Isn't $frac{ax+bi}{ax+bi}$ equal to $1$? [closed]
$begingroup$
Isn’t $dfrac{ax+bi}{ax+bi}$ equal to $1$?
Here, $i=sqrt{-1}$, & $a$,$b$ & $x$ $in$ $R$
abstract-algebra complex-numbers
asked Dec 27 '18 at 13:07
KruthikKruthik
133
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closed as off-topic by Jyrki Lahtonen, Lord Shark the Unknown, user91500, Namaste, Jennifer Dec 31 '18 at 15:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Lord Shark the Unknown, user91500, Namaste, Jennifer
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Isn’t $dfrac{ax+bi}{ax+bi}$ equal to $1$?
Here, $i=sqrt{-1}$, & $a$,$b$ & $x$ $in$ $R$
abstract-algebra complex-numbers
asked Dec 27 '18 at 13:07
KruthikKruthik
133
$endgroup$
closed as off-topic by Jyrki Lahtonen, Lord Shark the Unknown, user91500, Namaste, Jennifer Dec 31 '18 at 15:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Lord Shark the Unknown, user91500, Namaste, Jennifer
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Yes, unless $ax+bi = 0$.
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– Card_Trick
Dec 27 '18 at 13:21
1
$begingroup$
I suppose $a,$ $b,$ and $x$ all are meant to be real numbers, but I think it would be better to make that explicit. I don't know of any ironclad rule that says one can never use any of those symbols to represent a complex number.
$endgroup$
– David K
Dec 27 '18 at 13:53
add a comment |
$begingroup$
Isn’t $dfrac{ax+bi}{ax+bi}$ equal to $1$?
Here, $i=sqrt{-1}$, & $a$,$b$ & $x$ $in$ $R$
abstract-algebra complex-numbers
asked Dec 27 '18 at 13:07
KruthikKruthik
133
$endgroup$
Isn’t $dfrac{ax+bi}{ax+bi}$ equal to $1$?
Here, $i=sqrt{-1}$, & $a$,$b$ & $x$ $in$ $R$
abstract-algebra complex-numbers
abstract-algebra complex-numbers
asked Dec 27 '18 at 13:07
KruthikKruthik
133
asked Dec 27 '18 at 13:07
KruthikKruthik
133
edited Dec 27 '18 at 14:03
Kruthik
asked Dec 27 '18 at 13:07
KruthikKruthik
133
asked Dec 27 '18 at 13:07
KruthikKruthik
133
asked Dec 27 '18 at 13:07
KruthikKruthik
133
133
closed as off-topic by Jyrki Lahtonen, Lord Shark the Unknown, user91500, Namaste, Jennifer Dec 31 '18 at 15:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Lord Shark the Unknown, user91500, Namaste, Jennifer
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jyrki Lahtonen, Lord Shark the Unknown, user91500, Namaste, Jennifer Dec 31 '18 at 15:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Lord Shark the Unknown, user91500, Namaste, Jennifer
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Yes, unless $ax+bi = 0$.
$endgroup$
– Card_Trick
Dec 27 '18 at 13:21
1
$begingroup$
I suppose $a,$ $b,$ and $x$ all are meant to be real numbers, but I think it would be better to make that explicit. I don't know of any ironclad rule that says one can never use any of those symbols to represent a complex number.
$endgroup$
– David K
Dec 27 '18 at 13:53
add a comment |
3
$begingroup$
Yes, unless $ax+bi = 0$.
$endgroup$
– Card_Trick
Dec 27 '18 at 13:21
1
$begingroup$
I suppose $a,$ $b,$ and $x$ all are meant to be real numbers, but I think it would be better to make that explicit. I don't know of any ironclad rule that says one can never use any of those symbols to represent a complex number.
$endgroup$
– David K
Dec 27 '18 at 13:53
3
3
$begingroup$
Yes, unless $ax+bi = 0$.
$endgroup$
– Card_Trick
Dec 27 '18 at 13:21
$begingroup$
Yes, unless $ax+bi = 0$.
$endgroup$
– Card_Trick
Dec 27 '18 at 13:21
1
1
$begingroup$
I suppose $a,$ $b,$ and $x$ all are meant to be real numbers, but I think it would be better to make that explicit. I don't know of any ironclad rule that says one can never use any of those symbols to represent a complex number.
