SOLUTIONS - Springer
SOLUTIONSMethods of the contrary, namely that 2+ 3+ 5=r, whereris a rational the equality 2+ 3=r 5 to obtain 5+2 6=r2+5 2r 5. It followsthat 2 6+2r 5 is itself rational. Squaring again, we find that 24+20r2+8r 30is rational, and hence 30 is rational, too. Pythagoras method for proving that 2isirrational can now be applied to show that this is not true. Write 30=mnin lowestterms; then transform this intom2=30n2. It follows thatmis divisible by 2 and because2(m2)2=15n2it follows thatnis divisible by 2 as well. So the fraction was not in lowestterms, a contradiction.
(Canadian Mathematical Olympiad, 2002) 9.AssumethatA, B, andasatisfyA∪B =[0,1], A∩B =∅, B =A+a. Wecanassume that a is positive; otherwise, we can exchange A and B. Then (1 −a,1]⊂B; hence (1 −2a,1 −a]⊂A. An inductive argument shows that for any positive integer n, the interval (1−(2n+1)a,1−2na]is in B, while the interval (1 ...
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