Is it valid to multiply both sides of an equation by a complex number?
$begingroup$
In case I'd like to get rid of some complex denominators, could I multiply both sides of the equation by the common denominator?
I understand there's an other way, but i'd like to know if the above mentioned would be a valid operation, and if not, why.
algebra-precalculus complex-numbers
edited Dec 29 '18 at 23:14
Eric Wofsey
187k14215344
asked May 24 '17 at 19:49
boribori
62
$endgroup$
add a comment |
$begingroup$
In case I'd like to get rid of some complex denominators, could I multiply both sides of the equation by the common denominator?
I understand there's an other way, but i'd like to know if the above mentioned would be a valid operation, and if not, why.
algebra-precalculus complex-numbers
edited Dec 29 '18 at 23:14
Eric Wofsey
187k14215344
asked May 24 '17 at 19:49
boribori
62
$endgroup$
$begingroup$
Look up the definition of a FIELD.... $mathbb Q, mathbb R, mathbb C$ are fields. In a field if $zne 0$ then $x=yiff zx=zy.$... And in a field, if $d_1,...,d_n$ are all non-zero then their product $z$ is non-zero.
$endgroup$
– DanielWainfleet
May 24 '17 at 20:16
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Thanks @DanielWainfleet, this helps me a lot. I had empiric results, but I wasn't sure how to prove it.
$endgroup$
– bori
May 24 '17 at 20:48
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You can multiply both sides by an elephant as far as anyone cares. If $a = b$ then "doing anything you like to $a$" = "doing the exact same thing to $b$" because $a$ and $b$ are the exact same thing. And if you do something to $a$ you ARE do it to $b$.
$endgroup$
– fleablood
Dec 29 '18 at 23:23
add a comment |
$begingroup$
In case I'd like to get rid of some complex denominators, could I multiply both sides of the equation by the common denominator?
I understand there's an other way, but i'd like to know if the above mentioned would be a valid operation, and if not, why.
algebra-precalculus complex-numbers
edited Dec 29 '18 at 23:14
Eric Wofsey
187k14215344
asked May 24 '17 at 19:49
boribori
62
$endgroup$
In case I'd like to get rid of some complex denominators, could I multiply both sides of the equation by the common denominator?
I understand there's an other way, but i'd like to know if the above mentioned would be a valid operation, and if not, why.
algebra-precalculus complex-numbers
algebra-precalculus complex-numbers
edited Dec 29 '18 at 23:14
Eric Wofsey
187k14215344
asked May 24 '17 at 19:49
boribori
62
edited Dec 29 '18 at 23:14
Eric Wofsey
187k14215344
asked May 24 '17 at 19:49
boribori
62
edited Dec 29 '18 at 23:14
Eric Wofsey
187k14215344
edited Dec 29 '18 at 23:14
Eric Wofsey
187k14215344
edited Dec 29 '18 at 23:14
Eric Wofsey
187k14215344
187k14215344
asked May 24 '17 at 19:49
boribori
62
asked May 24 '17 at 19:49
boribori
62
asked May 24 '17 at 19:49
boribori
62
62
$begingroup$
Look up the definition of a FIELD.... $mathbb Q, mathbb R, mathbb C$ are fields. In a field if $zne 0$ then $x=yiff zx=zy.$... And in a field, if $d_1,...,d_n$ are all non-zero then their product $z$ is non-zero.
$endgroup$
– DanielWainfleet
May 24 '17 at 20:16
$begingroup$
Thanks @DanielWainfleet, this helps me a lot. I had empiric results, but I wasn't sure how to prove it.
$endgroup$
– bori
May 24 '17 at 20:48
$begingroup$
You can multiply both sides by an elephant as far as anyone cares. If $a = b$ then "doing anything you like to $a$" = "doing the exact same thing to $b$" because $a$ and $b$ are the exact same thing. And if you do something to $a$ you ARE do it to $b$.
$endgroup$
– fleablood
Dec 29 '18 at 23:23
add a comment |
$begingroup$
Look up the definition of a FIELD.... $mathbb Q, mathbb R, mathbb C$ are fields. In a field if $zne 0$ then $x=yiff zx=zy.$... And in a field, if $d_1,...,d_n$ are all non-zero then their product $z$ is non-zero.
$endgroup$
– DanielWainfleet
May 24 '17 at 20:16
$begingroup$
Thanks @DanielWainfleet, this helps me a lot. I had empiric results, but I wasn't sure how to prove it.
