prove that there is no rational number whose square is 3. detailed explanation please!
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Solution: Suppose there was a rational number x whose square was divisible by 3.
Then there would be integers p and q with no common divisors so that x = p/q and
x
2 = 3.
Thus
p
2
q
2
= 3, and so p
2 = 3q
2
which means p is divisible by 3, that is, there is an integer a so that p = 3a. Hence
3q
2 = p
2 = (3a)
2 = 9a
2
,
and so q
2 = 3a
2
. This means q is also divisible by 3, which contradicts our assump-
tion that p and q had no common divisors.
Then there would be integers p and q with no common divisors so that x = p/q and
x
2 = 3.
Thus
p
2
q
2
= 3, and so p
2 = 3q
2
which means p is divisible by 3, that is, there is an integer a so that p = 3a. Hence
3q
2 = p
2 = (3a)
2 = 9a
2
,
and so q
2 = 3a
2
. This means q is also divisible by 3, which contradicts our assump-
tion that p and q had no common divisors.
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