Question: we mentioned that gcd( m, n ) = gcd( n, m( mod n ) ) for all positive integers m, and n, where m >

we mentioned that gcd( m, n ) = gcd( n, m( mod n ) ) for all positive integers m, and n, where m > n. This is the basis of Euclid's algorithm for computing the gcd. Show that Euclid's algorithm always "halts".

Hint: Use the fact that m mod n n for positive m and n, and also the fact that gcd(k, 0) = k for any integer k.

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