Show that gcd(a,b)=|a|⇔a∣b?

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gunithd5 · Accepted Answer

Step 1Assume that a, b∈N. I use the following definition of D=gcd(a,b):1) D  is a common divisor of a and b;2) Every integer d∈N which is a common divisor of a and b divides D.Proposition 1: If gcd(a,b)=a then a∣bProposition 2: If a∣b, then gcd(a,b)=aProof: Since a∣a and a∣b, then any integer d such that d∣a and d∣b satisfies also the condition d|a.From a|a, a|b and d∣a we conclude that gcd(a,b)=a

ophelialee4xn · Accepted Answer

Step 1It is true that the final assertion holds if s=0 and t=1, but it is not true that that as+bt=az=b can only hold if s=0 and t=1 is false. Take, for example, a=2 and b=4. Then (−5)a+3(b)=b.Remember that, by definition, gcd(x,y)∣x and gcd(x,y)∣y. That gives one implication.For the converse, remember that d|x and d|y implies d∣gcd(x,y).

Show that \gcd(a,b)=|a| \iff a | b?

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