Prove that when a=qb+r, gcd(a,b)=gcd(b,r)

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||Lecture#27||If a=bq+r then gcd(a,b)=gcd(b,r) ||Prof.Latif Sajid ||Lecture#27||If a=bq+r then gcd(a,b)=gcd(b,r) ||Prof.Latif Sajid
Number Theory:If a=qb+r then gcd(a,b)=gcd(b,r) Number Theory:If a=qb+r then gcd(a,b)=gcd(b,r)
Lemma. If a= qb + r, then gcd(a, b) = gcd(b, r). Chapter #2 Lemma. If a= qb + r, then gcd(a, b) = gcd(b, r). Chapter #2
Ahhh! 💀 - oh hi jake #Shorts Ahhh! 💀 - oh hi jake #Shorts
gcd(a+bm,b)=gcd(a,b) gcd(a+bm,b)=gcd(a,b)
If gcd(a,b)=1 and c|a, then gcd(b,c)=1 If gcd(a,b)=1 and c|a, then gcd(b,c)=1
Number Theory, Lecture-32, If a = bq + r, then show that (a, b)= (b, r) Number Theory, Lecture-32, If a = bq + r, then show that (a, b)= (b, r)
Bézout Bézout's identity: ax+by=gcd(a,b)