Tagged Questions
0
votes
1answer
69 views
Why is the second equality wrong?
Here's a "proof" of $e^x=1$ for all $x$: $$\exp(x)=\exp\left(i2π⋅\frac{x}{i2π}\right)=\bigl(\exp(i2π)\bigr)^{x/(i2π)}=1^{x/(i2π)}=1$$ Why is the second equality wrong?
16
votes
1answer
389 views
What is wrong with this fake proof $e^i = 1$?
$$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$
Obviously, one of my algebraic manipulations is not valid.
9
votes
3answers
444 views
Why this proof $0=1$ is wrong?(breakfast joke)
We have $$e^{2\pi i n}=1$$
So we have $$e^{2\pi in+1}=e$$
which implies $$(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$$
Thus we have $$e^{-4\pi^{2}n^{2}+4\pi in+1}=e$$
This implies ...
4
votes
1answer
290 views
What's wrong with this proof that $e^{i\theta} = e^{-i\theta}$?
I recently learned that $\cos{\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and $\sin{\theta} = \frac{e^{i\theta} - e^{-i\theta}}{2}$ Based on this, I managed to "prove" that:
$$e^{i\theta} = ...
10
votes
4answers
426 views
Which step in this process allows me to erroneously conclude that $i = 1$
I was playing around with imaginary numbers and exponents and came up with this:
$$ i = \sqrt{-1} $$
$$ \sqrt{-1} = (-1)^{1/2} $$
$$ (-1)^{1/2} = (-1)^{2/4} $$
$$ (-1)^{2/4} = ((-1)^{2})^{1/4} ...
1
vote
1answer
188 views
paradoxical answers using 'i' [duplicate]
Possible Duplicate:
-1 is not 1, so where is the mistake?
Significance of $\displaystyle\sqrt[n]{a^n} $?
$i = \sqrt{-1} = \sqrt{\frac{-1}{1}} = \sqrt{\frac{1}{-1}} = \frac{1}{i}$
hence,
...
1
vote
3answers
244 views
A contradiction involving exponents
Where is the error in the following statement:
$i^2=(i^2)^{\frac{4}{4}}=(i^4)^{\frac{2}{4}}=(1)^{\frac{1}{2}}=1$?
I feel the error is in the first equality, because $(i^2)^{\frac{4}{4}}$ is in fact ...
18
votes
9answers
1k views
$i^2$ why is it $-1$ when you can show it is $1$?
We know $$i^2=-1 $$then why does this happen?
$$
i^2 = \sqrt{-1}\times\sqrt{-1}
$$
$$
=\sqrt{-1\times-1}
$$
$$
=\sqrt{1}
$$
$$
= 1
$$
EDIT: I see this has been dealt with before but at least with ...
11
votes
2answers
946 views
-1 is not 1, so where is the mistake?
I know there must be something unmathematical in the following but I don't know where it is:
\begin{align}
\sqrt{-1} &= i \\ \\
\frac1{\sqrt{-1}} &= \frac1i \\ \\
\frac{\sqrt1}{\sqrt{-1}} ...
