The Funniest Proof I’ve Seen

Although last time I said that I would delve deeper into thoughts about my last post, I can’t help but postpone it for my post after this one to make way for this. It’s a proof that my engineering teacher shared with my class.

Thesis:
Engineers and scientists will never make as much money as business executives. Now a rigorous mathematical proof that explains why this is true.

Postulate 1:
Knowledge is Power. (can’t argue against that)

Postulate 2:
Time is Money. (very true in life)

Proof:
As every engineer knows,

    Work
   ——— = Power
     Time

since Knowledge = Power, and Time = Money, we have 

Work
   ——— = Knowledge
     Money

solving for Money, we get

     Work
   ——— = Money
   Knowledge

Thus, as Knowledge approaches zero, Money approaches infinity regardless of the Work done.

Conclusion:                                                                                                                                         The less you know, the more money you make. ( Therefore implying that businessmen are more stupid than engineers, and thus they make much more money than engineers.)

Perhaps it’s not as funny as it seems, but when my engineering teacher was reading it out loud while showing us, I was laughing like crazy. I especially like the pun on the word “power”, where power can mean the power the US president holds or power in scientific terms. Either way, this is a one-of-a-kind proof. It’s like a math and science joke, except this time there is actual science and math. Most math and science jokes are like “What do you get if you add two apples with three apples? A high school math problem!” There’s no actual math and science involved. In this case, however, there is. The substitution and the workings of the equations are all math, and the power=work/time equation is science.

If you get across any jokes like these, please send them to me! Thanks.

A Sierpinski Triangle Problem

Mathematics is a strange and mysterious thing. It is as if it is almost magical, able to do things that our physical beings cannot percieve. Perhaps a very good example of this is a problem that I encountered just yesterday, having to do with the Sierpinski Triangle.

First of all, you might ask, what in the world is a Sierpinski triangle? Well, it is a fractal, in which it creates a self-similar pattern within itself. Perhaps a picture is better-

As you can see, the pattern here is just forming triangles inside every dark triangle on and on and on. In fact, this pattern can go on for infinity, as it will in the problem I will be showcasing below:

The segments joining the midpoints of the sides of an equilateral triangle are drawn, and the interior of the triangle they form is removed from the interior of the original triangle. The segments connecting the midpoint of the remaining triangles are joined, and the interiors of the triangles they form are removed from the interior of the original triangle. If this process is repeated forever, how much of the orginial interior will be left?

Let’s take a look back at the picture I showed you above. Suppose the black spaces were the interior of the triangle. If we were to remove the triangles, they would then leave white triangles behind. In other words, the black spaces are the interior, and the white spaces are the places where triangles have been removed. Now let’s go back to the problem. What are we trying to find? Well, we are trying to find out how much of the original interior (aka how much of the first triangle in the picture above) is left. So how do we approach this?

Perhaps we can find the total amount of area that was removed from the original triangle, or in other words, the total area of white space. We would then get that and subtract it from the area of the original triangle. This difference (aka the amount of black space left after all the removing) would then be divided by the area of the original triangle to get a fraction of how much is left of the original triangle. That’s our process.

Let’s begin. First thing- how do we calculate the total amount of space removed? In other words, how do we calculate the total amount of white space? Well,  let’s take this as an infinite geometric sequence. In our first cut, ¼x is removed, where x is the original triangle’s area. This is indicated by the second triangle in the picture above.  In our second cut, 3 (1/16) x is removed, as indicated by the third triangle above. In our third cut, 9 (1/64) x is removed, indicated by the fourth triangle. This pattern then continues on infinitely. Again, it is a sequence- (1/4)x, 3(1/16)x, 9(1/64)x,…… So in order to find the total amount of white space, we have to add all the terms here up. You might ask, well this is infinite, so how in the world do we add this up? Note that the first term is (1/4)x and the ratio is 3/4. We now use the formula for an infinite geometric series:

where a₁ is the first term, and r is the common ratio. Both of these we know. So let’s substitute. S= [(1/4)x] / [(1- 3/4)] = [(1/4)x] / [1/4] = x. Thus, a total of x is removed from the triangle.

But wait, isn’t x the area of the original triangle, too? This means that if a total of x is removed, then in essence, the whole triangle has disappeared. Thus, the answer is that zero percent of the original interior is left. Everything has disappeared. Now picture this in your head. You are given a paper triangle. You are to cut out a triangle whose edges are the midpoints of the original triangle. You do this for every triangle that you see after you cut every time. And not only that, this is repeated infinitely. If so, then in the end, you get nothing.

To me, that’s really hard to imagine. How in the world do you get nothing if you keep on cutting a paper? It’s crazy and bizarre. Yet, at the same time, this is what makes math beautiful. And in this case, seemingly magical.

I Took The AMC!

Recently, I have just been talking about love, love, and love. And I am boy gettin tired of it. So for today, I will switch my view to mathematics. Mostly because yesterday, I took the AMC 10 B, which I’d like to say I did poor on it. The beginning half I felt good, but the later half, there were some I left blank and some I just assumed it should be right. Man, the AMC is harder than it seems!

Here’s one question (no. 14) from that test I took:

Define . Which of the following describes the set of points (x,y) for which .

Here was my thought process, which eventually turned out to be wrong:

, so this also equals ab (a-b). Similarly, for , it would be                       xy(x-y) = yx (y-x) thus x-y = y-x  thus 2x = 2y and thus x=y. That would satisfy answer B.

