of Squareroot of 2
Irrational numbers are numbers that cannot be written in the form of `p/q`, where p and q are integers and q ≠ 0. Example `sqrt3,sqrt2`, π.
1)Theorem: let p be a prime number. If P divides `a^2` the p divides a, where a is a positive integer.
Proof: If `a^2/p` true then `a/p` is also ture.
a=`(p_1p_2...p_n)` as per fundamental theorem, this means
= `p_1^2 p_2^2..p_n^2`
given: p divides `a^2`
that means p must belong to pi ,where i lies between 1 to n
Thus, is a factor of a also.
2)Theorem: `sqrt2` is irrational
Proof: Let's assume `sqrt2` is rational. If `sqrt2` is rational the `sqrt2= a/b`, where a and b are co prime.
b `sqrt2`= a
`2b^2`= `a^2` That means 2 divides `a^2`. And hence 2 is prime number it also divides a.
2×k=`a^2` (k is constant)
2×k= a .......eq1
2`b^2`= 2`k^2`= 4`k^2`
`b^2`= 2`k^2`, Therefore 2 divides `b^2`, and since 2 is prime number it also divides b.
That means 2 is factor of a,b but according to our assumption a and b are co prime which
means they dont have common factors. Here our assumption that `sqrt2` is rational is
proven incorrect, thus sqrt is a irrational number.