are the unique real roots.
Also note that $x^6ge 0$, so $x^6-1ge -1$.
Now note that $f(-2)=64>0$, $f(2)=64>0$, and $f(0)=-1<0$. Then by the Intermediate Value Theorem, $f(x)>0$ on $(-infty,-1)cup (1,infty)$ and $f(x)<0$ on $(-1,1)$.
Then $$lvert xrvert<1 iff -1<x<1\ iff -1<x^6-1<0\ iff 0<x^6<1 $$
answered Jul 23 at 5:04
rnrstopstraffic
80747
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up vote
1
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up vote
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First, note that $f(x)=x^6-1$ only has real roots at $-1$ and $1$. If it would be necessary to prove this, you could factor into $(x^3-1)(x^3+1)$, apply the sum/difference of cubes formulas, show that the quadratic factors yield non-real roots, and then appeal to the Fundamental Theorem of Algebra to say that $-1$ and $1$ are the unique real roots.
Also note that $x^6ge 0$, so $x^6-1ge -1$.
Now note that $f(-2)=64>0$, $f(2)=64>0$, and $f(0)=-1<0$. Then by the Intermediate Value Theorem, $f(x)>0$ on $(-infty,-1)cup (1,infty)$ and $f(x)<0$ on $(-1,1)$.
Then $$lvert xrvert<1 iff -1<x<1\ iff -1<x^6-1<0\ iff 0<x^6<1 $$
answered Jul 23 at 5:04
rnrstopstraffic
80747
First, note that $f(x)=x^6-1$ only has real roots at $-1$ and $1$. If it would be necessary to prove this, you could factor into $(x^3-1)(x^3+1)$, apply the sum/difference of cubes formulas, show that the quadratic factors yield non-real roots, and then appeal to the Fundamental Theorem of Algebra to say that $-1$ and $1$ are the unique real roots.
Also note that $x^6ge 0$, so $x^6-1ge -1$.
Now note that $f(-2)=64>0$, $f(2)=64>0$, and $f(0)=-1<0$. Then by the Intermediate Value Theorem, $f(x)>0$ on $(-infty,-1)cup (1,infty)$ and $f(x)<0$ on $(-1,1)$.
Then $$lvert xrvert<1 iff -1<x<1\ iff -1<x^6-1<0\ iff 0<x^6<1 $$
answered Jul 23 at 5:04
rnrstopstraffic
80747
answered Jul 23 at 5:04
rnrstopstraffic
80747
answered Jul 23 at 5:04
rnrstopstraffic
80747
answered Jul 23 at 5:04
rnrstopstraffic
80747
80747
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The first part for $0<x<1$ is OK.
Then I would consider $−1<x<0$. But now multiplying by $x$ (which is negative) gives $−x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$. There is probably a more elegant way of going about this?
Note that $-1<x<0$ multiplied by $x$ (which is negative) results in $-xcolorred>x^2$ (because an inequality multiplied by a negative number will have reversed inequality sign).
Now multiply $-1<x<0$ by $-1$ to get: $1>-x$ (again note the inequality sign reversal).
Can you merge the two points and finish?
Also note, the case $x=0$ is trivial, which could be included in the first part as $0le x<1$.
answered Jul 23 at 8:06
farruhota
13.8k2632
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The first part for $0<x<1$ is OK.
Then I would consider $−1<x<0$. But now multiplying by $x$ (which is negative) gives $−x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$. There is probably a more elegant way of going about this?
Note that $-1<x<0$ multiplied by $x$ (which is negative) results in $-xcolorred>x^2$ (because an inequality multiplied by a negative number will have reversed inequality sign).
Now multiply $-1<x<0$ by $-1$ to get: $1>-x$ (again note the inequality sign reversal).
Can you merge the two points and finish?
Also note, the case $x=0$ is trivial, which could be included in the first part as $0le x<1$.
answered Jul 23 at 8:06
farruhota
13.8k2632
add a comment |Â
up vote
1
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up vote
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The first part for $0<x<1$ is OK.
Then I would consider $−1<x<0$. But now multiplying by $x$ (which is negative) gives $−x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$. There is probably a more elegant way of going about this?
