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Prove that $\sqrt{5.5}$ is irrational

My attempt:

Suppose $$\sqrt{5.5}=\frac p q\bigg/\quad()^2$$

$$\Longrightarrow 5.5=\frac{p^2}{q^2}\quad \text{gcd}(p,q)=1$$

$$\Longrightarrow q^2 \cdot 5.5=p^2$$

$$\Longrightarrow q\mid p^2\Longrightarrow \text{gcd}(q,p^2)\ge q$$

$$\Longrightarrow q\mid p\Longrightarrow \text{gcd}(q,p)\ge q$$

$$\Longrightarrow q=1$$

and also

$$p\mid q^2\Longrightarrow p\mid q\Longrightarrow p\mid 1 \Longrightarrow p=1$$

$$\Longrightarrow 1=1\cdot5.5$$

Сontradiction.

Is my attempt correct? If not how can I prove that?

3SAT
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    $\frac p q\bigg/\quad()^2$ means? –  Mar 08 '16 at 11:28
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    To do power of two for both sides – 3SAT Mar 08 '16 at 11:31
  • I'd recommend trying to get to the integers as soon as possible, so instead of writing $5.5$, try $\frac{11}{2}$. The equation in $p$ and $q$ then becomes $11q^2=2p^2$. Can you get a contradiction from that? –  Mar 08 '16 at 11:39
  • Sorry, completely wrong. $q|p^2$ doesn't imply $q|p$ and even less $q=1$, among others. –  Mar 08 '16 at 11:51
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    This is a garbled version of an almost identical question from the same user, posted three hours ago. Nehorai, what are you playing at? – TonyK Mar 08 '16 at 11:55
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    Although this is implicit in the other answers, I think it is good to say it out loud: the implication $q^2 \cdot 5.5 = p^2 \Rightarrow q|p^2$ is not as obvious as the analogous implication $q^2 \cdot 5 = p^2 \Rightarrow q|p^2$ since 5.5 is not an integer. – Vincent Mar 08 '16 at 11:56

3 Answers3

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Write $5.5 =\frac{11}{2}$, and suppose $\sqrt{\frac{11}{2}} = \frac{p}{q}$ with $\gcd(p,q)=1$. Then we have $$\frac{11}{2} = \frac{p^2}{q^2} \implies 11q^2 = 2p^2.$$ From this we obtain that $q^2$ is even, and so $q=2m$. Thus we have $$11(2m)^2=4 \dot\ 11m^2 = 2 p^2 \implies p^2 = 2 \dot\ 11 m^2.$$ So $p^2$ is even and thus $p$ is even. So $2 \ | \ \gcd(p,q)$, which is a contradiction.

Your attempt is not correct as from $5.5q^2 = p^2$ you cannot conclude that $q$ divides $p^2$ as $5.5$ is not a integer.

Hetebrij
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It isn't quite tru, we have $5.5=\frac{11}{2}$, what you should do more that is cleared that after $\gcd(p,q)=1$ we have that $$11q^2=2p^2$$ as we have that $11$ clearly doesn't divide $2$ we have taht $11|p^2$ which gives $11|p$ and $p=11n$ and $11^2n^2=p^2$ and as such

$$11q^2=2 \cdot 11^2 n^2$$ giving $$q^2=2 \cdot 11 n^2$$ which gives through similar reasoning that $11|q$ and as such our initial assumption is broken and it's irrational.

The key is you focus on the fixed known part, namely the fraction $\frac{11}{2}$ and it's constitutients rather than trying to have $p$ and $q$ dividing one another, which we cannot say from the initial as we can have them divide the constant parts.

Zelos Malum
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For $p,q$ not both even,

$$55q^2=10p^2$$ $\equiv\{ \text{simplify by }5\}$ $$11q^2=2p^2$$ $\implies\{ 11\text{ odd, RHS even}\}$ $$2|q^2$$ $\equiv\{ q\text{ odd}\implies q^2\text{ odd}\}$ $$\color{red}{2|q}$$ $\implies\{\text{ odd }11\text{, LHS multiple of }4\}$ $$4|2p^2$$ $\equiv\{\text{ simplify by } 2\}$ $$2|p^2$$ $\equiv\{ p\text{ odd}\implies p^2\text{ odd}\}$ $$\color{red}{2|p}$$

Contradiction.