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My question is whether proving $\forall m,n,p\in \mathbb{Z}:m=p\implies m+n=p+n$ is equivalent to prove $\forall n,p\in \mathbb{Z}:p+n=p+n$ and if it is why?

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    Yes, because if we prove p+n=p+n, and we know that m=p, then p+n=m+n. – Ilovemath Aug 29 '21 at 01:58
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    @Ilovemath The question rised because proving $p+n=p+n$ turned out to be easier for me than proving $m=p\implies m+n=p+n$ and found it funny. Is that normal (such a difficulty)? –  Aug 29 '21 at 02:00
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    Yes it's normal....as far as I think. – Ilovemath Aug 29 '21 at 02:02

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We can define the integers as equivalent classes of pairs of natural numbers.

Precisely, $d\in\mathbb{Z}$ iff there are natural numbers $n\in\mathbb{N}$ and $m\in\mathbb{N}$ such that $d = (m,n)$.

We say two integers $m = (m_{1},m_{2})$ and $n = (n_{1},n_{2})$ are equivalent iff $m_{1} + n_{2} = m_{2} + n_{1}$.

Moreover, we define the addition as follows:

$$m + n = (m_{1} + n_{1},n_{2} + m_{2})$$

Hence, if we assume $m = (m_{1},m_{2})\in\mathbb{Z}$, $n = (n_{1},n_{2})\in\mathbb{Z}$ and $p = (p_{1},p_{2})\in\mathbb{Z}$ where $m = p$, we get \begin{align*} m + n & = (m_{1} + n_{1}, m_{2} + n_{2})\\\\ & = (m_{1} + n_{1} + p_{2}, m_{2} + n_{2} + p_{2})\\\\ & = (m_{2} + n_{1} + p_{1}, m_{2} + n_{2} + p_{2})\\\\ & = (n_{1} + p_{1},n_{2} + p_{2})\\\\ & = p + n \end{align*}

and we are done.

Hopefully this helps!

user0102
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