Question

prove that 3+√2 is an irrational number.

Answer 1

let 3+√2 be a rational number

Therefore it can be written in p/q form where q is not equal to 0

So,3+√2=p/q

√2=p/q-3

√2=p-3q/q

so,here p-3q/q is a rational number than √2 will also be a rational number but in nature √2 is an irrational number.

So, our assumption was wrong..

By contradiction 3+√2 is an irrational number..

Hence Proved!!

HOPE..IT WILL HELP!!

Answer 2

Let us assume to the contrary that root 2 is irrational.Thus it can be presented in the form p/q where p and q are rational numbers and q not equal to 0 and they are co prime

p/q=root 2

squaring on both sides we get

p^2/q^2=2

p^2/2=q^2  .........(1)

2 divides p^2 and 2 divides p

put p=2c

4c^2/2=q^2

2c^2-q^2

q^2/2 = c^2

2 divides q^2

2 divides q

therefore p and q have a comon factor that is 2 and our assumption that root 2 is rational is wrong therefore root 2 is irrational

when we add rational and irrational the sum is irrational therefore 3 which is a rational number when added with a irrational number root 2 then the sum is irrational therefore 3+ root 2 is irrational

hence proved