MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document I suggested the...

myntfalskj4

myntfalskj4

Answered

2022-07-01

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document I suggested the following problem to my friend: prove that there exist irrational numbers a and b such that ab is rational.
Now, his inital solution was like this: let's take a rational number r and an irrational number i. Let's assume a = r i b = 1 i
So we have
a b = ( r i ) 1 i = r
which is rational per initial supposition. b is obviously irrational if i is. My friend says that it is also obvious that if r is rational and i is irrational, then r i is irrational. I quickly objected saying that r = 1 is an easy counterexample. To which my friend said, OK, for any positive rational number r, other than 1 and for any irrational number i r i is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?

Answer & Explanation

soosenhc

soosenhc

Expert

2022-07-02Added 16 answers

Consider 2 log 2 3
babyagelesszj

babyagelesszj

Expert

2022-07-03Added 7 answers

Let r be a positive rational number and i a positive irrational number. If r i is rational, then r i = a b for a, a , b Z such that b 0. In particular, i = log r ( a b ) .

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