IIT JEE , AIEEE Forum - Mathematics , Algebra

rudra.panda

A question: Prove that is irrational.

hsbhatt

Suppose $\frac{p}{q} = 2^{\sqrt 3}; p, q \in \mathbb{Z}; gcd (p,q) = 1$

Now $\frac{p}{q} = 2^{\sqrt 3} \Rightarrow p = 2^{\sqrt 3} q$

Since p is an integer it has a unique canonical factorisation into its prime factors in which the power to which 2 appears is an integer m.

So, if the above equation holds true m = $\sqrt 3$ . (Question: Can q have a power of 2 as its factor?)

But $1<\sqrt 3 < 2$ and we know that no integer can lie between 1 and 2. Thus we have obtained a contradiction with our original assumption that $\sqrt 3$ is a rational number

animal

my soln is not as perfect as hsbhatt sir's but it can b useful for mcq

by the binomial sereis

(1+x)ⁿ=1+nx+n(n-1)x²/2 + ........

now taking 2^{root 3} (1+1)^{root 3}we get that in the 2nd term of the expansion we get a multiple of root 3 and we will get it in many other terms and they will surely not cancel each other

also we know that rational +irrational=irrational so we it will be irrational.

hope u got it........

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