$endgroup$
– bori
May 24 '17 at 20:48
$begingroup$
You can multiply both sides by an elephant as far as anyone cares. If $a = b$ then "doing anything you like to $a$" = "doing the exact same thing to $b$" because $a$ and $b$ are the exact same thing. And if you do something to $a$ you ARE do it to $b$.
$endgroup$
– fleablood
Dec 29 '18 at 23:23
$begingroup$
Look up the definition of a FIELD.... $mathbb Q, mathbb R, mathbb C$ are fields. In a field if $zne 0$ then $x=yiff zx=zy.$... And in a field, if $d_1,...,d_n$ are all non-zero then their product $z$ is non-zero.
$endgroup$
– DanielWainfleet
May 24 '17 at 20:16
$begingroup$
Look up the definition of a FIELD.... $mathbb Q, mathbb R, mathbb C$ are fields. In a field if $zne 0$ then $x=yiff zx=zy.$... And in a field, if $d_1,...,d_n$ are all non-zero then their product $z$ is non-zero.
$endgroup$
– DanielWainfleet
May 24 '17 at 20:16
$begingroup$
Thanks @DanielWainfleet, this helps me a lot. I had empiric results, but I wasn't sure how to prove it.
$endgroup$
– bori
May 24 '17 at 20:48
$begingroup$
Thanks @DanielWainfleet, this helps me a lot. I had empiric results, but I wasn't sure how to prove it.
$endgroup$
– bori
May 24 '17 at 20:48
$begingroup$
You can multiply both sides by an elephant as far as anyone cares. If $a = b$ then "doing anything you like to $a$" = "doing the exact same thing to $b$" because $a$ and $b$ are the exact same thing. And if you do something to $a$ you ARE do it to $b$.
$endgroup$
– fleablood
Dec 29 '18 at 23:23
$begingroup$
You can multiply both sides by an elephant as far as anyone cares. If $a = b$ then "doing anything you like to $a$" = "doing the exact same thing to $b$" because $a$ and $b$ are the exact same thing. And if you do something to $a$ you ARE do it to $b$.
$endgroup$
– fleablood
Dec 29 '18 at 23:23
add a comment |
3 Answers
3
active
oldest
votes
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If $left(mbox{some expression}right)=left(mbox{another expression}right)$, this means that both expressions evaluate to the same complex number. Let's call this number $z$. This means that:
$$z=left(mbox{some expression}right)=left(mbox{another expression}right)$$
Well, if you want to multiply both sides by another complex, say $w$, the equality will still hold, because both $wcdotleft(mbox{some expression}right)$ and $wcdotleft(mbox{another expression}right)$ are just the same as $wcdot z$.
Beware that if $w$ is $0$ (that is, if you want to multiply both sides by zero), then the equality will still hold: it will be just $0=0$. Of course, this is not useful---you lost all the information you had in your original equation, because you can't divide both sizes by $0$ to get back where you started.
answered May 24 '17 at 20:01
foninifonini
1,78911038
$endgroup$
add a comment |
$begingroup$
It is perfectly valid as long as the multiplier you use isn't zero.
It's even valid to multiply both sides by a variable expression, but again you must include the proviso that variable multiplier cannot be zero. Many are under the mistaken impression that this is not valid, but it is -- you just have to include the additional assumption to the end.
For example, suppose you have the equation $x=1$. It is perfectly valid to multiply both sides by $x$, with the proviso that $xneq 0$:
$$x=1$$
$$x(x)=x(1)tag{$xneq 0$}$$
$$x^2=xtag{$xneq 0$}$$
$$x^2-x = 0tag{$xneq 0$}$$
$$x(x-1)=0tag{$xneq 0$}$$
This equation appears to have an additional solution -- but don't forget that $x=0$ is not a solution since it is excluded by the proviso.
answered May 24 '17 at 19:53
MPWMPW
30.3k12157
$endgroup$
add a comment |
$begingroup$
This is valid. You can still multiply both sides of an equation by a scalar and still have it hold. It's just like the real numbers.
answered May 24 '17 at 19:53
Sean RobersonSean Roberson
6,39031327
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If $left(mbox{some expression}right)=left(mbox{another expression}right)$, this means that both expressions evaluate to the same complex number. Let's call this number $z$. This means that:
$$z=left(mbox{some expression}right)=left(mbox{another expression}right)$$
Well, if you want to multiply both sides by another complex, say $w$, the equality will still hold, because both $wcdotleft(mbox{some expression}right)$ and $wcdotleft(mbox{another expression}right)$ are just the same as $wcdot z$.
