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I was just bored and started practicing even more the exponentiations, and as I was working, I went ahead and did this:

$(-1)^{2} = ((-1)^{\sqrt{2}})^{\sqrt{2}}$

So, I entered in my calculator of what is negative one to the power of the square root of two and it gave me negative one, so it goes like:

$(-1)^{2} = ((-1)^{\sqrt{2}})^{\sqrt{2}} = (-1)^{\sqrt{2}} = (-1)$

But in other hand, we have:

$(-1)^2 = (-1)(-1) = 1$

And that implies that $-1 = +1$... Is there something wrong in it? If not, could someone explain me how can that happen?

Thanks for the answers.

serkan
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  • How do you get $(-1)^{\sqrt{2}} = -1$? – Joshua Wang Mar 01 '21 at 17:40
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    It seems entirely possible that the designers of the calculator had not anticipated this. I would not infer intention or consistency from the results of the calculator. – leslie townes Mar 01 '21 at 17:45
  • How would you define $(-1)^{\sqrt2}$? – Kenta S Mar 01 '21 at 17:45
  • @leslietownes Designers of calculators would certainly have not anticipated the number of questions their calculators have spawned on MSE! Not to take away from curiosity, but there should be a calculator tag on MSE for just these questions (there is a calculator tag!) – Sarvesh Ravichandran Iyer Mar 01 '21 at 17:49
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    @TeresaLisbon :) I am not a huge fan of Stewart's calculus, but https://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%20Early%20Transcendentals/upfiles/ecet-liesmycomp_stu.pdf is surprisingly well written for a section in a mass marketed textbook. – leslie townes Mar 01 '21 at 17:51
  • @leslietownes I actually quite love the exercises in that note! Thank you for pointing it out, I will take a good look at it in the future. – Sarvesh Ravichandran Iyer Mar 01 '21 at 17:53
  • Typically calculators don't store values like $\sqrt 2$ exactly, and what very likely happened is that the calculator interpreted $\sqrt 2$ as being approximately rational, so $\sqrt 2\approx p/q$. When $p,q$ are both odd, it makes sense to have $(-1)^{p/q}=-1$, and so the calculator makes the mistate to get to $(-1)^{\sqrt 2}= -1$. –  Mar 01 '21 at 22:05
  • A "safer" way to calculate negative exponentials is to have $a^b=\exp(b\ln a)$, which uses the two well-defined functions exp and ln. Mathematica gives $(-1)^\sqrt{2}\approx -0.266-0.964 i$, which makes more sense. –  Mar 01 '21 at 22:07

1 Answers1

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Here you have $2$,$3$ mistakes:

$(-1)^{2} = ((-1)^{\sqrt{2}})^{\sqrt{2}} = (-1)^{\sqrt{2}} = (-1)$

The first equality is true, since $\sqrt 2\cdot \sqrt 2=2$. The expression $(-1)^{\sqrt{2}}$ is not real and certainly not $-1$.

vitamin d
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    This should be a comment, not an answer. – Rushabh Mehta Mar 01 '21 at 17:45
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    "Is there something wrong in it? If not, could someone explain me how can that happen?" That's the question. I pointed out what was right and what was wrong. Where is the problem? – vitamin d Mar 01 '21 at 17:50