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Refuting the assertion that any given number can be rewritten to be undefined in a domain.
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Consider the expression:
$$dfrac(x - c_1) cdot x(x - c_1)$$
This is often simplified as
$$x textfor x neq c_1$$
This simplification step can also be done an arbitrary number of times for
$$dfrac(x - c_1)(x - c_2) dots (x - c_n) cdot x(x - c_1)(x - c_2) dots (x - c_n)$$
In which case $x neq c_1, c_2, dots, c_n$.
Given that it is generally valid to simplify such expressions by repeated steps of elimination of the terms, does that not imply that it is equally valid to introduce arbitrarily many such terms? And if arbitrarily many such terms are introduced, how do we know that $x$ can be defined at all?
proof-verification
asked Aug 2 at 11:34
Oscar
18410
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Consider the expression:
$$dfrac(x - c_1) cdot x(x - c_1)$$
This is often simplified as
$$x textfor x neq c_1$$
This simplification step can also be done an arbitrary number of times for
$$dfrac(x - c_1)(x - c_2) dots (x - c_n) cdot x(x - c_1)(x - c_2) dots (x - c_n)$$
In which case $x neq c_1, c_2, dots, c_n$.
Given that it is generally valid to simplify such expressions by repeated steps of elimination of the terms, does that not imply that it is equally valid to introduce arbitrarily many such terms? And if arbitrarily many such terms are introduced, how do we know that $x$ can be defined at all?
proof-verification
asked Aug 2 at 11:34
Oscar
18410
Are you asserting that, going that way ad infinitum, we have to conclude that $x$ is different from every number ? If so, you have to take into account taht (considering for simplicity natural numbers), in order to conclude that $x ne 1,2,ldots$ we have to write an infinte fractional expression.
– Mauro ALLEGRANZA
Aug 2 at 11:46
But consider e.g. $dfrac (3-1)(3-1) cdot 3$; it is a "correct" expression and its value is simply $3$.
– Mauro ALLEGRANZA
Aug 2 at 11:49
@MauroALLEGRANZA Yes, that was my assertion, that since x could be rewritten as itself multiplied with a fraction of arbitrary length up to infinity, x would have to be different from all possible values and thus undefined. I have tried to explain to myself why this is nonsense, but the second fraction you gave is valid.
– Oscar
Aug 2 at 11:58
But consider the simple case with only $c_1$: it has at least the value $0$ that satisfies it.
– Mauro ALLEGRANZA
Aug 2 at 12:01
I am sorry, could you articulate that? Do you mean $dfracx - c_1x - c_1 cdot x$ with $c_1 = 0$? Then I would understand that this is a valid rewriting of x, as it hasn't changed the possible values that x can be
– Oscar
Aug 2 at 12:06
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the expression:
$$dfrac(x - c_1) cdot x(x - c_1)$$
This is often simplified as
$$x textfor x neq c_1$$
This simplification step can also be done an arbitrary number of times for
$$dfrac(x - c_1)(x - c_2) dots (x - c_n) cdot x(x - c_1)(x - c_2) dots (x - c_n)$$
In which case $x neq c_1, c_2, dots, c_n$.
Given that it is generally valid to simplify such expressions by repeated steps of elimination of the terms, does that not imply that it is equally valid to introduce arbitrarily many such terms? And if arbitrarily many such terms are introduced, how do we know that $x$ can be defined at all?
proof-verification
asked Aug 2 at 11:34
Oscar
18410
Consider the expression:
$$dfrac(x - c_1) cdot x(x - c_1)$$
This is often simplified as
$$x textfor x neq c_1$$
This simplification step can also be done an arbitrary number of times for
$$dfrac(x - c_1)(x - c_2) dots (x - c_n) cdot x(x - c_1)(x - c_2) dots (x - c_n)$$
In which case $x neq c_1, c_2, dots, c_n$.
Given that it is generally valid to simplify such expressions by repeated steps of elimination of the terms, does that not imply that it is equally valid to introduce arbitrarily many such terms? And if arbitrarily many such terms are introduced, how do we know that $x$ can be defined at all?
proof-verification
asked Aug 2 at 11:34
Oscar
18410
edited Aug 2 at 12:09
asked Aug 2 at 11:34
Oscar
18410
asked Aug 2 at 11:34
Oscar
18410
asked Aug 2 at 11:34
Oscar
18410
18410
Are you asserting that, going that way ad infinitum, we have to conclude that $x$ is different from every number ? If so, you have to take into account taht (considering for simplicity natural numbers), in order to conclude that $x ne 1,2,ldots$ we have to write an infinte fractional expression.
– Mauro ALLEGRANZA
Aug 2 at 11:46
But consider e.g. $dfrac (3-1)(3-1) cdot 3$; it is a "correct" expression and its value is simply $3$.
