Question

Prove that sqrt2is irrational.

Answer 1

ok so what is an irrational number?

Answer 2

@higgs Proof by contradiction!

Answer 3

Irrational number is a number which can not be represented in the form of p/q where p and q are integers and q is not equal to zero

Answer 4

Prove not explain

Answer 5

First, assume that \(\sqrt2\) is rational. Therefore,\[ \sqrt{2} = {p \over q} \qquad[\gcd({p,q = 0)]}\]

Answer 6

Let sqrt 2 be a rational number

since sqrt 2 is taken as a rational number hence it must be represented in the form of p/q

Answer 7

Square both sides.\[2 = {p^2 \over q^2}\]Multiply both sides by \(q^2\).\[ 2q^2 = p^2\]We already know that \(2q^2\) is an even number.

Answer 8

We also know that \(p\) is even from the above right?

Answer 9

oh ... wait :
\[\large{\frac{a}{b}\times \frac{a}{b}=\frac{a^2}{b^2} = 2}\]
\[\large{a^2=2b^2}\]

since a^2 is even so a must be even

let a = 2c

or 4c^2 = 2b^2

Answer 10

So, assume that \(p = 2k\).

Answer 11

\[2 = {4k^2 \over q^2}\]

Answer 12

\[2q^2 = 4k^2 \]

Answer 13

or 2c^2=b^2

hence b must be even

so a and b both are even .... but a/b was : reduced to lowest form

hence sqrt{2} is not a rational number ... though it is an irrational number

Answer 14

proved

Answer 15

We already knew that \(p\) is even, but now that \(q\) is also even, we know that the gcd is 2!

Answer 16

Note: there we assumed that HCF of a and b is 1

Answer 17

got it @higgs ?

Answer 18

But we first assumed that \(\gcd = 1\), but here's a contradiction!\[\gcd=2\Longleftrightarrow\gcd = 1\]

Answer 19

I get it now thanks guys

Answer 20

No problem

Answer 21

lol, google!