Question

Prove that 3a2 - 1 is never a perfect square.

Answer 1

takes a as 1 then answer is 5 so 5 is not a perfect square

Answer 2

Answer:

let a = 1 in root(3a^2-1) we get root(2)

let a = 2 in root(3a^2-1) we get root(11)

both are not perfect square let this be true for some random integer a = k

so, root(3k^2-1) is not a perfect square   #

let a=k+1

now root(3(k+1)^2-1)=root(3k^2-6k+2)

can be written as root(3k^2-1+3-6k)=root(3k^2+3(1-2k)-1)

=root(3(k^2+1-2k)-1) since k,1,2 are integers  k^2+1-2k=m^2

where m is some integer root(3m^2-1) we observe this is of the

form # so  root(3(k+1)^2-1) is not a perfect square so by principle of

m induction  3a^2-1 is not a perfect square

Step-by-step explanation:

just Induction bro