Is it wrong to deduce the premise from the conclusion?

Question

My professor gave me the following argument to prove:

Let $x$ be an even number. Then $x+x^2$ must be an even number.

I wrote down the following:

Hypothesis: $x = 2n$ for some $n\in \mathbb{Z}$
Thesis: $x + x^2 = 2m$ for some $m \in \mathbb{Z}$

Let us consider a direct proof.

$x+x^2=2m$ (replacing $x=2n$)
$2n + 4n^2=2m$
$2(n + 2n^2) = 2m$ (since $n\in\mathbb{N}$, we can replace $m=n+2n^2$)
2m = 2m

My professor says I only proved the converse (that if $x+x^2$ is even, then $x$ must be even) I claimed that showing that I can reach an identity from any equation using the act that $x=2n$ suffices as a formal proof, but he wasn't convinced. How can I prove I'm right?

Answer 1

You have 1) Assumed the Conclusion $x+ x^2 = 2m$ 2)Assumed the Hypothesis $x = 2n$ and 3) Concluded an identity. $2m = 2m$.

Your professor is actually wrong. You have not proven the converse (which isn't true). You have actually done nothing.

You can't start from the Conclusion and prove a premise.

If $x + x^2 = 2m$ then $x(x+1) = 2m$ so one of $x$ or $x+1$ is even but we don't know which one. For example. if $x = 3$ then $x + x^2 =3+9 =12$ is even.

You HAVE to start from the premise:

$x = 2m$ and you can !!!!!!!NOT!!!!!! assume the conclusion. You must prove the conclusion.

$x + x^2 = 2m + (2m)^2 = 2m + 4m^2 = 2(m + 2m^2)$ so $x+x^2$ is even. QED.

Note, that if $x$ is odd then

$x = 2m + 1$ for some $m$ and

$x + x^2 = (2m + 1) + (2m + 1)^2 = 2m + 1 +4m^2 + 4m + 1 = 4m^2 + 6m + 2 = 2(2m^2 + 3m + 1)$ is ... even.

$x + x^2$ is always even whether $x$ is even or odd.

So you can certainly NOT assume the conclusion and prove the premise as the premise need not be true.

And even if it were. You didn't prove anything.

Consider the false statment: All even number are squares of even numbers.

So we start off assuming the conclusion: Let $M = (x)^2= (2m)^2$ a square of an even number. We take the square root so $x = \pm 2m$. And therefore $x$ is an even number.

QED?

No. The statement was the exact opposite.

If $x = 2m$ is an even number then prove that $x= 2m = (2n)^2$ is a square of an even number.

We can not prove that because it is not true.

Answer 2

sorry it's not clear if the universe of discourse is natural numbers or integers since you reference both.

Assuming integers:

if $ x $ even -> $ (x + x^2) $ even.

because you're given that x is even, pretend $ \exists i \in \Bbb Z (x=2i) $

you need to show that $ \exists j \in \Bbb Z (x^2 + x = 2j) $

$$ x = 2i $$ $$ x^2 = 4i^2 $$ $$ (x^2 + x) = 4i^2 + 2i = 2j $$ $$ j = 2i^2 + i $$

Answer 3

Consider the invalid statement: "Let $x$ be a bird. Then $x$ is a duck."

Now, your approach is to assume the consequent, that $x$ is a duck, then demostrate that the duck is a bird, and decide that the statement was valid. Opps.

So, no, affirming the consequent is a falacy.