Bezout’s Identity for Integrs

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We prove the greatest common divisor of a and b can be written as a linear combination of a and b.
  1. Ranthony Clark
  2. Doceri
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  4. 1.2
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Great video: really good explanations and aesthetics!

On a side note, in my abstract algebra class, they proved r had to equal zero another way:

their proof included the assumption that a and b cannot equal zero and am+bn>0, and because r is an element of s and r<t, the supposed least element, if r had been assumed to be greater than 0, there would be a contradiction because r cannot be less than t.

Was there a reason they said a cannot equal b or why you didn't limit this? Thanks!

LarissaFord
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At 0:45, I think you meant S={a, b, m, n in Z | am+bn} instead of defining the set with two a terms.

jaredgoldman