False proof by contradiction?

Question

$P(x)$ is a polynomial, whose degree is either $1$ or $0$ (we do not know exactly which).

$Q(x)$ is a polynomial, whose degree is unknown.

Assuming that the degree of $P(x)$ is $1$, I prove that

Degree of $Q(x) = 1$ $\implies$ Degree of $Q(x) = A > 1$.

Can I conclude that the degree of $P(x)$ is $0$?

I think this is an example of an invalid proof by contradiction, but I wanted to double-check.

The reason is that a statement of the form

$$p \implies NOT(p) $$ is not a contradiction.

score 1 · Answer 1 · 2020-10-26T01:28:56.500

1

It would seem that you just proved $\deg P=0\lor \deg Q\ne 1$. Technically, I would say that you proved $(\deg P=0\lor \deg Q\ne 1\lor Q=0)$, but that's just my convention of choice of $0$ not having a degree.

edited Oct 26 '20 at 01:28

answered Oct 25 '20 at 23:03

You have said that "Degree of Q(x)=1 ⟹W Degree of Q(x)=A>1" – user247327 Oct 25 '20 at 23:25
@user247327 Are you talking to me? – Oct 26 '20 at 01:29
No, I was talking to Nickkatz2018 who posted that. – user247327 Oct 26 '20 at 14:07

False proof by contradiction?

1 Answers1