$endgroup$
– David K
Dec 27 '18 at 13:53
$begingroup$
I suppose $a,$ $b,$ and $x$ all are meant to be real numbers, but I think it would be better to make that explicit. I don't know of any ironclad rule that says one can never use any of those symbols to represent a complex number.
$endgroup$
– David K
Dec 27 '18 at 13:53
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Division in the field of complex numbers is defined in the following manner:
$frac{ax+bi}{ax+bi}=frac{(ax+bi)(ax-bi)}{(ax+bi)(ax-bi)}$, so
$frac{ax+bi}{ax+bi}=frac{a^2x^2+b^2}{a^2x^2+b^2}$, which is a real number and equal to 1. Hence your intuition is correct. But note that this is not true for $b=0$ and either $a=0$ or $x=0$ or both. Because $frac{a^2x^2+b^2}{a^2x^2+b^2}$ is not defined when $b=0$ and either $a=0$ or $x=0$ or both.
answered Dec 27 '18 at 13:15
toric_actionstoric_actions
1088
$endgroup$
3
$begingroup$
What's the reason for multiplying by the conjugate? The fact that $z/z=1$ (for $zne0$) is simply the definition of inverse element.
$endgroup$
– egreg
Dec 27 '18 at 14:35
$begingroup$
In this case, it definitely is but if you see I have mentioned that process as division and not merely as inversion.
$endgroup$
– toric_actions
Dec 27 '18 at 14:40
add a comment |
$begingroup$
If we assume that at least one of $ax$ and $b$ is non-zero, then yes. The complex numbers under the usual operations of addition and multiplication form a field.
answered Dec 27 '18 at 13:12
Hans HüttelHans Hüttel
3,3572921
$endgroup$
2
$begingroup$
This is incorrect as written. What if $a=1,x=0,b=0$?
$endgroup$
– MPW
Dec 27 '18 at 14:00
add a comment |
$begingroup$
In the complex field, every nonzero complex number $z$ has an inverse $z^{-1}$. It is customary to write
$$
wz^{-1}=frac{w}{z}
$$
By definition, $zz^{-1}=1$, so in the alternative notation
$$
frac{z}{z}=1
$$
for every complex number $zne0$.
Now take $z=ax+bi$ and you have your result. Of course, if $ax+bi=0$ (that is, $ax=0$ and $b=0$), the expression $(ax+bi)/(ax+bi)$ is undefined.
answered Dec 27 '18 at 14:39
egregegreg
184k1486205
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Division in the field of complex numbers is defined in the following manner:
$frac{ax+bi}{ax+bi}=frac{(ax+bi)(ax-bi)}{(ax+bi)(ax-bi)}$, so
$frac{ax+bi}{ax+bi}=frac{a^2x^2+b^2}{a^2x^2+b^2}$, which is a real number and equal to 1. Hence your intuition is correct. But note that this is not true for $b=0$ and either $a=0$ or $x=0$ or both. Because $frac{a^2x^2+b^2}{a^2x^2+b^2}$ is not defined when $b=0$ and either $a=0$ or $x=0$ or both.
answered Dec 27 '18 at 13:15
toric_actionstoric_actions
1088
$endgroup$
3
$begingroup$
What's the reason for multiplying by the conjugate? The fact that $z/z=1$ (for $zne0$) is simply the definition of inverse element.
$endgroup$
– egreg
Dec 27 '18 at 14:35
$begingroup$
In this case, it definitely is but if you see I have mentioned that process as division and not merely as inversion.
$endgroup$
– toric_actions
Dec 27 '18 at 14:40
add a comment |
$begingroup$
Division in the field of complex numbers is defined in the following manner:
$frac{ax+bi}{ax+bi}=frac{(ax+bi)(ax-bi)}{(ax+bi)(ax-bi)}$, so
$frac{ax+bi}{ax+bi}=frac{a^2x^2+b^2}{a^2x^2+b^2}$, which is a real number and equal to 1. Hence your intuition is correct. But note that this is not true for $b=0$ and either $a=0$ or $x=0$ or both. Because $frac{a^2x^2+b^2}{a^2x^2+b^2}$ is not defined when $b=0$ and either $a=0$ or $x=0$ or both.
answered Dec 27 '18 at 13:15
toric_actionstoric_actions
1088
$endgroup$
3
$begingroup$
What's the reason for multiplying by the conjugate? The fact that $z/z=1$ (for $zne0$) is simply the definition of inverse element.
