Prove that √3 is a irrational number.

Guest Apr 5, 2017

Here's one way :


Let us suppose that we have positive integers  a and b   such that


√3  = a / b     where a/b is in lowest terms  and a > b ......     square both sides


3  = a^2 / b^2


3b^2   =  a^2 


2b^2 + b^2   = a^2


2b^2   =  a^2 - b^2        (1)


We have three cases to consider


Case A   ....a and b are both even......but this is impossible because a/b would not be reduced to lowest terms if this were true since both would be divisible by 2


Case B  .......a is even and b is odd....but, in (1),  2b^2 is even but a^2 - b^2   is odd....and this is impossible  [Note the same result is produced if a is even and b is odd]


Case C ......the hardest to prove......let a, b  be odd....then we have that


2b^2 =  a^2  - b^2

2b^2 = (a + b) (a - b)


However.....if a and b are both odd......then (a + b) and (a - b) are both even....so....let (a + b)  = 2m and (a - b)  = 2n ......and we have that


2b^2  =  2m * 2n

2b^2  = 4mn 

b^2  = 2mn


But if b is odd, then so is b^2.......but 2mn is even......so this is a contradiction


Hence.......there are no positive integers a, b   of the same or different parity such that


√3  = a / b



cool cool cool

CPhill  Apr 5, 2017
edited by CPhill  Apr 5, 2017

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