Question

Note: This is NOT a question, but a tutorial.
An example on how to prove by contrapositive.

Answer 1

Prove that \(x^2\) is even when \(x\) is even.

First we assume that \(x\) is not even, so it is odd.
Any odd number is in the form \(2n + 1\) where \(n \in \mathbb{Z}\).
Let's take an integer \(k\). The square of the odd number \(2k + 1\) will be \(4k^2 + 4k + 1\).
Now if we give this thing a little tweak, we get \(2(2k^2 + 2k) + 1\).
Note how this is in the form \(2n + 1\) where \(n = 2k^2 + 2k\).

We get a statement in the 'if-then' form.
\(\textbf{If a square is odd, then the number is odd.}\)

The contrapositive of a statement is always true.
Let's try to find the contrapositive of this statement.

\(\textbf{If a number is not odd, then its square is not odd.}\)

As I pointed out in the beginning - 'Not even' means odd.
Following the same concept - 'Not odd' means 'even'.
Let us substitute 'Not odd' with 'even' in the given statement.

\(\textbf{If a number is even, then its square is even.}\)

Hence proved.

Answer 2

The definition of an even number is that an even number is divisible by 2. For example 2,4,6,8..... and so on.

This is why we say that an even number is in the form \(2n\) which means 2 times something where 'something' is \(n\).

If we notice, an odd number is after an even number in counting.
The next number to any number \(k\) is \(k + 1\).

The next number to an even number is \(2n + 1\) where \(2n\) is even.
As I said, a number next to an even number is ALWAYS odd.
So, \(2n + 1\) is odd where \(n \in \mathbb{Z}\).

Answer 3

I have been asked another question.
Why is \(2(2k^2 + 2k)\) even? Because even numbers are in the form \(2n\) and here \(n = 2k^2 + 2k \).

Why is \(2k^2 + 2k\) an integer? Because an integer squared multiplied by two is an integer, and the same integer multiplied by 2 is also an integer.

By the closer property, we can say that Integer + Integer = Integer

Answer 4

As the tutorial mentioned — \(n \in \mathbb{Z}\), we just said that \(2k^2 + 2k \) is an integer so we can take \(2k^2 + 2k = n\).

Answer 5

test

Answer 6

closure*

Answer 7

Nice tutorial :)

Answer 8

Thanks, Lou! =)

Answer 9

No prob. :) Any time.

Answer 10

very nice work Parth

Answer 11

lol I was expecting some criticism from Rhaukus :P

Answer 12

where are the pictures/

Answer 13

I have one criticism actually. You first assume that \(x\) is odd, and show that \(x^2\) is odd. You use this to prove \[\bf{\text{If a square is odd, then the number is odd.}}\]However, all it actually shows is \[\bf\text{If a number is odd, then the square is odd.}\]

I could be misreading something however.

Answer 14

lol both ways work, @kg

Answer 15

Oh I see what you mean, see the last part. We find the contrapositive of that statement.

Answer 16

Strictly speaking, both ways do not work. Your starting assumption should be "Assume \(x^2\) is odd." Using this, you need to show that \(x\) must be odd.

And the contrapositive of what I wrote is "If \(x^2\) is even, then \(x\) is even." You are attempting to prove the converse.

Answer 17

I see what you mean.

Answer 18

You have proven (very nicely) that if \(x^2\) is even, then \(x\) is even though.

Answer 19

Thanks ParthKohli, the contrapositive approach seems like a positive thing.

Answer 20

Should I say 'Thank you' or 'You are welcome'? Haha

Answer 21

@ganeshie8 Why are you looking at old tutorials? I don't even know when I made this one.

Answer 22

I'm reading this, and it's not even that good.

Answer 23

You made it on:
Fri Jun 29 2012 03:25:38 GMT-0600 (Mountain Standard Time)

Simple Inspect Element cheat :P

Btw, it is great! Very informative :)

Answer 24

Yeah, I know how to use inspect element on timestamps. I just don't know when I made this.