$endgroup$
– egreg
Dec 27 '18 at 14:35
$begingroup$
In this case, it definitely is but if you see I have mentioned that process as division and not merely as inversion.
$endgroup$
– toric_actions
Dec 27 '18 at 14:40
add a comment |
$begingroup$
Division in the field of complex numbers is defined in the following manner:
$frac{ax+bi}{ax+bi}=frac{(ax+bi)(ax-bi)}{(ax+bi)(ax-bi)}$, so
$frac{ax+bi}{ax+bi}=frac{a^2x^2+b^2}{a^2x^2+b^2}$, which is a real number and equal to 1. Hence your intuition is correct. But note that this is not true for $b=0$ and either $a=0$ or $x=0$ or both. Because $frac{a^2x^2+b^2}{a^2x^2+b^2}$ is not defined when $b=0$ and either $a=0$ or $x=0$ or both.
answered Dec 27 '18 at 13:15
toric_actionstoric_actions
1088
$endgroup$
Division in the field of complex numbers is defined in the following manner:
$frac{ax+bi}{ax+bi}=frac{(ax+bi)(ax-bi)}{(ax+bi)(ax-bi)}$, so
$frac{ax+bi}{ax+bi}=frac{a^2x^2+b^2}{a^2x^2+b^2}$, which is a real number and equal to 1. Hence your intuition is correct. But note that this is not true for $b=0$ and either $a=0$ or $x=0$ or both. Because $frac{a^2x^2+b^2}{a^2x^2+b^2}$ is not defined when $b=0$ and either $a=0$ or $x=0$ or both.
answered Dec 27 '18 at 13:15
toric_actionstoric_actions
1088
answered Dec 27 '18 at 13:15
toric_actionstoric_actions
1088
answered Dec 27 '18 at 13:15
toric_actionstoric_actions
1088
answered Dec 27 '18 at 13:15
toric_actionstoric_actions
1088
1088
3
$begingroup$
What's the reason for multiplying by the conjugate? The fact that $z/z=1$ (for $zne0$) is simply the definition of inverse element.
$endgroup$
– egreg
Dec 27 '18 at 14:35
$begingroup$
In this case, it definitely is but if you see I have mentioned that process as division and not merely as inversion.
$endgroup$
– toric_actions
Dec 27 '18 at 14:40
add a comment |
3
$begingroup$
What's the reason for multiplying by the conjugate? The fact that $z/z=1$ (for $zne0$) is simply the definition of inverse element.
$endgroup$
– egreg
Dec 27 '18 at 14:35
$begingroup$
In this case, it definitely is but if you see I have mentioned that process as division and not merely as inversion.
$endgroup$
– toric_actions
Dec 27 '18 at 14:40
3
3
$begingroup$
What's the reason for multiplying by the conjugate? The fact that $z/z=1$ (for $zne0$) is simply the definition of inverse element.
$endgroup$
– egreg
Dec 27 '18 at 14:35
$begingroup$
What's the reason for multiplying by the conjugate? The fact that $z/z=1$ (for $zne0$) is simply the definition of inverse element.
$endgroup$
– egreg
Dec 27 '18 at 14:35
$begingroup$
In this case, it definitely is but if you see I have mentioned that process as division and not merely as inversion.
$endgroup$
– toric_actions
Dec 27 '18 at 14:40
$begingroup$
In this case, it definitely is but if you see I have mentioned that process as division and not merely as inversion.
$endgroup$
– toric_actions
Dec 27 '18 at 14:40
add a comment |
$begingroup$
If we assume that at least one of $ax$ and $b$ is non-zero, then yes. The complex numbers under the usual operations of addition and multiplication form a field.
answered Dec 27 '18 at 13:12
Hans HüttelHans Hüttel
3,3572921
$endgroup$
2
$begingroup$
This is incorrect as written. What if $a=1,x=0,b=0$?
$endgroup$
– MPW
Dec 27 '18 at 14:00
add a comment |
$begingroup$
If we assume that at least one of $ax$ and $b$ is non-zero, then yes. The complex numbers under the usual operations of addition and multiplication form a field.
answered Dec 27 '18 at 13:12
Hans HüttelHans Hüttel
3,3572921
$endgroup$
2
$begingroup$
This is incorrect as written. What if $a=1,x=0,b=0$?
