Is taking modulus on both sides of an equation valid?Square root of a squared number changes sign, which to apply first?Why is it valid to multiply both sides of an equation by its complex conjugate?Will there be two square roots for a Complex number?Taking Mod on both sides, mathematically correct?Solving and graphing all values of $z$.When do you take into account the +2kpi for complex numbers arguments in complex equations$z^2 = sqrt3+ 3i$ (complex equation)Range of sum of complex numberssolve $(x+iy)^2= a+ ib$Question on the logic of a proof involving complex numbers
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Is taking modulus on both sides of an equation valid?
Square root of a squared number changes sign, which to apply first?Why is it valid to multiply both sides of an equation by its complex conjugate?Will there be two square roots for a Complex number?Taking Mod on both sides, mathematically correct?Solving and graphing all values of $z$.When do you take into account the +2kpi for complex numbers arguments in complex equations$z^2 = sqrt3+ 3i$ (complex equation)Range of sum of complex numberssolve $(x+iy)^2= a+ ib$Question on the logic of a proof involving complex numbers
$begingroup$
This might look like a copy of another question, but what I'm about to propose here is new. There's this question,
Find the least positive integral value of n for which $(frac1+i1-i)^n = 1$
While solving, if we multiply what is within the bracket, by the conjugate of denominator and divide by the same thing, we get $i$ in the bracket, that means, the question boils down to
$i ^ n= 1$
We know that the least positive value $n$ can have is $4$, for $i^n$ to be $1$.
Done.
Now,
IF I were to solve it by taking mod on both sides of the given equation,
I would get
$Big(fracBig)^n = |1|$
$Big(fracsqrt2sqrt2Big)^n = 1$
$1^n = 1$
NOTE that the least positive value of $n$ changes from $4$ to $1$.
Why is it so?
I read an answer on stack exchange, that it is valid to do anything with the equation until it maintains the equality. I didn't destroy the equality, so why does the answer vary?
Is there any restriction as to where to use the "taking-mod-both-sides" thing?
complex-numbers
edited May 12 at 0:35
Ryan Shesler
41712
asked May 12 at 0:26
user231094user231094
837
$endgroup$
add a comment |
$begingroup$
This might look like a copy of another question, but what I'm about to propose here is new. There's this question,
Find the least positive integral value of n for which $(frac1+i1-i)^n = 1$
While solving, if we multiply what is within the bracket, by the conjugate of denominator and divide by the same thing, we get $i$ in the bracket, that means, the question boils down to
$i ^ n= 1$
We know that the least positive value $n$ can have is $4$, for $i^n$ to be $1$.
Done.
Now,
IF I were to solve it by taking mod on both sides of the given equation,
I would get
$Big(fracBig)^n = |1|$
$Big(fracsqrt2sqrt2Big)^n = 1$
$1^n = 1$
NOTE that the least positive value of $n$ changes from $4$ to $1$.
Why is it so?
I read an answer on stack exchange, that it is valid to do anything with the equation until it maintains the equality. I didn't destroy the equality, so why does the answer vary?
Is there any restriction as to where to use the "taking-mod-both-sides" thing?
complex-numbers
edited May 12 at 0:35
Ryan Shesler
41712
asked May 12 at 0:26
user231094user231094
837
$endgroup$
16
$begingroup$
If $a=b$, then $|a|=|b|$. In this sense, it is "valid" to take the modulus of both sides. It is not necessarily the case, however, that if $|a|=|b|$, then $a=b$. Assuming this latter statement is what is leading you to incorrectly conclude that $n=1$ is a solution to the original problem.
$endgroup$
– jawheele
May 12 at 0:37
3
$begingroup$
Whenever you have $a = b$ (whatever things $a$, $b$ might be!), and a function applicable on $a$ and $b$, you always have: $a = b Rightarrow f(a) = f(b)$. Typically, this comes from basic rules in your logic, e.g. first-order logic with natural deduction.
$endgroup$
– ComFreek
May 12 at 7:11
1
$begingroup$
One way to approach this is to grok that the modulo operation "destroys" information. Say, if you take modulo10 and get 2 -- is it 2, 12, 22, or perhaps 222? Similarily x2 destroys information (the sign) hence the +/- value -- how could you compensate for it with modulo operation? (You could define algebras where the modulo operation does not destroy information, but that is something completely different)
$endgroup$
– ghellquist
May 12 at 20:27
$begingroup$
Question: What is the least non-negative value of $x$ such that $x = 2$? That's right, it's $2$. But if we multiply both sides by $0$, as in $0x = 0cdot 2$, then $0$ also works, why?
