22(x)Ifaandbare natural numbers, there there are natural numberssandtwith gcd(a,b)=sa+tb.Solution.False.1.47Given integersaandb(possibly negative) witha=0, prove that thereexist unique integersqandrwithb=qa+rand 0≤r<|a|.Solution.We have already proved this whena>0 andb≥0. Assumenow thata>0 andb<0. Now−b>0, and so there are integersqandrwith−b=qa+rand 0≤r<a; it follows thatb= −qb−r. Ifr=0,we are done; ifr>0, thenb=(−q−1)a+(a−r)and 0<a−r<a(by Proposition A.2(ii),−r<0 impliesa−r<a.Now assume thata<0, so that−a>0 (and so|a| = −a. By what wehave proved so far, there are integersqandrwithb=q(−a)+r, where0≤r<−a; that is,b=(−q)a+r, where 0≤r<|a|.1.48Prove that√2 is irrational using Proposition 1.14 instead of Euclid’s lemma.Solution.Assume, on the contrary, that√2=a/b, whereaandbare inte-gers. By Proposition 1.14, we havea=2kmandb=2n, wherek,≥0andm,nare odd. Ifk≥, then we may cancel to obtain√2=2
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