Tutoring math, this idea rarely surfaces.  However, it’s essential to number theory, a favourite of math professors.  The tutor discusses it for interest’s sake.

The square root of n is the number x such that x(x)=n. Perhaps more to the point: if x is the square root of n, then

x^2=n

Put in everyday terms, the square root is the number you “multiply by itself” to arrive at the original one. Therefore, 5 is the square root of 25.

I’ve written many posts that concern square roots: here and here are just two examples.

If you key √(2) into your calculator, you’ll find it’s 1.414213562…., which is an irrational number. An irrational number is a decimal that neither ends, nor follows a repeating pattern. Such a number cannot be written as a fraction of integers; in contrast, a rational number can.

Indirect proof

Let’s imagine we want to prove that √(2) is irrational. We can begin by assuming the opposite: that it is rational. Exploring the implications, we might arrive at a contradiction. Said contradiction will prove that √(2) can’t be rational. Then, the remaining conclusion will be that it’s irrational. Such an approach – where you assume the opposite, then prove it can’t be true – is called indirect proof.

Let’s begin, therefore, by assuming √(2) is rational. Then,

√(2)=a/b

where a and b are integers.

We’ll assume that a/b is in reduced form (that is, that a and b have no factors in common).

We can follow along with

2=a^2/b^2

Which leads to

2b^2=a^2

Since a^2 is 2 times b^2, a^2 must be even.

If a^2 is even, then a must be. After all, you can’t get an even number by multiplying two odds. Therefore,

a=2k,

for some integer k.

Furthermore, b must be odd, since a and b are assumed to have no factors in common.

Now, following along with a=2k,

2=(2k)^2/b^2=4k^2/b^2

However,

2=4k^2/b^2

leads to

2b^2=4k^2

which gives, when we divide both sides by two,

b^2=2k^2

Now we see that b^2 is even, implying that b must also be. Yet, we know a is even and that a and b share no common factors. Therefore, b can’t be even; otherwise, it will have 2 as a common factor with a.

The contradiction that b is proven to be even , while our earlier assumption prohibits its being so, proves that √(2) is not rational. It must, instead, be irrational.

As the school term develops, I’ll no doubt be drawn to more practical topics. Cheers:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.