$endgroup$
– immibis
May 12 at 22:42
add a comment |
$begingroup$
This might look like a copy of another question, but what I'm about to propose here is new. There's this question,
Find the least positive integral value of n for which $(frac1+i1-i)^n = 1$
While solving, if we multiply what is within the bracket, by the conjugate of denominator and divide by the same thing, we get $i$ in the bracket, that means, the question boils down to
$i ^ n= 1$
We know that the least positive value $n$ can have is $4$, for $i^n$ to be $1$.
Done.
Now,
IF I were to solve it by taking mod on both sides of the given equation,
I would get
$Big(fracBig)^n = |1|$
$Big(fracsqrt2sqrt2Big)^n = 1$
$1^n = 1$
NOTE that the least positive value of $n$ changes from $4$ to $1$.
Why is it so?
I read an answer on stack exchange, that it is valid to do anything with the equation until it maintains the equality. I didn't destroy the equality, so why does the answer vary?
Is there any restriction as to where to use the "taking-mod-both-sides" thing?
complex-numbers
edited May 12 at 0:35
Ryan Shesler
41712
asked May 12 at 0:26
user231094user231094
837
$endgroup$
This might look like a copy of another question, but what I'm about to propose here is new. There's this question,
Find the least positive integral value of n for which $(frac1+i1-i)^n = 1$
While solving, if we multiply what is within the bracket, by the conjugate of denominator and divide by the same thing, we get $i$ in the bracket, that means, the question boils down to
$i ^ n= 1$
We know that the least positive value $n$ can have is $4$, for $i^n$ to be $1$.
Done.
Now,
IF I were to solve it by taking mod on both sides of the given equation,
I would get
$Big(fracBig)^n = |1|$
$Big(fracsqrt2sqrt2Big)^n = 1$
$1^n = 1$
NOTE that the least positive value of $n$ changes from $4$ to $1$.
Why is it so?
I read an answer on stack exchange, that it is valid to do anything with the equation until it maintains the equality. I didn't destroy the equality, so why does the answer vary?
Is there any restriction as to where to use the "taking-mod-both-sides" thing?
complex-numbers
complex-numbers
edited May 12 at 0:35
Ryan Shesler
41712
asked May 12 at 0:26
user231094user231094
837
edited May 12 at 0:35
Ryan Shesler
41712
asked May 12 at 0:26
user231094user231094
837
edited May 12 at 0:35
Ryan Shesler
41712
edited May 12 at 0:35
Ryan Shesler
41712
edited May 12 at 0:35
Ryan Shesler
41712
41712
asked May 12 at 0:26
user231094user231094
837
asked May 12 at 0:26
user231094user231094
837
asked May 12 at 0:26
user231094user231094
837
837
16
$begingroup$
If $a=b$, then $|a|=|b|$. In this sense, it is "valid" to take the modulus of both sides. It is not necessarily the case, however, that if $|a|=|b|$, then $a=b$. Assuming this latter statement is what is leading you to incorrectly conclude that $n=1$ is a solution to the original problem.
$endgroup$
– jawheele
May 12 at 0:37
3
$begingroup$
Whenever you have $a = b$ (whatever things $a$, $b$ might be!), and a function applicable on $a$ and $b$, you always have: $a = b Rightarrow f(a) = f(b)$. Typically, this comes from basic rules in your logic, e.g. first-order logic with natural deduction.
$endgroup$
– ComFreek
May 12 at 7:11
1
$begingroup$
One way to approach this is to grok that the modulo operation "destroys" information. Say, if you take modulo10 and get 2 -- is it 2, 12, 22, or perhaps 222? Similarily x2 destroys information (the sign) hence the +/- value -- how could you compensate for it with modulo operation? (You could define algebras where the modulo operation does not destroy information, but that is something completely different)
$endgroup$
– ghellquist
May 12 at 20:27
$begingroup$
Question: What is the least non-negative value of $x$ such that $x = 2$? That's right, it's $2$. But if we multiply both sides by $0$, as in $0x = 0cdot 2$, then $0$ also works, why?
$endgroup$
– immibis
May 12 at 22:42
add a comment |
16
$begingroup$
If $a=b$, then $|a|=|b|$. In this sense, it is "valid" to take the modulus of both sides. It is not necessarily the case, however, that if $|a|=|b|$, then $a=b$. Assuming this latter statement is what is leading you to incorrectly conclude that $n=1$ is a solution to the original problem.
$endgroup$
– jawheele
May 12 at 0:37
3
$begingroup$
Whenever you have $a = b$ (whatever things $a$, $b$ might be!), and a function applicable on $a$ and $b$, you always have: $a = b Rightarrow f(a) = f(b)$. Typically, this comes from basic rules in your logic, e.g. first-order logic with natural deduction.
