If sqrt(3) was rational, then by definition it could be written as a/b, whereaand andbare both integers. To prove indirectly that sqrt(3) is irrational, we must prove thataandbcan not exist.

Suppose: sqrt(3) = a/b, a and b are integers (it is rational), and a/b is in lowest terms

1) Squaring both sides yields: 3 = a^2/b^2

2) Multiplying both sides by b^3 yields: 3b^2 = a^2

- Suppose b is even. Then b^2 is even and 3b^2 is also even. This means that a^2 and a must also be even. If both a and b are even numbers, then the ratio a/b can be simplified by cancelling a common factor of two, and contradicts our assumption that a/b is in lowest terms. For the sake of this proof, we must then assume that a and b are therefore both odd and that a/b is written in simplest terms.

3) Since a and b are odd, we can write the equations:

a = 2x+1

b = 2y+1

Where x and y are integers, thus restricting a and b to be any odd integer.

Substituting these into 3b^2 = a^2, we get:

3(2y+1)^2 = (2x+1)^2 (simplify)

3(4y^2+4y+1) = 4x^2+4x+1 (distribute)

12y^2+12y+3 = 4x^2+4x+1 (subtract 1)

12y^2+12y+2 = 4x^2+4x (factor)

2(6y^2+6y+1) = 2(2x^2+2x) (divide by 2)

6y^2+6y+1 = 2x^2+2x (factor right side)

6y^2+6y+1 = 2(x^2+1)

Now, for any choice of "y", the left hand side will be odd. For any choice of "x", the right hand side will be even. Thus, there exist no solutions to this equation, and proof that sqrt(3) is irrational is evident by contradiction.

Let me know if you have any questions!