Question

show that √3 is irrational ......
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ans it yourself................
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Answer 1

let us assume that √3 is rational
therefore,
√3=p/q
squaring on both sides, we get
3=p^2/q^2
3q^2=p^2
q^2=p^2/3
therefore, 3 divides p
let p is 3m
p^2 = 9m^2
but, p^2= 3q^2
so, 9m^2=3q^2
m^2=q^2/3
therefore 3 divides q
it's contradict our assumption and √3 is irrational..

Answer 2

hello... thx for points and here is your answer
let us assume √3 is rational
√3=a/b
√3b=a
squaring on both sides,
(√3b)²=a²
3b²=a²
a²/3=b²
There fore 3 will divide a²and it also divide a..... 1
a/3=c(where c Will be integer)
a=3c
There fore
3b²=3c²
3b²=9c²
b²=1/3×9c²
b²=3c²
b²/3=c²
here 3 divides b²and it also divide b.......2
By eq-1 and eq-2
3 divides both a and b
hence 3 is a Factor of a and b
so a and b are having a Factor
there fore a and b are not co-primes



So our contradiction is not correct
and √3 is also irrational.


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