Lets try and prove this. I am going to do it by contradiction.
Assume√10isrationalthen√10canbewrittenaspq$wherepandqarerelativelyprimeintegers(theyhavenocommonfactors)$
√10=pq10=p2q2p2=10q2Sop2$isamultipleof10$so p is a multiple of 10Letp=10g(10g)2=10q2100g2=10q210g2=q2Soq2$isamultipleof10$$soqisamultipleof10$$Ihavediscoveredthatpandqarebothmultiplesof10$
Therefore p and q are not relatively prime
therefore the original statement is contradicted
therefore √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
Whatifp2=900p2=30×30p2=3×5×2×3×5×2p2=3×3×2×2×5×5$seehowthefactorshavetobeinpairs?$p=3×2×5
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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?