# Under root

Q1. Prove that square root 6 is irrational.

Q2. Prove that cube root 6 is irrational.(without assuming the property that if a prime divides a cube of a no. it divides the no.itself)

Note by Cool Math
1 year, 5 months ago

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Let √6 = p/q (where p and q are integers and are co-prime ) Squaring both sides , 6=p²/q² $$\Rightarrow$$ p²=6q² (Since , 6 is a factor of p² then it is a factor of p also) Let p=6m (where m is an integer) Squaring both sides , p²=36m² $$\Rightarrow$$ 6q²=36m² or q²=6m². (Since , p²=6q²) (Since , 6 is a factor of q² then it will be a factor of q also) Hence , 6 is a $$**common**$$ factor of p and q. This contradicts our fact that p and q are co-prime Hence , √6 cannot be expressed in the form of p/q $$\boxed{\therefore \space \sqrt{6} \space is \space irrational}$$

- 1 year, 5 months ago

you may have forgotten that the theorem that you used to prove that 6 is a factor of p applies only for primes !! for example . 12 divides 36 . but 12 does not divide 6

- 1 year, 5 months ago

nice name by the way!!!

- 1 year, 5 months ago

:)

- 1 year, 5 months ago