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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 $n$.

Bart Michels
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JOX
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3 Answers3

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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 (:

Joshua
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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.

David
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  • 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?! – Timbuc Oct 01 '14 at 02:30
  • Because when op was talking about irrational numbers he mentioned the exercise – JOX Oct 01 '14 at 02:36
  • The OP is you, @DanielOrtizCosta ... and you don't mention at all what number is to be proved irrational. – Timbuc Oct 01 '14 at 02:46
  • @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. – David Oct 01 '14 at 02:53
  • Thanks @David. Somebody else now added the question to the OP. – Timbuc Oct 01 '14 at 02:58
  • @Timbuc ...and I have made the correction more explicit in my answer. Thanks for your comment. – David Oct 01 '14 at 02:59
  • Thanks, I just tried this, but unfortunately can't get far, how can I infer the denom is less than n? – JOX Oct 02 '14 at 00:38
  • What denominator did you get? – David Oct 02 '14 at 01:58
  • m-2n @David...., although by dividing by m, I can also get 1 - (2n/m) – JOX Oct 02 '14 at 03:31
  • By assumption $\frac mn=\sqrt5$, so by doing some simple algebra you should be able to prove that $m-2n>0$ and $m-2n<n$. – David Oct 02 '14 at 03:41
  • @David by Working with m - 2n I ended up with a new formula n(√5-2), and by using the fact that √5 = 2.*** I can obviously conclude n(√5-2) < n; however, I can't seem to find a more general way to do it without considering √5 = 2.***; can you hint me? – JOX Oct 02 '14 at 03:54
  • The inequality $m-2n>0$ is equivalent to $\frac mn>2$, can you see why this is true? – David Oct 02 '14 at 03:59
  • Definitely, but the problem is how can I prove m-2n is > 0 @David – JOX Oct 02 '14 at 06:15
  • m > 2n so if you subtract m - 2n, then it's definitely going to be greater than 0. – user125535 Oct 02 '14 at 13:32
  • @user125535 thank you – JOX Oct 02 '14 at 21:51
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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

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marty cohen
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