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The Irrationality of square root of three

Problem:
Prove that square root of three is an irrational number.

Solution:
The number, square root of three, 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 square root of three is rational so that we may write

square root of three = 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                                      4.

where we require m and n to be integers (to ensure integer values of a and b). When these expressions are substituted into eq. 2a, we obtain

3(4n2 + 4n + 1) = 4m2 + 4m + 1     5.

Upon performing some algebra, we acquire the further expression

6n2 + 6n + 1 = 2(m2 + m)               6.

The Left Hand Side of eq. 3a is an odd integer. The Right Hand Side, on the other hand, is an even integer. There are no solutions for eq. 3a. Therefore, integer values of a and b which satisfy the relationship square root of three = a/b cannot be found. We are forced to conclude that square root of three is irrational.

The other way: from Yahoo anser

We prove that by contradiction.
This means, we suppose √3 is rational and then get to something that contradicts this assumption.
If √3 is rational, we can express it as a/b, a fraction in its lowest terms (i.e., gcd(a,b) = 1). Also, b is not 1, because that would mean √3 is an integer, which it obviously is not.
Now square both sides.
√3 = a/b
3 = a²/b²
since gcd(a,b) = 1, then gcd(a²,b²) = 1 (has to be).
But from 3 = a²/b² we can express a² = 3b².
So a²/b² = 3b²/b² -> now this fraction is obviously NOT in its lowest terms (remember, b is not 1) -> hence the contradiction.
=> √3 is not rational. It is irrational.

The third way:like sqrt two

sqrt (3) = m/n   (a fraction, reduced to its lowest terms)
     3 = m²/n²   m² = 3n²   (So, m is a multiple of 3, call it 3q)
     9q² = 3n²   3q² = n²    (So, n is a multiple of 3)