View Full Version : irrational numbers

tagnostic
October 4th, 2009, 09:04 PM
Scikidus,

can you break it down for me
I just stumbled on a site trying to explain
why the sqr rt 2 is irrational

I know that it is

I've got the sqr rt -1
that makes sense

but the root of 2 is counterintuitive

sudikics
October 5th, 2009, 03:30 AM
And of course, right after I leave this nice PM in your account, I see this. Ah well, here goes nothing.

A rational number can by definition be expressed as an integer divided by another integer. When that fraction is reduced, then the two integers have no common factors left.

So let's say that sqrt(2) is rational for the moment, and that it equals a/b. a and b are reduced so that they share no common factors.

Now square both sides. You get

2 = a^2 / b^2.

Multiply bother sides by b^2, and you get

2 * b^2 = a^2.

Now any integer squared is an integer, and any integer times an even number is an even number. Therefore, since b^2 is an integer which is then multiplied by an even number, we know that 2*b^2 must be an even integer, and therefore a^2 is an even integer as well.

So a^2 is even. What does that make a? Well, an even*even is even, while an odd*odd is odd. Therefore, a must be even as well.

If a is even, then we know that one of the factors of a is 2. And so we create a new variable k such that a = 2*k. We know k is a factor of a as well as 2, and is therefore an integer.

Let's go back to 2 * b^2 = a^2. Substitute in 2k for a and you get:

2*b^2 = (2k)^2 = 4*k^2.

Divide bother sides by 2 and you get:

b^2 = 2*k^2

Does this situation look familiar? Through the same steps as above, we deduce that b must be even as well, and therefore have 2 as a factor.

But we already said that a has 2 as a factor, and that a and b shared no factors! We have reached a contradiction in our logic, and therefore must work backwards to see what assumptions we have made that led to this conclusion. We have in fact only made one assumption, that sqrt(2) is rational. Therefore, this assumption must be false, and the sqrt(2) cannot be rational.

Therefore, sqrt(2) is irrational. QED.

And with that I bid you good night.

tagnostic
October 5th, 2009, 05:26 AM
thanks
l8r
my brother