Question

Prove that √3 is irrational by contradiction method.

Answer 1

We will prove whether √3 is irrational by contradiction method. 
Let √3 be rational 
It can be expressed as √3 = a/b ( where a, b are integers and co-primes. 
√3 = a/b
3= a²/b² 
3b² = a²
3 divides a²
By the Fundamental theorem of Arithmetic 
so, 3 divides a .

a = 3k (for some integer) 

a² = 9k² 
3b² = 9k² 
b² = 3k² 

3 divides b²
3 divides b. 

Now 3 divides both a & b this contradicts the fact that they are co primes. 
this happened due to faulty assumption that √3 is rational. Hence, √3 is irrational. 

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Answer 2

Hey..!!!!

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I know I have to use a lemma to establish that if x is divisible by 3,

then x2 is divisible by 3.
The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of 6?

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I Hope It's help you...!!!!

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