Mixing
Why is `e^(i2À) = e^0` true while `i2À = 0` is false?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I came across a perplexing thing while testing some assumptions in Wolfram|Alpha, and as I don't have a math education beyond college algebra, I thought this would be a good place to ask. I would just like to emphasize that this is not a homework question, and is purely curiosity.
Common faulty math proofs tend to use logic like this:
1^0 = 1
2^0 = 1
therefore 1 = 2
This is obviously false because x^0 is defined as 1 for all real values of x.
But in the case of e^(i2À) = e^0, the base is the same: e. So if a^x = a^y is true, shouldn't it follow, then, that x = y and therefore i2À = 0? If not, why?
complex-numbers eulers-constant
asked 1 hour ago
Braden Best
1092
New contributor
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
I came across a perplexing thing while testing some assumptions in Wolfram|Alpha, and as I don't have a math education beyond college algebra, I thought this would be a good place to ask. I would just like to emphasize that this is not a homework question, and is purely curiosity.
Common faulty math proofs tend to use logic like this:
1^0 = 1
2^0 = 1
therefore 1 = 2
This is obviously false because x^0 is defined as 1 for all real values of x.
But in the case of e^(i2À) = e^0, the base is the same: e. So if a^x = a^y is true, shouldn't it follow, then, that x = y and therefore i2À = 0? If not, why?
complex-numbers eulers-constant
asked 1 hour ago
Braden Best
1092
New contributor
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Seems to be the same question as "why is $sin(0)=sin(2pi)$ true but $0 = 2pi$ false?
– Michael
1 hour ago
5
@Michael It is the exact same question, by equating imaginary parts of both sides. :P
– Frpzzd
1 hour ago
3
$(-1)^2=(1)^2$ but $-1ne 1$.
– hamam_Abdallah
1 hour ago
2
@Frpzzd : How come your comment that affirms my comment gets 2 upvotes, but my comment gets none? =)
– Michael
1 hour ago
3
@Michael I have no clue. But if this comment starts getting upvotes, then we'll know something funny is going on. XD
– Frpzzd
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I came across a perplexing thing while testing some assumptions in Wolfram|Alpha, and as I don't have a math education beyond college algebra, I thought this would be a good place to ask. I would just like to emphasize that this is not a homework question, and is purely curiosity.
Common faulty math proofs tend to use logic like this:
1^0 = 1
2^0 = 1
therefore 1 = 2
This is obviously false because x^0 is defined as 1 for all real values of x.
But in the case of e^(i2À) = e^0, the base is the same: e. So if a^x = a^y is true, shouldn't it follow, then, that x = y and therefore i2À = 0? If not, why?
complex-numbers eulers-constant
asked 1 hour ago
Braden Best
1092
New contributor
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I came across a perplexing thing while testing some assumptions in Wolfram|Alpha, and as I don't have a math education beyond college algebra, I thought this would be a good place to ask. I would just like to emphasize that this is not a homework question, and is purely curiosity.
Common faulty math proofs tend to use logic like this:
1^0 = 1
2^0 = 1
therefore 1 = 2
This is obviously false because x^0 is defined as 1 for all real values of x.
But in the case of e^(i2À) = e^0, the base is the same: e. So if a^x = a^y is true, shouldn't it follow, then, that x = y and therefore i2À = 0? If not, why?
complex-numbers eulers-constant
complex-numbers eulers-constant
asked 1 hour ago
Braden Best
1092
New contributor
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Braden Best
1092
New contributor
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Braden Best
1092
New contributor
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Braden Best
1092
asked 1 hour ago
Braden Best
1092
1092
New contributor
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Braden Best is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Seems to be the same question as "why is $sin(0)=sin(2pi)$ true but $0 = 2pi$ false?
– Michael
1 hour ago
5
@Michael It is the exact same question, by equating imaginary parts of both sides. :P
– Frpzzd
1 hour ago
3
$(-1)^2=(1)^2$ but $-1ne 1$.
– hamam_Abdallah
1 hour ago
2
@Frpzzd : How come your comment that affirms my comment gets 2 upvotes, but my comment gets none? =)
– Michael
1 hour ago
3
@Michael I have no clue. But if this comment starts getting upvotes, then we'll know something funny is going on. XD
– Frpzzd
1 hour ago
add a comment |Â
3
Seems to be the same question as "why is $sin(0)=sin(2pi)$ true but $0 = 2pi$ false?
– Michael
1 hour ago
5
@Michael It is the exact same question, by equating imaginary parts of both sides. :P
– Frpzzd
1 hour ago
3
$(-1)^2=(1)^2$ but $-1ne 1$.
– hamam_Abdallah
1 hour ago
2
@Frpzzd : How come your comment that affirms my comment gets 2 upvotes, but my comment gets none? =)
– Michael
1 hour ago
3
@Michael I have no clue. But if this comment starts getting upvotes, then we'll know something funny is going on. XD
– Frpzzd
1 hour ago
3
3
Seems to be the same question as "why is $sin(0)=sin(2pi)$ true but $0 = 2pi$ false?
