Question

can someone help me how to show that there is no rational number x such that xsquared =2

Answer 1

assume by contradiction sqrt(2) is rational
then sqrt(2) = q/p where q and p are relative prime integers (they are reduced), and p is not 0.
square both sides
2=p^2/q^2
2q^2=p^2
so p^2 is even
(it is not hard to show that p is then even)
so p is even
so
2q^2=(2k)^2 for some k in integer
so
2q^2=4k^2
q^2=2k^2
so q is even
so p and q are even
this is a contradiction because (p,q) = 1
thus sqrt(2) is irrational

Answer 2

if you need me to show that when a^2 is even a must be even I will

Answer 3

(p,q) = 1 because they are the same even integer?

Answer 4

no (p,q) = 1 means the are relatively prime, they share no common factor.

Answer 5

in other words p/q is in "lowest terms" so p and q cant both be even, because then they are not in lowest terms

Answer 6

and by the fundamental theorem of arithmetic we know that every ratio can be written in lowest terms....(but this is obvious).

Answer 7

understand @gachauJM @blurbendy

Answer 8

to show a^2 implies a is even

assume a^2 is even but a is odd
then a = 2k+1 for some integer k
so
a^2=(2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1 and since 2k^2+2k is an integer,
call it z, we have a^2 =2z+1 which is of course odd, but a^2 is even, and thus again by contradiction a must be even

Answer 9

@zzrOck3r i got it.thanks