Which of these numbers are irrational, `sqrt2` , `3/4` , 2, -1, or `sqrt3` ?

1 Answer

txmedteach's profile pic

txmedteach | High School Teacher | (Level 3) Associate Educator

Posted on

In order to determine whether or not a number is rational, we much see whether or not the number can be expressed as a fraction with an integer numerator and denominator.

Clearly, `3/4`, 2, and -1 are rational because each can easily be expressed as a fraction.

There are plenty of ways to show that the square root of 2 is irrational. We'll summarize one here:

Suppose `sqrt(2)` is rational and can be expressed as a fully-reduced fraction `n/m`. Now, the following must hold true:

`sqrt(2)^2 = n^2/m^2`

`2 = n^2/m^2`

`2m^2 = n^2`

Because `n^2` is equivalent to an even number, we know that n, itself, must be even. Therefore, we can express n as another number, `2k.`

Now, we can restate the fraction:

`sqrt(2) = (2k)/m`

Again, let's take the square of both sides and solve for `m^2` :

`sqrt(2)^2 = (2k)^2/m^2`

`2 = (4k^2)/m^2`

`2m^2 = 4k^2`

`m^2 = 2k^2`

Now, we come up on the fact that `m` must be even, just like `n`, and it can be expressed as 2l.

However, now we come up on a contradiction! Remember how we defined `n/m` as the most-reduced form of the fraction that could represent `sqrt2`. However, the fact that both n and m are even allow `n/m` to be further-reduced. contradicting our assumption. Therefore, `sqrt(2)` must be irrational.

There are other proofs at the link below.

In a similar way, you can prove that `sqrt(3)` is also irrational.

So, our final answer is `sqrt2` and `sqrt3`.