$endgroup$
– ComFreek
May 12 at 7:11
1
$begingroup$
One way to approach this is to grok that the modulo operation "destroys" information. Say, if you take modulo10 and get 2 -- is it 2, 12, 22, or perhaps 222? Similarily x2 destroys information (the sign) hence the +/- value -- how could you compensate for it with modulo operation? (You could define algebras where the modulo operation does not destroy information, but that is something completely different)
$endgroup$
– ghellquist
May 12 at 20:27
$begingroup$
Question: What is the least non-negative value of $x$ such that $x = 2$? That's right, it's $2$. But if we multiply both sides by $0$, as in $0x = 0cdot 2$, then $0$ also works, why?
$endgroup$
– immibis
May 12 at 22:42
16
16
$begingroup$
If $a=b$, then $|a|=|b|$. In this sense, it is "valid" to take the modulus of both sides. It is not necessarily the case, however, that if $|a|=|b|$, then $a=b$. Assuming this latter statement is what is leading you to incorrectly conclude that $n=1$ is a solution to the original problem.
$endgroup$
– jawheele
May 12 at 0:37
$begingroup$
If $a=b$, then $|a|=|b|$. In this sense, it is "valid" to take the modulus of both sides. It is not necessarily the case, however, that if $|a|=|b|$, then $a=b$. Assuming this latter statement is what is leading you to incorrectly conclude that $n=1$ is a solution to the original problem.
$endgroup$
– jawheele
May 12 at 0:37
3
3
$begingroup$
Whenever you have $a = b$ (whatever things $a$, $b$ might be!), and a function applicable on $a$ and $b$, you always have: $a = b Rightarrow f(a) = f(b)$. Typically, this comes from basic rules in your logic, e.g. first-order logic with natural deduction.
$endgroup$
– ComFreek
May 12 at 7:11
$begingroup$
Whenever you have $a = b$ (whatever things $a$, $b$ might be!), and a function applicable on $a$ and $b$, you always have: $a = b Rightarrow f(a) = f(b)$. Typically, this comes from basic rules in your logic, e.g. first-order logic with natural deduction.
$endgroup$
– ComFreek
May 12 at 7:11
1
1
$begingroup$
One way to approach this is to grok that the modulo operation "destroys" information. Say, if you take modulo10 and get 2 -- is it 2, 12, 22, or perhaps 222? Similarily x2 destroys information (the sign) hence the +/- value -- how could you compensate for it with modulo operation? (You could define algebras where the modulo operation does not destroy information, but that is something completely different)
$endgroup$
– ghellquist
May 12 at 20:27
$begingroup$
One way to approach this is to grok that the modulo operation "destroys" information. Say, if you take modulo10 and get 2 -- is it 2, 12, 22, or perhaps 222? Similarily x2 destroys information (the sign) hence the +/- value -- how could you compensate for it with modulo operation? (You could define algebras where the modulo operation does not destroy information, but that is something completely different)
$endgroup$
– ghellquist
May 12 at 20:27
$begingroup$
Question: What is the least non-negative value of $x$ such that $x = 2$? That's right, it's $2$. But if we multiply both sides by $0$, as in $0x = 0cdot 2$, then $0$ also works, why?
$endgroup$
– immibis
May 12 at 22:42
$begingroup$
Question: What is the least non-negative value of $x$ such that $x = 2$? That's right, it's $2$. But if we multiply both sides by $0$, as in $0x = 0cdot 2$, then $0$ also works, why?
$endgroup$
– immibis
May 12 at 22:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.
When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.
answered May 12 at 0:38
Morgan RodgersMorgan Rodgers
10.2k31541
$endgroup$
$begingroup$
It is a many to one map
$endgroup$
– Arjang
May 12 at 0:41
$begingroup$
So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
$endgroup$
– user231094
May 12 at 0:41
2
$begingroup$
@user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
$endgroup$
– jawheele
May 12 at 0:47
add a comment |
$begingroup$
I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.
Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.
When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each solution to see if it satisfies the original equation.
answered May 12 at 0:48
Lee MosherLee Mosher
53.6k43893
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.
When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.
answered May 12 at 0:38
Morgan RodgersMorgan Rodgers
10.2k31541
$endgroup$
$begingroup$
It is a many to one map
$endgroup$
– Arjang
May 12 at 0:41
$begingroup$
So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
$endgroup$
– user231094
May 12 at 0:41
2
$begingroup$
@user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
$endgroup$
– jawheele
May 12 at 0:47
add a comment |
$begingroup$
The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.