$endgroup$
– MPW
Dec 27 '18 at 14:00
add a comment |
$begingroup$
If we assume that at least one of $ax$ and $b$ is non-zero, then yes. The complex numbers under the usual operations of addition and multiplication form a field.
answered Dec 27 '18 at 13:12
Hans HüttelHans Hüttel
3,3572921
$endgroup$
If we assume that at least one of $ax$ and $b$ is non-zero, then yes. The complex numbers under the usual operations of addition and multiplication form a field.
answered Dec 27 '18 at 13:12
Hans HüttelHans Hüttel
3,3572921
edited Dec 27 '18 at 20:45
answered Dec 27 '18 at 13:12
Hans HüttelHans Hüttel
3,3572921
answered Dec 27 '18 at 13:12
Hans HüttelHans Hüttel
3,3572921
answered Dec 27 '18 at 13:12
Hans HüttelHans Hüttel
3,3572921
3,3572921
2
$begingroup$
This is incorrect as written. What if $a=1,x=0,b=0$?
$endgroup$
– MPW
Dec 27 '18 at 14:00
add a comment |
2
$begingroup$
This is incorrect as written. What if $a=1,x=0,b=0$?
$endgroup$
– MPW
Dec 27 '18 at 14:00
2
2
$begingroup$
This is incorrect as written. What if $a=1,x=0,b=0$?
$endgroup$
– MPW
Dec 27 '18 at 14:00
$begingroup$
This is incorrect as written. What if $a=1,x=0,b=0$?
$endgroup$
– MPW
Dec 27 '18 at 14:00
add a comment |
$begingroup$
In the complex field, every nonzero complex number $z$ has an inverse $z^{-1}$. It is customary to write
$$
wz^{-1}=frac{w}{z}
$$
By definition, $zz^{-1}=1$, so in the alternative notation
$$
frac{z}{z}=1
$$
for every complex number $zne0$.
Now take $z=ax+bi$ and you have your result. Of course, if $ax+bi=0$ (that is, $ax=0$ and $b=0$), the expression $(ax+bi)/(ax+bi)$ is undefined.
answered Dec 27 '18 at 14:39
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
In the complex field, every nonzero complex number $z$ has an inverse $z^{-1}$. It is customary to write
$$
wz^{-1}=frac{w}{z}
$$
By definition, $zz^{-1}=1$, so in the alternative notation
$$
frac{z}{z}=1
$$
for every complex number $zne0$.
Now take $z=ax+bi$ and you have your result. Of course, if $ax+bi=0$ (that is, $ax=0$ and $b=0$), the expression $(ax+bi)/(ax+bi)$ is undefined.
answered Dec 27 '18 at 14:39
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
In the complex field, every nonzero complex number $z$ has an inverse $z^{-1}$. It is customary to write
$$
wz^{-1}=frac{w}{z}
$$
By definition, $zz^{-1}=1$, so in the alternative notation
$$
frac{z}{z}=1
$$
for every complex number $zne0$.
Now take $z=ax+bi$ and you have your result. Of course, if $ax+bi=0$ (that is, $ax=0$ and $b=0$), the expression $(ax+bi)/(ax+bi)$ is undefined.
answered Dec 27 '18 at 14:39
egregegreg
184k1486205
$endgroup$
In the complex field, every nonzero complex number $z$ has an inverse $z^{-1}$. It is customary to write
$$
wz^{-1}=frac{w}{z}
$$
By definition, $zz^{-1}=1$, so in the alternative notation
$$
frac{z}{z}=1
$$
for every complex number $zne0$.
Now take $z=ax+bi$ and you have your result. Of course, if $ax+bi=0$ (that is, $ax=0$ and $b=0$), the expression $(ax+bi)/(ax+bi)$ is undefined.
answered Dec 27 '18 at 14:39
egregegreg
184k1486205
edited Dec 27 '18 at 14:42
answered Dec 27 '18 at 14:39
egregegreg
184k1486205
answered Dec 27 '18 at 14:39
egregegreg
184k1486205
answered Dec 27 '18 at 14:39
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
3
$begingroup$
Yes, unless $ax+bi = 0$.
$endgroup$
– Card_Trick
Dec 27 '18 at 13:21
1
$begingroup$
I suppose $a,$ $b,$ and $x$ all are meant to be real numbers, but I think it would be better to make that explicit. I don't know of any ironclad rule that says one can never use any of those symbols to represent a complex number.
$endgroup$
– David K
Dec 27 '18 at 13:53