I got this riddle that I just couldnt solve. It's simple, how can you prove that a:b will always be C ? (a, b and c are natural numbers) For example 12:3=4, 4 will be the only solution.
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1What do you mean by $a:b$? – Elchanan Solomon Mar 02 '14 at 06:50
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Natural number divided by another natural number – zarko Mar 02 '14 at 06:51
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1This is one of those questions that sound so simple it actually reduces to working out the construction of the natural numbers and their arithmetic. Are you interested in a resource that formally describes the natural numbers and how to prove things about them? – Elchanan Solomon Mar 02 '14 at 06:55
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If there's a refrence for this specific question, then yes, otherwise it wont help much. – zarko Mar 02 '14 at 06:56
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1What do you mean by "always"? Mathematical facts are not time dependent. – Andrés E. Caicedo Mar 02 '14 at 06:56
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If you will divide a number by abother number there will only be one answer, how can you prove that? – zarko Mar 02 '14 at 06:58
2 Answers
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Well, by writing $a/b$ as an expression, you're sort of assuming that it is "well-defined", that is, it has exactly one value. Otherwise, there might be a scenario where you could say $a/b \ne a/b$, which would be absurd. So I'll rephrase the question.
If $b \ne 0$ (can't divide by $0$), $a = bc$ and $a = bc'$, how do we know that $c = c'$?
Proof: $$ bc = bc' \implies bc - bc' = 0 \implies b(c - c') = 0$$ In the integers, if two things multiply to $0$, at least one is $0$. Since $b \ne 0$, we know that $c - c' = 0$, and so $c = c'$.
Note that this is essentially a proof that division is well-defined.
Henry Swanson
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