2
$\begingroup$

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 $n$.

$\endgroup$
  • 4
    $\begingroup$ Ok, what is the question? $\endgroup$ – Timbuc Oct 1 '14 at 2:16
  • 3
    $\begingroup$ Did you really mean to write $1/(m/n)+2$ (that is, $(1/(m/n))+2$) or did you mean $1/((m/n) + 2)$? $\endgroup$ – David K Oct 1 '14 at 2:29
4
$\begingroup$

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 (:

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ First, I think he has $$\frac mn=\left(\frac1{\frac mn}+2\right)+2=\frac nm+2+2=\frac{n+4m}m$$ so how did you get your expression? Second, how in the world did you guess the OP wanted to prove $\;\sqrt5\;$ is irrational?! $\endgroup$ – Timbuc Oct 1 '14 at 2:30
  • $\begingroup$ Because when op was talking about irrational numbers he mentioned the exercise $\endgroup$ – JOX Oct 1 '14 at 2:36
  • $\begingroup$ The OP is you, @DanielOrtizCosta ... and you don't mention at all what number is to be proved irrational. $\endgroup$ – Timbuc Oct 1 '14 at 2:46
  • $\begingroup$ @Timbuc it's true that the OP has$$\frac1{m/n}+2\ ,$$but it should be$$\frac1{\frac mn+2}$$so I "discreetly" corrected it. And I guessed that the question was about $\sqrt5$ because this is a standard method for proving irrationality of $\sqrt n$, though not as well known as some other methods. Of course it may well be that my guess was wrong. $\endgroup$ – David Oct 1 '14 at 2:53
  • $\begingroup$ Thanks @David. Somebody else now added the question to the OP. $\endgroup$ – Timbuc Oct 1 '14 at 2:58
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.