Give a proof that the square root of 3 is an irrational number?

Answer 1

Problem:

Prove that is an irrational number.

Solution:

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 (b not equal 0). 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.

or

As it is discussed above 3 = a2/b2 = even/even or odd/odd.

If 3 = a2/b2 = odd/odd, then a2 + b2 = even.

So that a2 + b2 = 2p, and we have:

(2m + 1)2 + (2n + 1)2 = 2p

4m2 + 4m + 1 + 4n2 + 4n + 1 = 2p

4m2 + 4m + 4n2 + 4n + 2 = 2p

2m2 + 2m + 2n2 + 2n + 1 = p where p is odd

3 = a2/b2

3 + 1 = a2/b2 + 1

4 = (a2 + b2)/b2

4 = 2p/b2

2 = p/b2, so p must be even, a contradiction!

Thus, the square root of 3 cannot be a rational number, so it is an irrational.