– Mauro ALLEGRANZA
Aug 2 at 11:49
@MauroALLEGRANZA Yes, that was my assertion, that since x could be rewritten as itself multiplied with a fraction of arbitrary length up to infinity, x would have to be different from all possible values and thus undefined. I have tried to explain to myself why this is nonsense, but the second fraction you gave is valid.
– Oscar
Aug 2 at 11:58
But consider the simple case with only $c_1$: it has at least the value $0$ that satisfies it.
– Mauro ALLEGRANZA
Aug 2 at 12:01
I am sorry, could you articulate that? Do you mean $dfracx - c_1x - c_1 cdot x$ with $c_1 = 0$? Then I would understand that this is a valid rewriting of x, as it hasn't changed the possible values that x can be
– Oscar
Aug 2 at 12:06
 |Â
show 4 more comments
Are you asserting that, going that way ad infinitum, we have to conclude that $x$ is different from every number ? If so, you have to take into account taht (considering for simplicity natural numbers), in order to conclude that $x ne 1,2,ldots$ we have to write an infinte fractional expression.
– Mauro ALLEGRANZA
Aug 2 at 11:46
But consider e.g. $dfrac (3-1)(3-1) cdot 3$; it is a "correct" expression and its value is simply $3$.
– Mauro ALLEGRANZA
Aug 2 at 11:49
@MauroALLEGRANZA Yes, that was my assertion, that since x could be rewritten as itself multiplied with a fraction of arbitrary length up to infinity, x would have to be different from all possible values and thus undefined. I have tried to explain to myself why this is nonsense, but the second fraction you gave is valid.
– Oscar
Aug 2 at 11:58
But consider the simple case with only $c_1$: it has at least the value $0$ that satisfies it.
– Mauro ALLEGRANZA
Aug 2 at 12:01
I am sorry, could you articulate that? Do you mean $dfracx - c_1x - c_1 cdot x$ with $c_1 = 0$? Then I would understand that this is a valid rewriting of x, as it hasn't changed the possible values that x can be
– Oscar
Aug 2 at 12:06
Are you asserting that, going that way ad infinitum, we have to conclude that $x$ is different from every number ? If so, you have to take into account taht (considering for simplicity natural numbers), in order to conclude that $x ne 1,2,ldots$ we have to write an infinte fractional expression.
– Mauro ALLEGRANZA
Aug 2 at 11:46
Are you asserting that, going that way ad infinitum, we have to conclude that $x$ is different from every number ? If so, you have to take into account taht (considering for simplicity natural numbers), in order to conclude that $x ne 1,2,ldots$ we have to write an infinte fractional expression.
– Mauro ALLEGRANZA
Aug 2 at 11:46
But consider e.g. $dfrac (3-1)(3-1) cdot 3$; it is a "correct" expression and its value is simply $3$.
– Mauro ALLEGRANZA
Aug 2 at 11:49
But consider e.g. $dfrac (3-1)(3-1) cdot 3$; it is a "correct" expression and its value is simply $3$.
– Mauro ALLEGRANZA
Aug 2 at 11:49
@MauroALLEGRANZA Yes, that was my assertion, that since x could be rewritten as itself multiplied with a fraction of arbitrary length up to infinity, x would have to be different from all possible values and thus undefined. I have tried to explain to myself why this is nonsense, but the second fraction you gave is valid.
– Oscar
Aug 2 at 11:58
@MauroALLEGRANZA Yes, that was my assertion, that since x could be rewritten as itself multiplied with a fraction of arbitrary length up to infinity, x would have to be different from all possible values and thus undefined. I have tried to explain to myself why this is nonsense, but the second fraction you gave is valid.
– Oscar
Aug 2 at 11:58
But consider the simple case with only $c_1$: it has at least the value $0$ that satisfies it.
– Mauro ALLEGRANZA
Aug 2 at 12:01
But consider the simple case with only $c_1$: it has at least the value $0$ that satisfies it.
– Mauro ALLEGRANZA
Aug 2 at 12:01
I am sorry, could you articulate that? Do you mean $dfracx - c_1x - c_1 cdot x$ with $c_1 = 0$? Then I would understand that this is a valid rewriting of x, as it hasn't changed the possible values that x can be
– Oscar
Aug 2 at 12:06
I am sorry, could you articulate that? Do you mean $dfracx - c_1x - c_1 cdot x$ with $c_1 = 0$? Then I would understand that this is a valid rewriting of x, as it hasn't changed the possible values that x can be
– Oscar
Aug 2 at 12:06
 |Â
show 4 more comments
1 Answer
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I think I can actually explain to myself why this is nonsense.
