Where is the fault in my proof?

Question

I had some spare time, so I was just doing random equations, then accidentally came up with a proof that showed that i was -1. I know this is wrong, but I can't find where I went wrong. Could someone point out where a mathematical error was made?

$$(-1)^{2.5}=-1\\ (-1)^{5/2}=-1\\ (\sqrt{-1})^5=-1\\ i^5=-1\\ i=-1$$

As a rule, once you have negative or complex $x$, it is no longer true that $(x^y)^z=x^{yz}$. — Thomas Andrews, Jul 08 '16 at 13:41
A little confusing. The -1 should be in parentheses (otherwise the power only applies to the 1, but judging by the third line, you want the power to apply to (-1). — paw88789, Jul 08 '16 at 13:42
@RK01 Nope. See: http://www.wolframalpha.com/input/?i=(-1)%5E(2.5) — amcalde, Jul 08 '16 at 13:44
Thank you! I assumed that because I put in -1^2.5 into my calculator without putting the parentheses around the -1. — RK01, Jul 08 '16 at 13:45

score 3 · Accepted Answer · answered Jul 08 '16 at 13:53

Your mistake is that you have "$(-1)^{5/2} = -1$". It actually holds that $(-1)^{5/2} = i$ since you get by euler identity that $$(-1)^{5/2} = {e^{i\pi}}^{5/2} = e^{5/2 i\pi} = i.$$ Furthermore you shouldn't write $\sqrt{-1} = i$ because the root isn't defined for negative values and you can get all sorts of wrong proofs by using the rules for square roots in combination with this notation.

score -1 · Answer 2 · edited Jul 08 '16 at 14:23

-1

$$(-1)^{2.5} =(-1)^2\times (-1)^{1/2}= \sqrt{(-1)} = i$$ is your mistake

edited Jul 08 '16 at 14:23

Majid

3,357

answered Jul 08 '16 at 13:39

user352874

19

2

Sorry, that line is nowhere in the OPs argument, so it can't be the OPs mistake. – Thomas Andrews Jul 08 '16 at 13:40
2

user352874 is asserting that $(-1)^{2.5}$ is equal to i, not -1 as RK01 said. The OP's mistake was in not saying that. Of course, user 352874 is also mistaken, $(-1)^{2.5}$ is multivalued. – user247327 Jul 08 '16 at 13:50

Where is the fault in my proof?

2 Answers2