Irrationality of the Square Root of Four?

OK, last time we proved that the square root of two is not one whole number divided by another whole number, because we could continue dividing for ever, and we know that we cannot do that because we would eventually run out of numbers. It does all sound a bit fishy though. Is this not just some trick of the maths? To double check, lets try the same process, but with the square root of four instead. The square root of four is easy - it is just two, because two multiplied by two is four.

Anyway, we are assuming that the square root of four can be written as one number over another number. So lets use the same letters as last time.

\[\frac{a}{b}=\sqrt{4}\]

Excellent, again. Some fraction equals the square root of four. That's the assumption that we want to make. Nothing wrong so far. Let's now do a mathematical operation on this equation.

What we want to do is get the right hand side to just be four. This means that we need to multiply it by itself to get four. Obviously the square root of four multiplied by the square root of four is four - because that's where it came from. Let's skip the not equals sign stuff this time around, you get the idea.

\[\frac{a^2}{b^2}=4\]

And again, we know that we can multiply both sides by \(b^2\) to get:

\[a^2=4\cdot b^2\]

Now at this point can we say that \(a\) is an even number? Yes, we can run through all of our odds and evens arguments and still say that, because four is just two multiplied by two, so the left hand side is still some other number on the right hand side multiplied by two. So that gives us:

\[(2\cdot c)^2=4\cdot b^2\]

And we can square out contents of the brackets on the left hand side:

\[2\cdot 2\cdot c\cdot c=4\cdot b^2\]
\[2\cdot 2\cdot c^2=4\cdot b^2\]
\[4\cdot c^2=4\cdot b^2\]

OK. Last time when hunting for the square root of two we had this:

\[4\cdot c^2=2\cdot b^2\]

What is the difference here? The right hand side is multiplied by the SAME NUMBER as the left hand side - four. What can we gather from this? Well we can divide both sides by four to get this:

\[c^2 = b^2\]

And we can square root each side to get this:

\[c = b\]

So we now know that \(c\) is the same as \(b\). We also know that \(a\) is the same as \(2\cdot c\). So we can now convert both \(a\) and \(b\) into their equivalents of \(c\) and stick those back in the very first equation:

\[\frac{2c}{c}=\sqrt{4}\]

And from our arguments last time we also know that is the same as:

\[2\cdot \frac{c}{c}=\sqrt{4}\]
\[2\cdot 1=\sqrt{4}\]
\[2=\sqrt{4}\]

So we have established that two is the square root of four. Which it is. So our assumption this time, that the square root of four can be written as one number divided by another number IS correct. In fact we know that it is always correct as long as the top number is the bottom number multiplied by two.