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Problem Regarding H.C.F.

Prove that \(\gcd(a+b, a-b) \geq \gcd(a,b) \), where \(a\) and \(b\) are two integers.

Probable direction: Lowest positive value of the equation {ax + by} will give the h.c.f of a & b.
While the lowest positive value { (a+b)x+(a-b)y } will give the h.c.f. of (a+b) & (a-b).
[ x and y are integer variables]
So, the proof comes down to : [the lowest positive value of { (a+b)x + (a-b)y } ] >= [ lowest positive value of { ax + by } ] .

Note by D K
1 year, 4 months ago

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Great! Do you now see why the last step is obvious?

Hint: Do not let your notation do double duty. It would be helpful to use \(x_1, y_1 \) in one of them, and \( x_2, y_2 \) in the other. This way, you can ask about the relationship of these terms.

Calvin Lin Staff - 1 year, 4 months ago

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Yes, it's solved :)

D K - 1 year, 4 months ago

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