#2**+5 **

**Lets try and prove this. I am going to do it by contradiction.**

$$\\Assume \;\;\sqrt{10}\; is\; rational\\\\

then \;\sqrt{10}\;can\;\; be\;\; written\;\; as \;\;\dfrac{p}{q}\\

$where p and q are relatively prime integers (they have no common factors)$\\\\$$

$$\\\sqrt{10}=\frac{p}{q}\\\\

10=\frac{p^2}{q^2}\\\\

p^2=10q^2\\\\

So\;\; p^2\;\;$ is a multiple of 10$\\\\

\mbox{so p is a multiple of 10}\\\\

Let\; p=10g\\\\

(10g)^2=10q^2\\\\

100g^2=10q^2\\\\

10g^2=q^2\\\\

So\;\; q^2 $ is a multiple of 10$\\\\

$so q is a multiple of 10$\\\\

$I have discovered that p and q are both multiples of 10$$$

Therefore p and q are not relatively prime

therefore the original statement is contradicted

**therefore $${\sqrt{{\mathtt{10}}}}$$ is irrational.**

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**Note:**

I have said that since p^{2 }is a multiple of 10, p must also be a multiple of 10. **WHY is so.**

Well the prime factors of squared numbers must come in pairs.

Lets look at an example

$$\\What\;\; if\;\; p^2=900\\

p^2=30\times 30\\

p^2=3\times 5\times 2 \quad \times \quad 3\times 5\times 2\\

p^2=3\times 3\times 2 \times 2\times 5\times 5\\

$see how the factors have to be in pairs?$\\

p=3\times 2 \times 5\\$$

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10 is the product of 2 prime numbers

Can you see how if p^2 is a product of 10 then p must also be the product of 10?

Melody
Jan 11, 2015