Question

```
Prove using the Division Algorithm that an integer is either even or odd, but never both.
```
Is mine a legitimate proof?

Answer 1

Okay, so I did this:

A number is either \(0\pmod{2}\) or \(1\pmod{2}\). Using the division algorithm, we may write both statements as \(n = 2k + 0\) or \(n = 2k + 1\) and \(k\in\mathbb{Z}\). As seen, these two are just the definitions of even and odd numbers.

Answer 2

@AccessDenied Is there a theorem that tells us that a number is either \(0\pmod{2}\) or \(1 \pmod{2}\)?

Answer 3

Or roughly, in the statement \(x\pmod{y}, \ \ x,y\in\mathbb{N}\), there are \(y\) number of possibilities of \(x\)?

Answer 4

What is the idea behind using the scroll bar there above..??

Answer 5

It made a scroll bar automatically.

Answer 6

Really??

Without using any code or command??

Answer 7

I intended to put it inside a quotation block.

Answer 8

Anyway, how do I go about proving that a number can never be both?

Answer 9

NO GOOD AT IT..

Answer 10

```
Parth Kohli
```

Answer 11

Just trying how you did that..