rblog

This story is being blogged upon request.

Today Matthew asked me if there was a number that when added to its neighbor was the same as one-half of that number multiplied by its neighbor. He can understand that we need to solve the equation:
x + x + 1 = x * (x + 1) / 2

This works out to solving
x^2 - 3x - 2 = 0 Note: x^2 is read as "x squared"

Quickly checking that 1, -1, 2, -2 are not solutions the rational roots theorem tells us that there are no rational roots. Applying the quadratic formula, we get that the two answers are:

(3 + squareRoot(17) ) / 2
(3 - squareRoot(17) )/2

Taking a look at the first one (the one that is positive) and recalling that the square root of 17 is 4.123 (easy to remember), we have:
x = (3 + 4.123) = 3.562
so x + 1 = 4.562
Together they are: 8.124
Multiplied and halved they are: 8.125

Matthew was very pleased by this.

Not to be outdone, Emily, who is taking the advanced math track - algebra in 7th grade, got an A+ on her most recent math test and was treated to a proof of the irrationality of the squareRoot(2). This was first discovered by the greeks (Pythagoras and crew) and the existence of numbers that could not be represented as fractions was so terrifying to them that it was a sworn secret. Below is my proof, not Pythagoras' (he would have used a geometric proof - algebra was not known to the Greeks).

Assume that there exist a and b such that a/b = squareRoot(2). Furthermore, assume that a/b is in lowest terms, that is that there are no common factors of a and b. Surely if the square root of 2 can be represented by a fraction then it can be represented by a fraction in lowest terms. We can also assume that a, b are not zero.

So, by assumption we have:
a/b = squareRoot(2)
Multiplying by b....
a = b * squareRoot(2)
Squaring both sides....
a^2 = b^2 * 2 = 2 * b^2

Since a^2 is a multiple of 2, a^2 is even. And since an odd number times itself is odd, if a^2 is even, that means a must be even.

Since a is even, we can write it as 2 * k, since all even numbers are a multiple of 2.

Rewriting we now have:
(2k)^2 = 2 * b^2 or
4k^2 = 2 * b^2
Now dividing both sides by 2...
2k^2 = b^2

Now we have b^2 equal to an even number (as 2 * k^2 is a multiple of 2 and hence even). In order for b^2 to be even, b must be even.

But if a and b are both even, then they have a common divisor - namely 2. But our assumption stated that a and b are in lowest terms and have no common divisors. So our assumption that there was a fraction a/b = squareRoot(2) has led to a contradiction. Thus, there is NO fraction that is equal to squareRoot(2) and thus squareRoot(2) is an irrational number.

This was the first algebraic proof that Emily had seen and her first exposure to the technique of "proof by contradiction" - also known as "Reductio ad absurdum".