Note that $-1<x<0$ multiplied by $x$ (which is negative) results in $-xcolorred>x^2$ (because an inequality multiplied by a negative number will have reversed inequality sign).
Now multiply $-1<x<0$ by $-1$ to get: $1>-x$ (again note the inequality sign reversal).
Can you merge the two points and finish?
Also note, the case $x=0$ is trivial, which could be included in the first part as $0le x<1$.
answered Jul 23 at 8:06
farruhota
13.8k2632
The first part for $0<x<1$ is OK.
Then I would consider $−1<x<0$. But now multiplying by $x$ (which is negative) gives $−x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$. There is probably a more elegant way of going about this?
Note that $-1<x<0$ multiplied by $x$ (which is negative) results in $-xcolorred>x^2$ (because an inequality multiplied by a negative number will have reversed inequality sign).
Now multiply $-1<x<0$ by $-1$ to get: $1>-x$ (again note the inequality sign reversal).
Can you merge the two points and finish?
Also note, the case $x=0$ is trivial, which could be included in the first part as $0le x<1$.
answered Jul 23 at 8:06
farruhota
13.8k2632
answered Jul 23 at 8:06
farruhota
13.8k2632
answered Jul 23 at 8:06
farruhota
13.8k2632
answered Jul 23 at 8:06
farruhota
13.8k2632
13.8k2632
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add a comment |Â
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0
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Since $6$ is even $x^6=|x|^6$ and since $|x|<1$ we have $ x^6=|x|^6<1$
Btw multiplying by something negative changes the order ($<$ becomes $>$)
answered Jul 18 at 1:18
H. Walter
536
(1) This only addresses the forward implication (i.e. $|x|<1$ implies that $|x|^6 < 1$), and (2) it seems to assume the result, as if you assume that $|x|<1$ implies that $x^6 < 1$, it kind of renders the exercise moot, no?
– Xander Henderson
Jul 18 at 3:46
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Since $6$ is even $x^6=|x|^6$ and since $|x|<1$ we have $ x^6=|x|^6<1$
Btw multiplying by something negative changes the order ($<$ becomes $>$)
answered Jul 18 at 1:18
H. Walter
536
(1) This only addresses the forward implication (i.e. $|x|<1$ implies that $|x|^6 < 1$), and (2) it seems to assume the result, as if you assume that $|x|<1$ implies that $x^6 < 1$, it kind of renders the exercise moot, no?
– Xander Henderson
Jul 18 at 3:46
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $6$ is even $x^6=|x|^6$ and since $|x|<1$ we have $ x^6=|x|^6<1$
Btw multiplying by something negative changes the order ($<$ becomes $>$)
answered Jul 18 at 1:18
H. Walter
536
Since $6$ is even $x^6=|x|^6$ and since $|x|<1$ we have $ x^6=|x|^6<1$
Btw multiplying by something negative changes the order ($<$ becomes $>$)
answered Jul 18 at 1:18
H. Walter
536
answered Jul 18 at 1:18
H. Walter
536
answered Jul 18 at 1:18
H. Walter
536
answered Jul 18 at 1:18
H. Walter
536
536
(1) This only addresses the forward implication (i.e. $|x|<1$ implies that $|x|^6 < 1$), and (2) it seems to assume the result, as if you assume that $|x|<1$ implies that $x^6 < 1$, it kind of renders the exercise moot, no?
– Xander Henderson
Jul 18 at 3:46
add a comment |Â
(1) This only addresses the forward implication (i.e. $|x|<1$ implies that $|x|^6 < 1$), and (2) it seems to assume the result, as if you assume that $|x|<1$ implies that $x^6 < 1$, it kind of renders the exercise moot, no?
– Xander Henderson
Jul 18 at 3:46
(1) This only addresses the forward implication (i.e. $|x|<1$ implies that $|x|^6 < 1$), and (2) it seems to assume the result, as if you assume that $|x|<1$ implies that $x^6 < 1$, it kind of renders the exercise moot, no?
