Prove that there is no rational number whose square is 3?

Answer 1

This is the same as proving the square root of 3, which I will write sqrt(3) is irrational. When I prove that, I am proving there is not rational number whose square is 3. So here goes: I put some extra help comments in (extra help) like this. Just ignore them if you don't need them.

Let sqrt(3) = a/b where a and b are integers and they are in lowest terms, meaning they have no common factors, WE know if they had common factors we can always find another expression that has no common factors by canceling out the common ones. So a/b does exist.

squaring both sides of sqrt(3) = a/b gives us 3 = a2/b2 so a2 = 3b2

Now a2 must have its factors as even powers of primes.(because of the exponent 2 on the left and both sides must have the same total power since they are equal). So it must
have 32 as one of its factors. (can't be just 3 since that is 31 which is an odd exponent and when added to the 2 in b2 gives us an odd exponents also) This means a must have 3 as a factor. ( if 32 is a factor then 3 must be also)

Now 3b2 has a factor 32 so b2 has a
factor 3. But if b2 has a factor 3, then since powers must be even,
it has 32 as a factor. This means that b must have 3 as a factor also.

So we have shown that both a and b have 3 as a factor. But we stated
that a/b has no common
factors and we just found one, namely 3. So we have a contradiction, and it follows that
we cannot express sqrt(3) as a rational fraction.

Here is another proof from WikiAnswers.com from someone who asked the same question.

The number, , is irrational, ie., it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us assume that is rational so that we may write
= a/b
1.
for a and b = any two integers. We must then show that no two such integers can be found. We begin by squaring both sides of eq. 1:
3 = a2/b2
2.
or
3b2 = a2
2a.
If b is odd, then b2 is odd; in this case, a2 and a are also odd. Similarly, if b is even, then b2, a2, and a are even. Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2, we must assume that a and b are odd, and that the ratio a/b is already reduced to smallest possible terms. With a and b both odd, we may write
a = 2m + 1
3.
and
b = 2n +1

Therefore,
(a)^2 + (b)^2 = (2m+1)^2 + (2n+1)^2
and
(a)^2 - (b)^2 = (2m+1)^2 - (2n+1)^2

Also,
[(a)^2 + (b)^2] / [(a)^2 - (b)^2] = 2

Which leads to:

[2(m^2 + n^2) + 2(m + n) + 1] / [2(m^2 - n^2) + 2(m - n)] = 2

Which is impossible since numerator is odd but denominator is even.

One more proof for you. Sometimes one proof makes more sense to a given person. Also, it is nice to see things proved in a slightly different way

If √3 is rational, we can express it as a/b, a fraction in its lowest terms this mean the greatest common divisor or gcd of a and b written (a,b) or gcd(a,b) = 1. Also it is important to note that b is not 1, because that would mean √3 is an integer, which we know it is not.

Now square both sides as we did in the other proofs.
√3 = a/b
3 = a²/b²

since gcd(a,b) = 1, then gcd(a²,b²) = 1 (This is because if the there are no common factors and you square it, you stilll have no common factors, look at 2/3 for example, gcd (2,3)=1 and 4/9 still has gcd of 1 because is is 2/3 x2/3)

But from 3 = a²/b² we can write a² = 3b².

we can write this as a²/b² = 3b²/b² (since b2 /b2 =1) This fraction is clearly NOT in its lowest terms (remember, b is not 1) hence the contradiction.

and we have that √3 is not rational.