When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.
answered May 12 at 0:38
Morgan RodgersMorgan Rodgers
10.2k31541
$endgroup$
$begingroup$
It is a many to one map
$endgroup$
– Arjang
May 12 at 0:41
$begingroup$
So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
$endgroup$
– user231094
May 12 at 0:41
2
$begingroup$
@user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
$endgroup$
– jawheele
May 12 at 0:47
add a comment |
$begingroup$
The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.
When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.
answered May 12 at 0:38
Morgan RodgersMorgan Rodgers
10.2k31541
$endgroup$
The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.
When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.
answered May 12 at 0:38
Morgan RodgersMorgan Rodgers
10.2k31541
edited May 12 at 0:49
answered May 12 at 0:38
Morgan RodgersMorgan Rodgers
10.2k31541
answered May 12 at 0:38
Morgan RodgersMorgan Rodgers
10.2k31541
answered May 12 at 0:38
Morgan RodgersMorgan Rodgers
10.2k31541
10.2k31541
$begingroup$
It is a many to one map
$endgroup$
– Arjang
May 12 at 0:41
$begingroup$
So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
$endgroup$
– user231094
May 12 at 0:41
2
$begingroup$
@user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
$endgroup$
– jawheele
May 12 at 0:47
add a comment |
$begingroup$
It is a many to one map
$endgroup$
– Arjang
May 12 at 0:41
$begingroup$
So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
$endgroup$
– user231094
May 12 at 0:41
2
$begingroup$
@user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
$endgroup$
– jawheele
May 12 at 0:47
$begingroup$
It is a many to one map
$endgroup$
– Arjang
May 12 at 0:41
$begingroup$
It is a many to one map
$endgroup$
– Arjang
May 12 at 0:41
$begingroup$
So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
$endgroup$
– user231094
May 12 at 0:41
$begingroup$
So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
$endgroup$
– user231094
May 12 at 0:41
2
2
$begingroup$
@user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
$endgroup$
– jawheele
May 12 at 0:47
$begingroup$
@user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
$endgroup$
– jawheele
May 12 at 0:47
add a comment |
$begingroup$
I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.
Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.
When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each solution to see if it satisfies the original equation.
answered May 12 at 0:48
Lee MosherLee Mosher
53.6k43893
$endgroup$
add a comment |
$begingroup$
I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.
Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.
When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each solution to see if it satisfies the original equation.
answered May 12 at 0:48
Lee MosherLee Mosher
53.6k43893
$endgroup$
add a comment |
$begingroup$
I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.
Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.
When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each solution to see if it satisfies the original equation.
answered May 12 at 0:48
Lee MosherLee Mosher
53.6k43893
$endgroup$
I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.
Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.
When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each solution to see if it satisfies the original equation.
answered May 12 at 0:48
Lee MosherLee Mosher
53.6k43893
edited May 13 at 0:07
answered May 12 at 0:48
Lee MosherLee Mosher
53.6k43893
answered May 12 at 0:48
Lee MosherLee Mosher
53.6k43893
answered May 12 at 0:48
Lee MosherLee Mosher
53.6k43893
53.6k43893
add a comment |
add a comment |
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$begingroup$
If $a=b$, then $|a|=|b|$. In this sense, it is "valid" to take the modulus of both sides. It is not necessarily the case, however, that if $|a|=|b|$, then $a=b$. Assuming this latter statement is what is leading you to incorrectly conclude that $n=1$ is a solution to the original problem.
$endgroup$
– jawheele
May 12 at 0:37
3
$begingroup$
Whenever you have $a = b$ (whatever things $a$, $b$ might be!), and a function applicable on $a$ and $b$, you always have: $a = b Rightarrow f(a) = f(b)$. Typically, this comes from basic rules in your logic, e.g. first-order logic with natural deduction.
$endgroup$
– ComFreek
May 12 at 7:11
1
$begingroup$
One way to approach this is to grok that the modulo operation "destroys" information. Say, if you take modulo10 and get 2 -- is it 2, 12, 22, or perhaps 222? Similarily x2 destroys information (the sign) hence the +/- value -- how could you compensate for it with modulo operation? (You could define algebras where the modulo operation does not destroy information, but that is something completely different)
$endgroup$
– ghellquist
May 12 at 20:27
$begingroup$
Question: What is the least non-negative value of $x$ such that $x = 2$? That's right, it's $2$. But if we multiply both sides by $0$, as in $0x = 0cdot 2$, then $0$ also works, why?
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– immibis
May 12 at 22:42