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I've been reading mathematical analysis written by B.A.Zorich recently and have some doubts.

Why is the mapping

$f:\mathbb{N\times N}\rightarrow\mathbb{N} \quad given\;by\quad (m,n)\rightarrow\frac{(m+n-2)(m+n-1)}{2}+m$

a bijection($\mathbb{N}\;$in this book does not contain the number zero)?

The question may be very simple, but I'm just a beginner of analysis so I would appreciate it if someone could answer it in detail.

Arturo Magidin
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Erutaner
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1 Answers1

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Hint: forget Algebra, and just focus on intuition.

In the mapping, the 1st expression is the sum of the numbers from $1$ through $(m+n-2)$.

Suppose that there exists $(A,B) \in \Bbb{N^2}$ that yields the same mapping, where $(A,B) \neq (m,n)$.

Hint: $(A+B) = (m+n)$ generates a contradiction.

Suppose $(A+B) \neq (m+n).$

Without loss of generality, $(A+B) > (m+n)$.

Hint: If you have two positive integers, $r$ and $s$, with $r > s$, and you compute the sums

$\displaystyle E = \sum_{i=1}^r (i) ~~~\text{and}~~~ F = \sum_{i=1}^s (i)$

Then what is the minimum possible value of $E - F~~$?

user2661923
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  • It's very enlightening. I've understood it. This means that as long as the function values are equal, the two pairs must be equal, otherwise a contradiction will be deduced. However, this method seems to only prove that the mapping is injective, and additional proof may be needed in the case of surjection. – Erutaner May 05 '22 at 14:14
  • @Erutaner Since I was not permitted to provide a real answer, I was boxed in. The difficulty is that your posting has a number of defects. See this article for details. The surjective property follows from considering that you have total control over the top end of the summation, which is $(m+n-2)$. Within this control, you have fine tuning control of being able to specify $m$. ...see next comment – user2661923 May 05 '22 at 17:00
  • @Erutaner So, for example, it is easy to see that each of the following numbers are in the range: ${0+1::1+1,1+2::3+1,3+2,3+3::\cdots}.$ – user2661923 May 05 '22 at 17:00