Question

If 2a^2 = b^2 where a, b belong to integers, then 2 is a common divisor of a and b
prove

Answer 1

yikes this one again

Answer 2

first of all it is not possible to have integer a and b with
\[2a^2=b^2\] but that is ok, we can pretend here we go

Answer 3

u cant really search for previous questions is sux

Answer 4

first note that
\[2a^2\] is even because it has a 2 in front

Answer 5

then since
\[2a^2=b^2\] that means
\[b^2\] is even

Answer 6

now if
\[b^2\] is even , that means b is even because if you square an even number you get an even number, and if you square an odd number you get an odd number. so not only is b^2 even but also b must be even

Answer 7

ok that was so easy idk why i didnt see it ty :)

Answer 8

now since b is even, you know that not only is b^2 even, but it is divisible by 4, because the square of any even number is divisible by 4 (easy proof if you need it)

Answer 9

well we are not quite done actually, because your question (although i don't actually see it) is to show that both b and a are even, so we have still to show that a is even

Answer 10

we know that b is even, and that b^2 is divisible by 4 and we know
\[2a^2=b^2\] which means
\[2a^2\] is divisible by 4, which means
\[a^2\] is even which means
\[a\] is even as above. done

Answer 11

ty

Answer 12

can u say that if 2|b, then there exist k such that 2k=b, so 2a^2 = (2k)^2 => a^2=2k^2, thus two is a common divisor of a and b?

Answer 13

yw
but please be advised it is not possible for
\[2a^2=b^2\] for whole umber a and b

Answer 14

@onceupon a time yes you are right, but you are assuming that b is even. you have to show that as well. i just said it in english. but the whole thing is stupid because you cannot have one whole number squared as twice another one squared. a picture will show you why