Beware that if $w$ is $0$ (that is, if you want to multiply both sides by zero), then the equality will still hold: it will be just $0=0$. Of course, this is not useful---you lost all the information you had in your original equation, because you can't divide both sizes by $0$ to get back where you started.
answered May 24 '17 at 20:01
foninifonini
1,78911038
$endgroup$
add a comment |
$begingroup$
If $left(mbox{some expression}right)=left(mbox{another expression}right)$, this means that both expressions evaluate to the same complex number. Let's call this number $z$. This means that:
$$z=left(mbox{some expression}right)=left(mbox{another expression}right)$$
Well, if you want to multiply both sides by another complex, say $w$, the equality will still hold, because both $wcdotleft(mbox{some expression}right)$ and $wcdotleft(mbox{another expression}right)$ are just the same as $wcdot z$.
Beware that if $w$ is $0$ (that is, if you want to multiply both sides by zero), then the equality will still hold: it will be just $0=0$. Of course, this is not useful---you lost all the information you had in your original equation, because you can't divide both sizes by $0$ to get back where you started.
answered May 24 '17 at 20:01
foninifonini
1,78911038
$endgroup$
add a comment |
$begingroup$
If $left(mbox{some expression}right)=left(mbox{another expression}right)$, this means that both expressions evaluate to the same complex number. Let's call this number $z$. This means that:
$$z=left(mbox{some expression}right)=left(mbox{another expression}right)$$
Well, if you want to multiply both sides by another complex, say $w$, the equality will still hold, because both $wcdotleft(mbox{some expression}right)$ and $wcdotleft(mbox{another expression}right)$ are just the same as $wcdot z$.
Beware that if $w$ is $0$ (that is, if you want to multiply both sides by zero), then the equality will still hold: it will be just $0=0$. Of course, this is not useful---you lost all the information you had in your original equation, because you can't divide both sizes by $0$ to get back where you started.
answered May 24 '17 at 20:01
foninifonini
1,78911038
$endgroup$
If $left(mbox{some expression}right)=left(mbox{another expression}right)$, this means that both expressions evaluate to the same complex number. Let's call this number $z$. This means that:
$$z=left(mbox{some expression}right)=left(mbox{another expression}right)$$
Well, if you want to multiply both sides by another complex, say $w$, the equality will still hold, because both $wcdotleft(mbox{some expression}right)$ and $wcdotleft(mbox{another expression}right)$ are just the same as $wcdot z$.
Beware that if $w$ is $0$ (that is, if you want to multiply both sides by zero), then the equality will still hold: it will be just $0=0$. Of course, this is not useful---you lost all the information you had in your original equation, because you can't divide both sizes by $0$ to get back where you started.
answered May 24 '17 at 20:01
foninifonini
1,78911038
answered May 24 '17 at 20:01
foninifonini
1,78911038
answered May 24 '17 at 20:01
foninifonini
1,78911038
answered May 24 '17 at 20:01
foninifonini
1,78911038
1,78911038
add a comment |
add a comment |
$begingroup$
It is perfectly valid as long as the multiplier you use isn't zero.
It's even valid to multiply both sides by a variable expression, but again you must include the proviso that variable multiplier cannot be zero. Many are under the mistaken impression that this is not valid, but it is -- you just have to include the additional assumption to the end.
For example, suppose you have the equation $x=1$. It is perfectly valid to multiply both sides by $x$, with the proviso that $xneq 0$:
$$x=1$$
$$x(x)=x(1)tag{$xneq 0$}$$
$$x^2=xtag{$xneq 0$}$$
$$x^2-x = 0tag{$xneq 0$}$$
$$x(x-1)=0tag{$xneq 0$}$$
This equation appears to have an additional solution -- but don't forget that $x=0$ is not a solution since it is excluded by the proviso.
answered May 24 '17 at 19:53
MPWMPW
30.3k12157
$endgroup$
add a comment |
$begingroup$
It is perfectly valid as long as the multiplier you use isn't zero.
It's even valid to multiply both sides by a variable expression, but again you must include the proviso that variable multiplier cannot be zero. Many are under the mistaken impression that this is not valid, but it is -- you just have to include the additional assumption to the end.
