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For any integer N, there is an integer P such that one of the following is true:

N = 10P

N = 10P + 1

N = 10P + 2

N = 10P + 3

N = 10P + 4

N = 10P + 5

N = 10P + 6

N = 10P + 7

N = 10P + 8

N = 10P + 9

We know that there are 9 integers between every multiple of 10. We know that any N % 10 will produce a remainder between 0 and 9.

How do we formalize an idea like this in a mathematical proof?

sammy
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  • It looks to me as though the idea that $(10P+9)+1 = 10P+10 = 10(P+1)$, relying as it does on associativity and distributivity of multiplication over addition, is key here. So you need an axiomatic basis that provides those things. What guidance are you looking for? – Brian Tung Jul 20 '16 at 21:50
  • Think about this: what does $N\equiv a\ (\mod 10)$ mean exactly? – Aweygan Jul 20 '16 at 21:54
  • There are probably many ways to show this, but one can do so constructively. Take N round down to the nearest integer divisible by 10 and divide by 10. This is your P. Show that it must be true that if you add mod(N,10) to 10P you recover N. – David Jul 20 '16 at 22:03

1 Answers1

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We start with a number $n$ and define the set of all $n-10k$, where $k$ is a non-negative integer and $n-10k\ge 0$.

This set is not empty because it contains $n$.

The smallest element cannot be greater than $9$ because we could subtract $10$ to get a smaller element of the set.

Hence, for some $k$ we have $0\le n-10k\le 9$ and therefore $n=10k+q$ with $0\le q\le 9$.

Peter
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