?

Abstract Algebra

Free Version

Upgrade subject to access all content

Difficult

Proof by Contradiction

Free
ABSALG-VGGNK9

Proof by contradiction is a common way that mathematicians justify theorems.

If you are trying to prove that some statement or property $P$ is true, you assume $P$ is false, and by mathematically logical argument deduce a contradiction. Therefore, your initial assumption that $P$ is false cannot hold, so that $P$ must be true.

In each choice of the following questions, do not worry about whether the italicized statements are true or not, especially if you have no idea about whether they are true or not! Just worry about whether they contradict each other.​

Which of the following statements (about $x$ contradicting $y$) is INCORRECT​? ​

A

$\sqrt{2}$ is not a rational number contradicts there are positive integers $p$,$q$ such that $\sqrt{2}=p/q$

B

$\sqrt{2}$ is not a rational number contradicts if $P$ is any non-zero polynomial with rational coefficients, and $P(\sqrt{2})=0$, then $P(-\sqrt{2})=0$.

C

$\sqrt{2}$ is a rational number contradicts every fraction can be written in the form $p/q$ where $p$ and $q\ge 1$ are integers with no factors $\ge 2$ in common

D

$\sqrt{2}$ is a rational number contradicts there is no non-zero polynomial $P$ of degree 1 and with coefficients rational numbers such that $P(\sqrt{2})=0$.

E

$\sqrt{2}$ is a rational number contradicts if $P$ is any non-zero polynomial with rational coefficients, and $P(\sqrt{2})\not=0$, then $P(-\sqrt{2})\not=0$.