For example, suppose you have the equation $x=1$. It is perfectly valid to multiply both sides by $x$, with the proviso that $xneq 0$:
$$x=1$$
$$x(x)=x(1)tag{$xneq 0$}$$
$$x^2=xtag{$xneq 0$}$$
$$x^2-x = 0tag{$xneq 0$}$$
$$x(x-1)=0tag{$xneq 0$}$$
This equation appears to have an additional solution -- but don't forget that $x=0$ is not a solution since it is excluded by the proviso.
answered May 24 '17 at 19:53
MPWMPW
30.3k12157
$endgroup$
add a comment |
$begingroup$
It is perfectly valid as long as the multiplier you use isn't zero.
It's even valid to multiply both sides by a variable expression, but again you must include the proviso that variable multiplier cannot be zero. Many are under the mistaken impression that this is not valid, but it is -- you just have to include the additional assumption to the end.
For example, suppose you have the equation $x=1$. It is perfectly valid to multiply both sides by $x$, with the proviso that $xneq 0$:
$$x=1$$
$$x(x)=x(1)tag{$xneq 0$}$$
$$x^2=xtag{$xneq 0$}$$
$$x^2-x = 0tag{$xneq 0$}$$
$$x(x-1)=0tag{$xneq 0$}$$
This equation appears to have an additional solution -- but don't forget that $x=0$ is not a solution since it is excluded by the proviso.
answered May 24 '17 at 19:53
MPWMPW
30.3k12157
$endgroup$
It is perfectly valid as long as the multiplier you use isn't zero.
It's even valid to multiply both sides by a variable expression, but again you must include the proviso that variable multiplier cannot be zero. Many are under the mistaken impression that this is not valid, but it is -- you just have to include the additional assumption to the end.
For example, suppose you have the equation $x=1$. It is perfectly valid to multiply both sides by $x$, with the proviso that $xneq 0$:
$$x=1$$
$$x(x)=x(1)tag{$xneq 0$}$$
$$x^2=xtag{$xneq 0$}$$
$$x^2-x = 0tag{$xneq 0$}$$
$$x(x-1)=0tag{$xneq 0$}$$
This equation appears to have an additional solution -- but don't forget that $x=0$ is not a solution since it is excluded by the proviso.
answered May 24 '17 at 19:53
MPWMPW
30.3k12157
edited May 24 '17 at 20:00
answered May 24 '17 at 19:53
MPWMPW
30.3k12157
answered May 24 '17 at 19:53
MPWMPW
30.3k12157
answered May 24 '17 at 19:53
MPWMPW
30.3k12157
30.3k12157
add a comment |
add a comment |
$begingroup$
This is valid. You can still multiply both sides of an equation by a scalar and still have it hold. It's just like the real numbers.
answered May 24 '17 at 19:53
Sean RobersonSean Roberson
6,39031327
$endgroup$
add a comment |
$begingroup$
This is valid. You can still multiply both sides of an equation by a scalar and still have it hold. It's just like the real numbers.
answered May 24 '17 at 19:53
Sean RobersonSean Roberson
6,39031327
$endgroup$
add a comment |
$begingroup$
This is valid. You can still multiply both sides of an equation by a scalar and still have it hold. It's just like the real numbers.
answered May 24 '17 at 19:53
Sean RobersonSean Roberson
6,39031327
$endgroup$
This is valid. You can still multiply both sides of an equation by a scalar and still have it hold. It's just like the real numbers.
answered May 24 '17 at 19:53
Sean RobersonSean Roberson
6,39031327
answered May 24 '17 at 19:53
Sean RobersonSean Roberson
6,39031327
answered May 24 '17 at 19:53
Sean RobersonSean Roberson
6,39031327
answered May 24 '17 at 19:53
Sean RobersonSean Roberson
6,39031327
6,39031327
add a comment |
add a comment |
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$begingroup$
Look up the definition of a FIELD.... $mathbb Q, mathbb R, mathbb C$ are fields. In a field if $zne 0$ then $x=yiff zx=zy.$... And in a field, if $d_1,...,d_n$ are all non-zero then their product $z$ is non-zero.
$endgroup$
– DanielWainfleet
May 24 '17 at 20:16
$begingroup$
Thanks @DanielWainfleet, this helps me a lot. I had empiric results, but I wasn't sure how to prove it.
$endgroup$
– bori
May 24 '17 at 20:48
$begingroup$
You can multiply both sides by an elephant as far as anyone cares. If $a = b$ then "doing anything you like to $a$" = "doing the exact same thing to $b$" because $a$ and $b$ are the exact same thing. And if you do something to $a$ you ARE do it to $b$.
$endgroup$
– fleablood
Dec 29 '18 at 23:23