– Xander Henderson
Jul 18 at 3:46
(1) This only addresses the forward implication (i.e. $|x|<1$ implies that $|x|^6 < 1$), and (2) it seems to assume the result, as if you assume that $|x|<1$ implies that $x^6 < 1$, it kind of renders the exercise moot, no?
– Xander Henderson
Jul 18 at 3:46
add a comment |Â
up vote
0
down vote
Note that $$|x|^6=x^6 $$
If you raise a positive number which is less than one to any positive power, the result is less than one.
Similarly if you raise a positive number which is greater than one to any positive power, the result is greater than one.
Thus $$x^6<1 iff |x|^6 <1 iff |x|<1$$
answered Jul 18 at 1:23
Mohammad Riazi-Kermani
27.5k41852
2
You assume that "if you raise a positive number which is less than one to any positive power, the result is less than one." It seems to me that if one can make that assumption, then the exercise is moot.
– Xander Henderson
Jul 18 at 1:51
add a comment |Â
up vote
0
down vote
Note that $$|x|^6=x^6 $$
If you raise a positive number which is less than one to any positive power, the result is less than one.
Similarly if you raise a positive number which is greater than one to any positive power, the result is greater than one.
Thus $$x^6<1 iff |x|^6 <1 iff |x|<1$$
answered Jul 18 at 1:23
Mohammad Riazi-Kermani
27.5k41852
2
You assume that "if you raise a positive number which is less than one to any positive power, the result is less than one." It seems to me that if one can make that assumption, then the exercise is moot.
– Xander Henderson
Jul 18 at 1:51
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that $$|x|^6=x^6 $$
If you raise a positive number which is less than one to any positive power, the result is less than one.
Similarly if you raise a positive number which is greater than one to any positive power, the result is greater than one.
Thus $$x^6<1 iff |x|^6 <1 iff |x|<1$$
answered Jul 18 at 1:23
Mohammad Riazi-Kermani
27.5k41852
Note that $$|x|^6=x^6 $$
If you raise a positive number which is less than one to any positive power, the result is less than one.
Similarly if you raise a positive number which is greater than one to any positive power, the result is greater than one.
Thus $$x^6<1 iff |x|^6 <1 iff |x|<1$$
answered Jul 18 at 1:23
Mohammad Riazi-Kermani
27.5k41852
edited Jul 18 at 1:34
answered Jul 18 at 1:23
Mohammad Riazi-Kermani
27.5k41852
answered Jul 18 at 1:23
Mohammad Riazi-Kermani
27.5k41852
answered Jul 18 at 1:23
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
2
You assume that "if you raise a positive number which is less than one to any positive power, the result is less than one." It seems to me that if one can make that assumption, then the exercise is moot.
– Xander Henderson
Jul 18 at 1:51
add a comment |Â
2
You assume that "if you raise a positive number which is less than one to any positive power, the result is less than one." It seems to me that if one can make that assumption, then the exercise is moot.
– Xander Henderson
Jul 18 at 1:51
2
2
You assume that "if you raise a positive number which is less than one to any positive power, the result is less than one." It seems to me that if one can make that assumption, then the exercise is moot.
– Xander Henderson
Jul 18 at 1:51
You assume that "if you raise a positive number which is less than one to any positive power, the result is less than one." It seems to me that if one can make that assumption, then the exercise is moot.
– Xander Henderson
Jul 18 at 1:51
add a comment |Â
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4
Use that $x^6=(-x)^6$
– Mateus Rocha
Jul 18 at 1:16
2
Hint: $;dfracxx=|x|^5+|x|^4+ldots+1 gt 0,$, so $x^6-1$ and $|x|-1$ have the same sign.
– dxiv
Jul 18 at 1:18
2
In general, if $|x| < 1$, then $|x|^n < 1$, as your logic in the $0 < x < 1$ case proves. Also, $x le |x|$ and $|x^n| = |x|^n$. Putting it together gives $$x^n le |x^n| = |x|^n < 1.$$
– Theo Bendit
Jul 18 at 1:23
Does anyone else want to write up an intermediate value theorem answer? I can't right now, but if I don't see one soon I'll put one up.
– rnrstopstraffic
Jul 23 at 1:55