Given the value x, I think it is intuitive that this could be rewritten as $dfracc1c1 cdot x$ for $cneq 0$. This is permissible because we haven't arbitrarily introduced an area in a domain where x is undefined. Multiplying by $dfraccc =1$ does not fundamentally change anything. On the other hand, in the case of $dfracx−cx−c cdot x , x neq c$ , rewriting this as $x$ is only permissible in the case that the restrictions on the domain are maintained. I.e, because the area in which $x$ is not defined is the same before and after the simplification step. Introducing such a term arbitrarily however breaks this rule, and is thus invalid.
answered Aug 2 at 12:08
Oscar
18410
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think I can actually explain to myself why this is nonsense.
Given the value x, I think it is intuitive that this could be rewritten as $dfracc1c1 cdot x$ for $cneq 0$. This is permissible because we haven't arbitrarily introduced an area in a domain where x is undefined. Multiplying by $dfraccc =1$ does not fundamentally change anything. On the other hand, in the case of $dfracx−cx−c cdot x , x neq c$ , rewriting this as $x$ is only permissible in the case that the restrictions on the domain are maintained. I.e, because the area in which $x$ is not defined is the same before and after the simplification step. Introducing such a term arbitrarily however breaks this rule, and is thus invalid.
answered Aug 2 at 12:08
Oscar
18410
add a comment |Â
up vote
0
down vote
I think I can actually explain to myself why this is nonsense.
Given the value x, I think it is intuitive that this could be rewritten as $dfracc1c1 cdot x$ for $cneq 0$. This is permissible because we haven't arbitrarily introduced an area in a domain where x is undefined. Multiplying by $dfraccc =1$ does not fundamentally change anything. On the other hand, in the case of $dfracx−cx−c cdot x , x neq c$ , rewriting this as $x$ is only permissible in the case that the restrictions on the domain are maintained. I.e, because the area in which $x$ is not defined is the same before and after the simplification step. Introducing such a term arbitrarily however breaks this rule, and is thus invalid.
answered Aug 2 at 12:08
Oscar
18410
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think I can actually explain to myself why this is nonsense.
Given the value x, I think it is intuitive that this could be rewritten as $dfracc1c1 cdot x$ for $cneq 0$. This is permissible because we haven't arbitrarily introduced an area in a domain where x is undefined. Multiplying by $dfraccc =1$ does not fundamentally change anything. On the other hand, in the case of $dfracx−cx−c cdot x , x neq c$ , rewriting this as $x$ is only permissible in the case that the restrictions on the domain are maintained. I.e, because the area in which $x$ is not defined is the same before and after the simplification step. Introducing such a term arbitrarily however breaks this rule, and is thus invalid.
answered Aug 2 at 12:08
Oscar
18410
I think I can actually explain to myself why this is nonsense.
Given the value x, I think it is intuitive that this could be rewritten as $dfracc1c1 cdot x$ for $cneq 0$. This is permissible because we haven't arbitrarily introduced an area in a domain where x is undefined. Multiplying by $dfraccc =1$ does not fundamentally change anything. On the other hand, in the case of $dfracx−cx−c cdot x , x neq c$ , rewriting this as $x$ is only permissible in the case that the restrictions on the domain are maintained. I.e, because the area in which $x$ is not defined is the same before and after the simplification step. Introducing such a term arbitrarily however breaks this rule, and is thus invalid.
answered Aug 2 at 12:08
Oscar
18410
answered Aug 2 at 12:08
Oscar
18410
answered Aug 2 at 12:08
Oscar
18410
answered Aug 2 at 12:08
Oscar
18410
18410
add a comment |Â
add a comment |Â
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Are you asserting that, going that way ad infinitum, we have to conclude that $x$ is different from every number ? If so, you have to take into account taht (considering for simplicity natural numbers), in order to conclude that $x ne 1,2,ldots$ we have to write an infinte fractional expression.
– Mauro ALLEGRANZA
Aug 2 at 11:46
But consider e.g. $dfrac (3-1)(3-1) cdot 3$; it is a "correct" expression and its value is simply $3$.
– Mauro ALLEGRANZA
Aug 2 at 11:49
@MauroALLEGRANZA Yes, that was my assertion, that since x could be rewritten as itself multiplied with a fraction of arbitrary length up to infinity, x would have to be different from all possible values and thus undefined. I have tried to explain to myself why this is nonsense, but the second fraction you gave is valid.
– Oscar
Aug 2 at 11:58
But consider the simple case with only $c_1$: it has at least the value $0$ that satisfies it.
– Mauro ALLEGRANZA
Aug 2 at 12:01
I am sorry, could you articulate that? Do you mean $dfracx - c_1x - c_1 cdot x$ with $c_1 = 0$? Then I would understand that this is a valid rewriting of x, as it hasn't changed the possible values that x can be
– Oscar
Aug 2 at 12:06