Proving that the square root of 5 is irrational

Question

Prove that $\sqrt{5}$ is irrational.

I begin with the identity $(\sqrt{5} + 2 )(\sqrt{5} - 2 ) = 1$.

Then I am told to extract $\sqrt{5}$ from the first or second factor and consider it to be $\frac{m}{n}$ so I should replace it in both sides.

I have $$\frac{m}{n} = (\frac{1}{\frac{m}{n}} + 2) + 2.$$

I am also told to work on the right side until I have a denominator less than $n$ and I have to explain the reasoning.

Then I have to prove this is false by contradiction.

Right now my main problem is I can't get a denominator less than $n$.

Answer 1

Hint: correcting and simplifying your RHS, $$\frac{m}{n}=\frac{1}{(m/n)+2}+2=\frac{2m+5n}{m+2n}\ ,$$ but there is no way the RHS denominator is less than the LHS denominator. Try doing something similar but starting with $$\sqrt5=\frac{1}{\sqrt5-2}-2\ .$$ This will give you a proof that $\sqrt5$ is irrational.

Answer 2

Once again:

Here is a proof of mine by contradiction that if $n$ is a positive integer that is not a perfect square then $\sqrt{n}$ is irrational: $\sqrt{17}$ is irrational: the Well-ordering Principle

Answer 3

Consider $\sqrt{15}$ to be rational. Then we can express it into the form $\frac{p}{q}$, where p and q are integers with $gcd(p, q) = 1.$

Now: $$\frac{p}{q}= \sqrt{15}$$ $$⇒\frac{p^2}{q^2} = 15$$ $$⇒p^2 = 15 q^2$$ $$⇒ 15|p^2$$ $$⇒ 15|p \tag{*}$$

---

Now let $p = 15m$, for some $m ∈ ℕ$ $$p= 15m$$ $$⇒p^2 = (15)^2 m^2$$ $$⇒15q^2 = (15^2) m^2 \text { since $p^2 = 15q^2$}$$ $$⇒ 15|q^2$$ $$⇒ 15|q \tag{**}$$

---

Hence, from $(*)$ and $(**)$, leads us to think that our original assumption that the $\gcd = 1$ is wrong. This is a contradiction. Thus, our original statement holds. Hope this helps (: