+118608 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 p2 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?
