Question

How to prove that √3 is an irrational number ?

Answer 1

Prove by contradiction - see below

Explanation:

Assume `sqrt3` is rational

`:. p/q =sqrt3 :{p,q} in ZZ, q!=0`

Also assume that `p and q` are in the simplest form (coprime) such that the Highest Common Factor `{p, q} =1`

Cross multiply:

`p=sqrt3 q`

Square both sides:

`p^2= 3q^2` [A]

`:. q^2 = p^2/3`

This implies that `3|p^2` (i.e. 3 divides `p^2`)

Since `3|p^2` then `3|p`

I.e `3` is a factor of `p`

So, we can write `p=3k` where `k` is some constant.

Replace `p=3k` in [A] above:

`(3k)^2 = 3q^2`

`9k^2 = 3q^2`

`3k^2 = q^2`

`:. k^2 = q^2/3 -> 3|q^2 -> 3|q`

Hence, `3` is also a factor of `q`

Now, since `3` is a factor of both `p and q` they cannot be coprime (as their HCF is 3 rather than 1)

Hence our assumption that `sqrt3` is rational has led to a contradiction.

Therefore we must conclude that `sqrt3` is irrational.

Answer 2

A Proof by Contradiction works with `sqrt(2)`, but not with `sqrt(3)`.

Explanation:

Here is a hint from " Mathematical Proofs " by Gary Chartrand, Albert Polimeni and Ping Zhang: page 90, Exercise 5.10 in what must be the 1st edition (green cover):

"Prove that `sqrt(3)` is irrational. [Hint: First prove that for an integer `a`, `3|a^2 iff 3|a`. Recall that every integer can be written as `3q, 3q+1` or `3q + 2` for some integer `q`.]"

Plainer English, "First prove that for an integer `a`, `3` divides `a^2` if and only if `3` divides `a`".

Answer 3

`sqrt(3) = [1;bar(1,2)] = 1+1/(1+1/(2+1/(1+1/(2+...))))` does not terminate, so is irrational.

Explanation:

Here's another proof, based on continued fractions.

First note that the continued fraction of any rational number will terminate.

Next, suppose `x > 0` satisfies:

`x = 1+1/(1+1/(1+x))`

`color(white)(x) = 1+(x+1)/(x+2)`

`color(white)(x) = (2x+3)/(x+2)`

Then multiplying both ends by `(x+2)` we see:

`x^2+2x = 2x+3`

Hence:

`x^2=3`

So:

`x = sqrt(3)`

So we have found:

`sqrt(3) = 1+1/(1+1/(1+sqrt(3)))`

`color(white)(sqrt(3)) = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+...))))))`

Then since the continued fraction does not terminate, it cannot represent a rational number.