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Show that root 3 is an irrational number.

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Let us assume, to the contrary, that √3 is rational. 
So, we can find coprime integers a and b(b ≠ 0) 
such that √3 = a/b 
=> √3b = a 
=> 3b2 = a2 ….(i) [Squaring both the sides] 
=> a2 is divisible by 3 
=> a is divisible by 3 
So, we can write a = 3c for some integer c. 
=> a2 = 9c2 ….[Squaring both the sides] 
=> 3b2 = 9c2 ….[From (i)] 
=> b2 = 3c2 
=> b2 is divisible by 3 
=> b is divisible by 3 
=> 3 divides both a and b. 
=> a and b have at least 3 as a common factor. 
But this contradicts the fact that a and b are coprime. 
This contradiction arises because we have assumed that √3 is rational.

Hence, √3 is irrational.

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