Why is the Distributive Law true?

Remember back in sixth grade or so, when the teacher said, "Okay, here's the deal. a(b + c) = ab + ac." Maybe not. However, this is called the Distributive Law, or the Distributive Property. My class had to do a whole project on it, so it really got jammed into our heads. However, in the midst of this big project, I always was thinking in the back of my head "Why is this true?" And since the proof is so incredibly simple, I had to share it.

First off, let me describe the Distributive Property in further detail. I've used it in the past without thoroughly going over it, so that must get done now. Take this problem:

8(5 + 3) =

We could do it this way:

8(5 + 3)
8(8)
64

Or, we could use the distributive property. This means we distribute the 8 to the 5 and the 3 to get the answer.

8(5 + 3)
8(5) + 8(3)
40 + 24
64

This doesn't seem useful right now, but let's try it with variables and you'll see. Try simplifying the following.

3(x + 4)

We can't combine the x and 4 because they are not "like terms." We can't add apples and oranges, so we can't add x and 4. However, we can use the distributive property.

3(x + 4)
3(x) + 3(4)
3x + 12

And there you go. You may be wondering what I was at that moment; why does it work? I'll bet you can prove it algebraically, geometrically, and every other way, but we're going to prove this logically. Let's take a story problem:

I have eight bags of fruit, each with five apples and three pears inside. How much fruit do I have in all?

One way we could do it is figure out how much fruit is in one bag. To do it, we would add the five apples to the three pears to get eight pieces of fruit. By multiplying by the eight bags, we get sixty-four.

We could also total up the number of apples and the number of pears. Eight bags times five apples is forty apples, and three pears times the eight bags is twenty-four pears. 40 + 24 = 64, which is the same answer as before.

If you think about it, this second method is the exact same thing as the distributive property; we multiplied each term in the parentheses by the number of bags, which is known as the "factor," then added those two totals together. It makes perfect sense, it just takes that little story problem to understand it. This property is so important in Algebra, from FOIL to combining like terms, from radicals to point-slope form. It is endless what you can do with the Distributive Property. Sometime leading up to the problem of the weeks, I will post about something else you can use this property for, which will most likely come in handy in the problems!

For the majority of February, we will be putting together mathematics and technology, and using a calculator to graph equations, scatter plots, and even play, what my friends call "games" on a calculator. Except they throw coins and dice, not live birds that have explosives inside of them...

Fibonacci Day: Proof of Binet's Formula

I don’t know if you noticed, but today’s a Fibonacci day! It is January 21, and 21 is the eighth Fibonacci number.

Several months back, I presented the explicit formula for Fibonacci numbers; F_n = (a^n –  b^n)/(a – b), assuming the following:

a = 1 + √(5)/2

b = 1 – √(5)/2

How on earth does that work? We’re going to use a technique called “proof by induction” to prove it. This is a difficult technique, but if you really try to follow each step, it will all come together in the end.

Fibonacci Identity	Fn+1 = Fn + Fn-1
Original Formula	Fn+1 = (a^n – b^n)/(a – b) + (a^n-1 – b^n-1)/(a – b)
Add like denominators	Fn+1 = (a^n – b^n + a^n-1 – b^n-1)/(a – b)
Communative Property	Fn+1 = (a^n + a^n-1 + b^n – b^n-1)/(a – b)
Distributive Property	Fn+1 = (a^n-1(a + 1) – b^n-1(b + 1))/(a – b)
Substitute for a and b	Fn+1 = (a^n-1((1 + √(5))/2) + 2/2) – b^n-1((1 - √(5))/2 + 2/2))/(a – b)
Add like denominators	Fn+1 = (a^n-1((3 + √(5))/2) – b^n-1((3 – √(5))/2))/(a – b)
Complicate Fractions	Fn+1 = (a^n-1((6 + 2√(5))/4) – b^n-1((6 – 2√(5))/4))/(a – b)
Factor	Fn+1 = (a^n-1(((1 + √(5))/2)^2) – b^n-1(((1 – √(5))/2)^2)/(a – b)
Substitution	Fn+1 = (a^n-1(a^2) – b^n-1(b^2))/(a – b)
Add Exponents	Fn+1 = (a^n+1 – b^n+1)/(a – b)

  Q.E.D

And there’s the proof. It is very confusing, but if you read it through a couple of times, it will begin to sink in. When I see proofs like this, it is massive chaos in the middle, but once it all comes together in the end, it just makes it such a cool proof. 

