# Thread: Proof - Implication signs

1. ## Proof - Implication signs

Hi all,

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

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

thanks sammy

2. (i) Consider $\displaystyle N=9$
(ii) Assume the last digit of N is 1. That is, $\displaystyle N = 1(mod \ 10)$. Use the fact that $\displaystyle a \cdot b (mod \ n) = (a(mod \ n)\cdot b(mod \ n)) (mod \ n)$.

3. Try to see if it works from $\displaystyle N$ to $\displaystyle N^2$ :

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

So, $\displaystyle N = 1 + 10k$

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

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

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

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

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

Now you try the other way round ($\displaystyle N^2$ to $\displaystyle N$), and see if it works. Then you will know what symbol to put ...

4. That's great . 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 $\displaystyle N=1(mod \ 10)$, and that the terms squared would also be $\displaystyle 1(mod \ 10)$ but does $\displaystyle 1 \neq 1(mod \ 10)$?

sorry if this is dumb question.

sammy

5. Hi sammy

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