Proof Question (1 Viewer)

[SB257426]

I am just wondering if anyone can have a look at my solution to the following proof question:

Question: "Prove by contradiction that there exists no 'n' that is an element of the natural numbers, such that n^2 + 2 is divsible by 4"

My Solution:

By way of contradiction assume their exists 'n', such that n^2 + 2 is divisible by 4.

When n is even, such that n = 2p,

(2p)^2 + 2
= 4p^2 + 2
= 4k+ 2 which is not divisible by 4 (due to the remainder of 2)

When n is odd such that n = 2p+1

(2p+1)^2 + 2
= 4p^2 + 4p + 3
= 4(p^2+p) +3
4k+3, which is not divisible by 4 (due to the remainder of 3)

Therefore, this contradicts the original statement.

Hence, there exists no 'n' such that n^2 + 2 is divisible by 4

I am fairly new to the topic of proof so am not too confident yet. That's why I am asking if there are any flaws in my proof

Cheers

 

[5uckerberg]

Very good stuff but you forgot which region you are looking at. Someone here might say, oh we did it with some irrational number like where . This is why instead of

Hence, there exists no such that is divisible by 4.

It should be

Hence, there exists no such that is divisible by 4 within the domain of natural numbers.

 

[SB257426]

Very good stuff but you forgot which region you are looking at. Someone here might say, oh we did it with some irrational number like where . This is why instead of

Hence, there exists no such that is divisible by 4.

It should be

Hence, there exists no such that is divisible by 4 within the domain of natural numbers.

Thanks for the advice man, I appreciate it