– Michael
1 hour ago
Seems to be the same question as "why is $sin(0)=sin(2pi)$ true but $0 = 2pi$ false?
– Michael
1 hour ago
5
5
@Michael It is the exact same question, by equating imaginary parts of both sides. :P
– Frpzzd
1 hour ago
@Michael It is the exact same question, by equating imaginary parts of both sides. :P
– Frpzzd
1 hour ago
3
3
$(-1)^2=(1)^2$ but $-1ne 1$.
– hamam_Abdallah
1 hour ago
$(-1)^2=(1)^2$ but $-1ne 1$.
– hamam_Abdallah
1 hour ago
2
2
@Frpzzd : How come your comment that affirms my comment gets 2 upvotes, but my comment gets none? =)
– Michael
1 hour ago
@Frpzzd : How come your comment that affirms my comment gets 2 upvotes, but my comment gets none? =)
– Michael
1 hour ago
3
3
@Michael I have no clue. But if this comment starts getting upvotes, then we'll know something funny is going on. XD
– Frpzzd
1 hour ago
@Michael I have no clue. But if this comment starts getting upvotes, then we'll know something funny is going on. XD
– Frpzzd
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
That's because the exponential isn't an injective function on $mathbb C$. If is an injective function on $mathbb R$, and so the single-valued logarithm is well-defined on $mathbb R^+$ but not on $mathbb C$.
This isn't true for the same reason that $(-1)^2=1^2$ doesn't imply that $-1=1$.
answered 1 hour ago
Frpzzd
18.4k63697
Would you care to elaborate/add examples? For example, I don't see how the(-1)^2 = 1^2example is equivalent, because-1 = 1is false, unlikee = e.
– Braden Best
1 hour ago
@BradenBest It is equivalent because $f(x)=x^2$ is not injective on $mathbb R$ just as $g(x)=e^x$ is not injective on $mathbb C$.
– Frpzzd
1 hour ago
add a comment |Â
up vote
3
down vote
Recall that
$$f(x)=f(y) implies x=y$$
require that $f$ is an injective function.
answered 1 hour ago
gimusi
77.6k73889
add a comment |Â
up vote
0
down vote
Simply because the exponential function is periodic with period $2pi i$.
That means that for any $x$, the values of $e^x$ and $e^x+2pi i k$ are equal for all integers $k$.
In particular, you are looking at $x=0$ and $k=0,1$; so $e^0=e^2pi i$, but of course $0$ and $2pi i$ are different numbers.
answered 1 hour ago
MPW
29.2k11855
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
That's because the exponential isn't an injective function on $mathbb C$. If is an injective function on $mathbb R$, and so the single-valued logarithm is well-defined on $mathbb R^+$ but not on $mathbb C$.
This isn't true for the same reason that $(-1)^2=1^2$ doesn't imply that $-1=1$.
answered 1 hour ago
Frpzzd
18.4k63697
Would you care to elaborate/add examples? For example, I don't see how the(-1)^2 = 1^2example is equivalent, because-1 = 1is false, unlikee = e.
– Braden Best
1 hour ago
@BradenBest It is equivalent because $f(x)=x^2$ is not injective on $mathbb R$ just as $g(x)=e^x$ is not injective on $mathbb C$.
– Frpzzd
1 hour ago
add a comment |Â
up vote
3
down vote
That's because the exponential isn't an injective function on $mathbb C$. If is an injective function on $mathbb R$, and so the single-valued logarithm is well-defined on $mathbb R^+$ but not on $mathbb C$.
This isn't true for the same reason that $(-1)^2=1^2$ doesn't imply that $-1=1$.
answered 1 hour ago
Frpzzd
18.4k63697
Would you care to elaborate/add examples? For example, I don't see how the(-1)^2 = 1^2example is equivalent, because-1 = 1is false, unlikee = e.
– Braden Best
1 hour ago
@BradenBest It is equivalent because $f(x)=x^2$ is not injective on $mathbb R$ just as $g(x)=e^x$ is not injective on $mathbb C$.
– Frpzzd
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
That's because the exponential isn't an injective function on $mathbb C$. If is an injective function on $mathbb R$, and so the single-valued logarithm is well-defined on $mathbb R^+$ but not on $mathbb C$.
This isn't true for the same reason that $(-1)^2=1^2$ doesn't imply that $-1=1$.
answered 1 hour ago
Frpzzd
18.4k63697
That's because the exponential isn't an injective function on $mathbb C$. If is an injective function on $mathbb R$, and so the single-valued logarithm is well-defined on $mathbb R^+$ but not on $mathbb C$.
This isn't true for the same reason that $(-1)^2=1^2$ doesn't imply that $-1=1$.
answered 1 hour ago
Frpzzd
18.4k63697
answered 1 hour ago
Frpzzd
18.4k63697
answered 1 hour ago
Frpzzd
18.4k63697
answered 1 hour ago
Frpzzd
18.4k63697
18.4k63697
Would you care to elaborate/add examples? For example, I don't see how the(-1)^2 = 1^2example is equivalent, because-1 = 1is false, unlikee = e.
