Prove that if n ? m > 0, then gcd (m,n) = gcd (m,n?m). Solution To prove : gcd(m

Question

Prove that if n ? m > 0, then gcd (m,n) = gcd (m,n?m).

Explanation / Answer

To prove : gcd(m,n) = gcd(m,n-m) if n>m>0 Solution : As we know gcd(m,n) is the gcdc of m and n. and we know n*m = gcd(m,n) Let's define these as follows: n?gcd(m,n) = m m?gcd(m,n) = n Let's subtract second equation from the first: (n?m)?gcd(m,n) = m?n Statement "gcd(m-n,n) = gcd(m,n)" is equivalent to "gcd((n-m)* gcd(m,n), n) = gcd(m,n)", which is true if (n-m) is not a multiple of any of divisors of n apart from 1. Let's assume (n-m) is a multiple of c, such that c != 1 && n is divisible by c. Because n is divisible by c and gcd(m,n) is a multiple of all the common divisors of m and n, m must not be divisible by c, implying that a must not be divisible by c, which leads to contradiction: (n-m) must be divisible by c and be a whole number, but: (n?m)/c = n/c ? b/c?N because a subtraction of a whole number from a fraction is a fraction, Thus, n-m is not a multiple of c and gcd(m,n) = gcd(m-n,n)

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