Proof - Implication signs

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November 12th 2009, 12:57 AM
sammy28

Proof - Implication signs

Hi all,

For these pairs of statements replace ... with $\Rightarrow \text{,} \Leftarrow \text{ or }\Leftrightarrow$ . Assume $N$ is a positive Integer.

The last digit of $N$ is 1 ... The last digit of $N^2$ is 1.

thanks sammy
November 12th 2009, 01:21 AM
Defunkt

(i) Consider $N=9$
(ii) Assume the last digit of N is 1. That is, $N = 1(mod \ 10)$ . Use the fact that $a \cdot b (mod \ n) = (a(mod \ n)\cdot b(mod \ n)) (mod \ n)$ .
November 12th 2009, 01:33 AM
Bacterius

Try to see if it works from $N$ to $N^2$ :

$N \equiv 1$ (mod $10$ )

So, $N = 1 + 10k$

Now, $N^2 = (1 + 10k)^2$ , which is equivalent to :

$N^2 = 1^2 + 20k + (10k)^2$
$N^2 = 1^2 + 20k + 100k^2$

Simple deductions :
$20k \equiv 0$ (mod $10$ )
$100k^2 \equiv 0$ (mod $10$ )

So, $N^2 \equiv 1$ (mod $10$ )

Therefore, $N \Rightarrow N^2$ (not formally written but you understand)

Now you try the other way round ( $N^2$ to $N$ ), and see if it works. Then you will know what symbol to put ...
November 12th 2009, 08:47 AM
sammy28

That's great (Hi). This question is in the introduction to proofs in my 'AS' level book and we have not covered modulus yet. It was aimed really at using the correct implication sign.

I could write out the first few terms and did see both ended in the digit one, but i couldnt work out how to give it a proof.

I can understand why terms 11, 21, 31, 41 would result in $N=1(mod \ 10)$ , and that the terms squared would also be $1(mod \ 10)$ but does $1 \neq 1(mod \ 10)$ ?

sorry if this is dumb question.

sammy
November 12th 2009, 10:51 AM
Defunkt

Hi sammy

$1 \equiv 1 (mod \ 10)$