Simple question about rational and irrational numbers
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From: Ludovicus on 21 Jul 2010 10:03 On 1 jun, 13:39, Craig Feinstein <cafei...(a)msn.com> wrote: > What about conditions in which both a, b are algebraic and irrational, > c is rational, and ab=c^2 is impossible? > Craig This question is a platiude. "The condition for a and b being irrational and c rational in a.b = c^2 , is that a.b must be a perfect square otherwise it is impossible. So, if a = sqrt(2) and b = sqrt(32) we can have a.b = c^2 because a.b = 64 = 8^2. Ludovicus
From: Michael Stemper on 22 Jul 2010 09:32
In article <d0a5b861-32e4-4a3a-9a3c-a20eab19b06e(a)v37g2000vbv.googlegroups.com>, Craig Feinstein <cafeinst(a)msn.com> wrote: > Is the following possible? > > a,b are irrational. c is rational. ab=c^2. Others have given specific examples, so I'll do the general case. Let c be any non-zero rational number. Express c as p/q, where p,q are non-zero integers. c^2 = (p^2)/(q^2), so c^2 is a non-zero rational by the closure of multiplication of non-zero integers. Let b be any irrational number. Since b is irrational, it is non-zero, and has a multiplicative inverse, which is also irrational. (proof left as an exercise) Let a = (c^2)/b Then, ab = c^2 All that remains is to prove that a is irrational. This is done by contradiction: Assume a is rational. Then, we can express it as r/s, where r and s are integers, and s is non-zero. With c = p/q as before, we have a = r/s = (p^2)/(q^2) * 1/b Since p is non-zero, this yields (r*q^2)/(s*p^2) = 1/b, with p, q, r, and s all integers, and ps != 0. This shows that 1/b is rational, which contradicts what was shown (as an exercise) above. -- Michael F. Stemper #include <Standard_Disclaimer> If we aren't supposed to eat animals, why are they made from meat? |