Proof by Contradiction: Converse of A(n) Holds for All n ∈ Z

[UOAMCBURGER]

Homework Statement

“If n = 3q + 1 or n = 3q + 2 for some q ∈ Z, then n 2 = 3t + 1 for some t ∈ Z.”
Use proof by contradiction to show that the converse of A(n) is true for all n ∈ Z.
For the proof by contradiction, on the answer sheet provided they have assumed n^2 = 3t+1 but n != 3q+1 and n != 3q+2.

Is it possible to have a valid proof by contradiction if you were to assume n = 3q+1 or n = 3q+2 is true for some n in Z, but n^2 != 3t+1 and then show that you can get n^2 into the form 3(.t..) + 1 ? where 3(.t..) is just obviously some multiple of 3... Hence a contradiction.?

 

[tnich]

Is it possible to have a valid proof by contradiction if you were to assume n = 3q+1 or n = 3q+2 is true for some n in Z, but n^2 != 3t+1 and then show that you can get n^2 into the form 3(.t..) + 1 ? where 3(.t..) is just obviously some multiple of 3... Hence a contradiction.?

This would not be a proof by contradiction. It would just be proving the statement “If n = 3q + 1 or n = 3q + 2 for some q ∈ Z, then n 2 = 3t + 1 for some t ∈ Z” directly. The assumption that n^2 $\ne$ 3t+1 would be superfluous.

 

[tnich]

This would not be a proof by contradiction. It would just be proving the statement “If n = 3q + 1 or n = 3q + 2 for some q ∈ Z, then n 2 = 3t + 1 for some t ∈ Z” directly. The assumption that n^2 $\ne$ 3t+1 would be superfluous.

Also, the problem is to prove the converse, not the original statement. So what you need to prove is
$\forall t, n \in \mathbb Z$ such that $n^2=3t+1, \exists q \in \mathbb Z$ such that $n=3q+1$ or $n=3q+2$

 

[UOAMCBURGER]

Also, the problem is to prove the converse, not the original statement. So what you need to prove is
$\forall t, n \in \mathbb Z$ such that $n^2=3t+1, \exists q \in \mathbb Z$ such that $n=3q+1$ or $n=3q+2$

oh that is why i was getting confused, because given statement A>B for a proof by contradiction you get A>~B and show some contradiction, hence my question about the order of the proof. I see what I've done, thanks.

 

[Ray Vickson]

Homework Statement

“If n = 3q + 1 or n = 3q + 2 for some q ∈ Z, then n 2 = 3t + 1 for some t ∈ Z.”
Use proof by contradiction to show that the converse of A(n) is true for all n ∈ Z.
For the proof by contradiction, on the answer sheet provided they have assumed n^2 = 3t+1 but n != 3q+1 and n != 3q+2.

Is it possible to have a valid proof by contradiction if you were to assume n = 3q+1 or n = 3q+2 is true for some n in Z, but n^2 != 3t+1 and then show that you can get n^2 into the form 3(.t..) + 1 ? where 3(.t..) is just obviously some multiple of 3... Hence a contradiction.?

Is n 2 supposed to be $n^2?$ If so, write it properly, as n^2.

 

[UOAMCBURGER]

Is n 2 supposed to be $n^2?$ If so, write it properly, as n^2.

yes its meant to be n², and obviously I did NOT intentionally write it as n 2 ...