What do you assume to prove \sqrt2 irrational?

\sqrt2=\frac pq with p,q integers, no common factor, q\neq0.

In the \sqrt2 proof, what follows from p^2=2q^2?

p^2 is even, so p is even; write p=2k.

Where is the contradiction in the \sqrt2 proof?

p and q are both even, which contradicts having no common factor.

Negation of 'for all integers n, n^2 is even'?

There exists an integer n such that n^2 is odd.

What assumption starts Euclid's proof?

There are finitely many primes, p1,\ldots,pn.

Why is no pi a factor of N?

N leaves remainder 1 when divided by any pi.

Must N be prime in Euclid's proof?

No. It only needs a prime factor that is not in the list, for example 30031=59\times509.

How do you show \sqrt2+\sqrt3 is irrational?

Assume it equals rational r, isolate and square to get \sqrt2=\frac{r^2-1}{2r}, which is rational: contradiction.

Proof by contradictionEdexcel International A Level Maths: Flashcards

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Outline proof by contradiction.

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Outline proof by contradiction.: Assume the statement is false, deduce something impossible, and conclude that the statement is true.
What do you assume to prove $\sqrt2$ irrational?: $\sqrt2=\frac pq$ with $p,q$ integers, no common factor, $q\neq0$ .
In the $\sqrt2$ proof, what follows from $p^2=2q^2$ ?: $p^2$ is even, so $p$ is even; write $p=2k$ .
Where is the contradiction in the $\sqrt2$ proof?: $p$ and $q$ are both even, which contradicts having no common factor.
Negation of 'x is irrational'?: ' $x$ is rational'.
Negation of 'for all integers $n$ , $n^2$ is even'?: There exists an integer $n$ such that $n^2$ is odd.
What assumption starts Euclid's proof?: There are finitely many primes, $p_1,\ldots,p_n$ .
Define N in Euclid's proof.: $N=p_1p_2\cdots p_n+1$ .
Why is no $p_i$ a factor of $N$ ?: $N$ leaves remainder 1 when divided by any $p_i$ .
Must N be prime in Euclid's proof?: No. It only needs a prime factor that is not in the list, for example $30031=59\times509$ .
True or false: if a product is even, at least one factor is even.: True.
How do you show $\sqrt2+\sqrt3$ is irrational?: Assume it equals rational $r$ , isolate and square to get $\sqrt2=\frac{r^2-1}{2r}$ , which is rational: contradiction.

Exam questions on Proof by contradiction

A student begins a proof by contradiction that $\sqrt2$ is irrational.
Given that $p^2=2q^2$ and so $p$ is even, complete the proof.2 marks
Euclid's proof that there are infinitely many primes begins by assuming that there are only finitely many primes, $p_1,p_2,\ldots,p_n$ . Let $N=p_1p_2\cdots p_n+1$ .
A student takes the first six primes, $2,3,5,7,11,13$ , so $N=30031$ . Show that $N$ is not prime, and state what this shows about Euclid's argument.2 marks
In a proof by contradiction you assume that the statement is false and show that this leads to an impossible result.
Prove by contradiction that there are no integers $m$ and $n$ such that $14m+21n=1$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).