Can you correctly add six numbers?

Since lately, the posts have been a little too complicated for some people, we are going to take it down a notch and do some simple addition. But it's going to be cool.

Okay, just follow along, and add up the numbers in your head. No calculators, or paper, just your brain. They won't be hard. Take 1000 and add 20. Now, add 1030. Got it? Add 1000. Now, add another 1030. And add 20. Did you get 5000? Congratulations!

What is so cool about this? The cool thing is that the answer isn't 5000, it's 4100. It's a problem that takes advantage of the fact that people hear the numbers, don't see the numbers, when they do this problem. Look at the pronunciation of the numbers:

one-thousand and twenty
two-thousand and fifty
three-thousand and fifty
four-thousand and eighty

What do you want to say next? Five-thousand, right! This is because you've said one-thousand, two-thousand, three-thousand, four-thousand. You naturally want to say five-thousand, and this urge is too strong to actually do it correctly.

Bonus: Since that post was a little shorter than the others, I will show you one more cool-ish thing.

Take the last two digits of the year you were born in
Add your age (as of December 31, 2011)
You are thinking of either 11 or probably 111

I was shocked when I heard someone saying how that was so cool, so I thought I may as well post it. But think about it, isn't the definition of age the current year minus the year you were born on? If you add the year of your birthday to your age, shouldn't you get the current year, or the year before it? I thought it wasn't that interesting, but you tell me if you thought it was cool.

How did they find the Quadratic Formula?

A few weeks ago, we were looking at quadratic equations, equations with an x^2 term in it. We learned how to complete the square, which is translating from standard form to vertex form.

Another cool thing I never mentioned that is cool about vertex form is that you can calculate x when the equation is in vertex form. For (x + 3)^2 – 1 = 0, you can say that x = -2 or -4, just by simply manipulating variables.

(x + 3)^2 – 1 = 0

(x + 3)^2 = 1                 (add one to both sides)

x + 3 = 1 or -1               (take the ± square root of both sides; use 1 and -1 because 1
                                      and -1 squared are both 1)

x = -2 or -4                    (subtract three from both sides, from the 1 and the -1 to get
                                      both answers)

What about for standard form, using the coefficients a, b, and c. Maybe there is a formula we can derive just by following the steps involved in completing the square. Remember our original equation: ax² + bx + c = 0. We’ll do it alongside an example utilizing every step, 2x² – 4x + 6 = 0.

Starting Equation	ax2 + bx + c = 0	2x2 – 4x – 6 = 0
Move the constant over to the right hand side, just to make things less messy.	ax2 + bx = -c	2x2 – 4x = 6
Divide through by a. Normally, we would factor it, but it is the same thing.	x2 + b/ax = -c/a	x2 – 2x = 3
Take half of the new b term and square it.	(b/2a)2 = b^2/4a^2	(-1)2 = 1
Add in and subtract out this number. For this, we’ll just add it to both sides, since c is on the other side.	x2 + b/ax + b^2/4a^2 = -c/a + b^2/4a^2	x2 – 2x + 1 = 3 + 1
Simplify the right hand side. For the first example, we will multiply the first fraction by 4a/4a to get a common denominator.	4a/4a(-c/a) + b^2/4a^2 -4ac/4a^2 + b^2/4a^2 b^2 – 4ac/4a^2 x2 + b/ax + b^2/4a^2 = b^2 – 4ac/4a^2	x2 – 2x + 1 = 4
Convert the left hand side into squared form.	(x + b/2a)2 = b^2 – 4ac/4a^2	(x – 1)2 = 4
Square root both sides. Don't forget to put our ± sign before the square root!	x + b/2a = ±√(b^2 – 4ac)/2a	x – 1 = ±2
Isolate the x by adding/subtracting the remaining constant.	x = -b/2a ± √(b^2 – 4ac)/2a x = -b ± √(b^2 – 4ac)/2a	x = 1 ± 2 x = -1 or 3

And we have a formula; x = -b ± √(b^2 – 4ac)/2a. This is known as the quadratic formula, the formula that takes ANY quadratic equation and generates its x-intercepts. Not only that, but it doesn't generate false intercepts, and it is very clear when there are no intercepts.

This formula is pretty well-known, but I think it's cool how it is derived. This is a formula that you want to memorize, because it is so important.