– Braden Best
1 hour ago
@BradenBest It is equivalent because $f(x)=x^2$ is not injective on $mathbb R$ just as $g(x)=e^x$ is not injective on $mathbb C$.
– Frpzzd
1 hour ago
add a comment |Â
Would you care to elaborate/add examples? For example, I don't see how the(-1)^2 = 1^2example is equivalent, because-1 = 1is false, unlikee = e.
– Braden Best
1 hour ago
@BradenBest It is equivalent because $f(x)=x^2$ is not injective on $mathbb R$ just as $g(x)=e^x$ is not injective on $mathbb C$.
– Frpzzd
1 hour ago
Would you care to elaborate/add examples? For example, I don't see how the
(-1)^2 = 1^2 example is equivalent, because -1 = 1 is false, unlike e = e.– Braden Best
1 hour ago
Would you care to elaborate/add examples? For example, I don't see how the
(-1)^2 = 1^2 example is equivalent, because -1 = 1 is false, unlike e = e.– Braden Best
1 hour ago
@BradenBest It is equivalent because $f(x)=x^2$ is not injective on $mathbb R$ just as $g(x)=e^x$ is not injective on $mathbb C$.
– Frpzzd
1 hour ago
@BradenBest It is equivalent because $f(x)=x^2$ is not injective on $mathbb R$ just as $g(x)=e^x$ is not injective on $mathbb C$.
– Frpzzd
1 hour ago
add a comment |Â
up vote
3
down vote
Recall that
$$f(x)=f(y) implies x=y$$
require that $f$ is an injective function.
answered 1 hour ago
gimusi
77.6k73889
add a comment |Â
up vote
3
down vote
Recall that
$$f(x)=f(y) implies x=y$$
require that $f$ is an injective function.
answered 1 hour ago
gimusi
77.6k73889
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Recall that
$$f(x)=f(y) implies x=y$$
require that $f$ is an injective function.
answered 1 hour ago
gimusi
77.6k73889
Recall that
$$f(x)=f(y) implies x=y$$
require that $f$ is an injective function.
answered 1 hour ago
gimusi
77.6k73889
answered 1 hour ago
gimusi
77.6k73889
answered 1 hour ago
gimusi
77.6k73889
answered 1 hour ago
gimusi
77.6k73889
77.6k73889
add a comment |Â
add a comment |Â
up vote
0
down vote
Simply because the exponential function is periodic with period $2pi i$.
That means that for any $x$, the values of $e^x$ and $e^x+2pi i k$ are equal for all integers $k$.
In particular, you are looking at $x=0$ and $k=0,1$; so $e^0=e^2pi i$, but of course $0$ and $2pi i$ are different numbers.
answered 1 hour ago
MPW
29.2k11855
add a comment |Â
up vote
0
down vote
Simply because the exponential function is periodic with period $2pi i$.
That means that for any $x$, the values of $e^x$ and $e^x+2pi i k$ are equal for all integers $k$.
In particular, you are looking at $x=0$ and $k=0,1$; so $e^0=e^2pi i$, but of course $0$ and $2pi i$ are different numbers.
answered 1 hour ago
MPW
29.2k11855
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Simply because the exponential function is periodic with period $2pi i$.
That means that for any $x$, the values of $e^x$ and $e^x+2pi i k$ are equal for all integers $k$.
In particular, you are looking at $x=0$ and $k=0,1$; so $e^0=e^2pi i$, but of course $0$ and $2pi i$ are different numbers.
answered 1 hour ago
MPW
29.2k11855
Simply because the exponential function is periodic with period $2pi i$.
That means that for any $x$, the values of $e^x$ and $e^x+2pi i k$ are equal for all integers $k$.
In particular, you are looking at $x=0$ and $k=0,1$; so $e^0=e^2pi i$, but of course $0$ and $2pi i$ are different numbers.
answered 1 hour ago
MPW
29.2k11855
answered 1 hour ago
MPW
29.2k11855
answered 1 hour ago
MPW
29.2k11855
answered 1 hour ago
MPW
29.2k11855
29.2k11855
add a comment |Â
add a comment |Â
Braden Best is a new contributor. Be nice, and check out our Code of Conduct.
Braden Best is a new contributor. Be nice, and check out our Code of Conduct.
Braden Best is a new contributor. Be nice, and check out our Code of Conduct.
Braden Best is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2962571%2fwhy-is-ei2%25cf%2580-e0-true-while-i2%25cf%2580-0-is-false%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
Seems to be the same question as "why is $sin(0)=sin(2pi)$ true but $0 = 2pi$ false?
– Michael
1 hour ago
5
@Michael It is the exact same question, by equating imaginary parts of both sides. :P
– Frpzzd
1 hour ago
3
$(-1)^2=(1)^2$ but $-1ne 1$.
– hamam_Abdallah
1 hour ago
2
@Frpzzd : How come your comment that affirms my comment gets 2 upvotes, but my comment gets none? =)
– Michael
1 hour ago
3
@Michael I have no clue. But if this comment starts getting upvotes, then we'll know something funny is going on. XD
– Frpzzd
1 hour ago