Divisibility Proof with the Sorcerer: a|b and a|c implies a|(b + c)

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Divisibility Proof with the Sorcerer: a|b and a|c implies a|(b + c)
  1. The Math Sorcerer
  2. divisibility proof
  3. a divides b and a divides c
  4. a divides the sum
  5. divisible
  6. number theory
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Комментарии
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sir i am from nepal and i saw that your understanding power is so good sir ..thank you

Shushil-
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If a|b, a|c then b = ax and c = ay

b+c = ax+ay = a(x+y) which is divisible by a. ◾️(QED)

duckymomo
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My attempted proof before watching the video:

If a|b and a|c:
then b=xa for some integer x;
and c=ya for some integer y;
so b+c = (x+y)a for some integers x and y;
substitute integer z for x+y; (sum of 2 integers is an integer)
then b+c = za for some integer z;
so (b+c)/a = z for some integer z;
therefore a|(b+c);
therefore, if a|b and a|c, then a|(b+c)

greenfinmusic
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can u do a proof if a, b, c are natural numbers?

lewurstol