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Consider: 1) sqrt2^sqrt2
2) (sqrt2^sqrt2)^sqrt2

2007-10-08 04:45:10 · 5 answers · asked by Luthien 4 in Science & Mathematics Mathematics

5 answers

Given a and b are irrational, a^b could be rational. The interesting thing about your second expression to consider is, given (sqrt 2 ^ sqrt 2) ^ sqrt 2,
we know it is equal to (sqrt 2) ^2 by multiplying exponents. This gives us a value of 2.

What we still don't know in the example is whether or not (squrt 2 ^ sqrt 2) is rational or irrational.

2007-10-08 04:56:47 · answer #1 · answered by Hiker 4 · 1 0

Hiker, you've just about got it. We know that expression (2) is rational, right? We don't really know if sqrt(2)^sqrt(2) is rational, as you mentioned. If it's irrational, then we have (2) is irrational^irrational=rational, which contradicts the hypothesis. But if it's rational, then expression (1) is a contradiction. So the answer to the original question would be no.

EDIT: steiner1745 answers the question of whether (1) is rational...nicely done. I remembered that it was indeed irrational, but had never seen a reason behind it. For the purposes of showing that a^b need not be irrational even when a and b are irrational though, we don't need to know this bit.

Oh, and steiner1745, when I was first presented with this proposition a few months ago, e^(i*pi) was my first response too. My instructor, a graph theorist, didn't like my use of complex numbers :)

2007-10-08 13:46:09 · answer #2 · answered by Ben 6 · 0 0

Consider
e^(iπ) = -1.
e and iπ are certainly irrational, so the answer to
your question is no.
BTW. A poster asked whether √2^√2 is rational.
This number is not only irrational, it is transcendental.
This follows from the theorem of Gelfond-Schneider,
which states that a^b is transcendental if
1). a is algebraic, a .ne. 0 or 1
and
2. b is algebraic and irrational.
Since √2 is a root of x²-2 = 0 and √2 is irrational,
the result follows.

2007-10-08 13:55:48 · answer #3 · answered by steiner1745 7 · 0 0

No. The first example proves that 2#s that are irrational can be multiplied to be a rational #. Therefore, it does not logically follow.

2007-10-08 11:49:39 · answer #4 · answered by jared_e42 5 · 0 2

sqrt(2)^sqrt(2) = 1.6325....
(sqrt(2)^sqrt(2))^sqrt(2) = 2 - that's what my calculator says, I don't know how to prove it.

2007-10-08 11:52:21 · answer #5 · answered by Helen B 5 · 0 2

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