The actual way to solve it, unfortunately to my dismay:

 and . Therefore, we have the equation  Factoring out a  gives  Factoring both sides further, . It follows that if , , or , both sides of the equation equal 0. By this, there are 3 lines (, , or ) so the answer is (E) three lines.

Building Where I Took My AMC

Well, look’s like this question I missed. There were also a few other problems in which I missed due to miscalculations. For instance, one question I saw 5-3 as 5+3 and put my answer as 8, when it was supposed to be 2. Next time I take the AMC, I better be careful of these silly mistakes.

Updated 9:40 pm: By the way, refer to my previous mathematics posts to know more about the AMC.

A More Difficult AMC Problem

In my January 31 post, I gave you an example of an American Mathematics Competition question. However, as I said in that post, that problem was just medium level. The AMC problem that I am about to show you now is a much harder problem. Its difficulty level would be an 8 out of 10, maybe.

The sum of the first  positive odd integers is 212 more than the sum of the first  positive even integers. What is the sum of all possible values of ?

Given an unlimited amount of time, anybody could solve this problem, whether through the hard way or easy way. However, one must remember that the AMC is very time consuming, and thus one must solve this in a quick way. In order for that to happen, one must have a few number theory concepts in mind, just simple ones. Or one must find that pattern by himself.

So solve it. After you do that, here’s the two ways you could have answered this question.

SOLUTION ONE: The sum of the first  odd integers is given by . The sum of the first  even integers is given by . These are two concepts you needed to solve the question easily.

Thus, . Since we want to solve for n, rearrange as a quadratic equation: . Use the quadratic formula: .  is clearly an integer, so  must be not only a perfect square, but also an odd perfect square. This is because the entire expression must be an integer, and for the numerator to be even (divisible by 2),  must be odd. Let  = . (Note that this means that .) This can be rewritten as  , which can then be rewritten to . Factor the left side by using the difference of squares. . Our goal is to find possible values for , then use the equation above to find . The difference between the factors is  We have three pairs of factors,  and . The differences between these factors are , , and  – those are all possible values for . Thus the possibilities for  are , , and . Now plug in these values into the equation .  can equal , , or . Add . The answer is .

SOLUTION TWO:  As above, start off by noting that the sum of the first  odd integers  and the sum of the first  even integers . Clearly , so let , where  is some positive integer. We have: . Expanding, grouping like terms and factoring, we get: . We know that  and  are both positive integers, so we need only check values of  from  to  (). Plugging in, the only values of  that give integral solutions are  and . These gives  values of  and , respectively. . Hence, the answer is .

I myself wasn’t able to solve this, because I didn’t know anything about the sum of the first m odd integers being represented by m^2. Or the one about the first n even integers equaling n(n+1). This question showed me that the AMC is not just about smartness, but also a vast amount of knowledge or at least a genius skill of pattern recognition.

An American Mathematics Competition Question

Recently, I have just registered for the American Mathematics Competition, or AMC for short. The AMC is split into two types- AMC 12, which requires pre-calculus and trig, and AMC 10, which tests regular 9th grade and 10th grade math. Yet, this test is anything but regular. In fact very challenging. (I registered for AMC 10).

I was just practicing this test, and I got really frustrated. This test is 25 multiple-choice under 75 minutes. You think that’s enough time? It’s anything but that; it’s definitely time-consuming. The beginning problems are plain simple, but as the test progresses more and more, so does the more and more difficult the tests become. For example, the first problem took me just 1 minute. Later, another problem took me 15 minutes. Anyway, here’s one problem you might want to try out. This is perhaps a medium-level problem:

Let a and b be relatively prime integers with a>b>zero and (a³-b³)/(a-b)³ = 73/3. Find a-b.

Choices: (a) 1     (b)2      (c)3     (d)4       (e)5

So, try out this problem by yourself. Once you’re done, here’s the solution:

Factor a^3 – b^3.

a^3 – b^3 = (a-b)(a^2 + ab + b^2)

……..

Thus, (a^3 – b^3)/((a-b)^3) = (a^2 + ab + b^2)/[(a-b)^2]

……..

Cross-multiply.

……..

73(a-b)^2 = 3(a^2 + ab + b^2)

73(a^2 – 2ab + b^2) = 3a^2 + 3ab + 3b^2

73a^2 – 146ab + 73b^2 = 3a^2 + 3ab + 3b^2

………

0 = 70a^2 – 149ab + 70b^2

0 = 70a^2 – 100ab – 49ab + 70b^2

0 = 10a(7a – 10b) – 7b(7a – 10b)

0 = (10a – 7b)(7a – 10b)

………

a = (7/10)*b OR a = (10/7)*b

………

Since a > b, a cannot be (7/10)*b.

………

Therefore, a = 10b/7, or a/b = 10/7.

Since a and b are relatively prime, a = 10, and b = 7. Thus, (a – b) = (10 – 7) = 3. Thus (c).

Japanese Multiplication

I was surfing around WordPress.com blogs, and I found a pretty interesting post on quipsandquarks.wordpress.com. Usually we think of multiplying as no big deal and just a boring function. In Japan, they do it in a somewhat exotic way that intrigued me. Here is a video from that post.

Pretty cool isn’t it? Like I said before, it seems to be a weird way of multiplying numbers. But if you think about it, it does make a lot of sense. Let’s say I’m multiplying 2 by 2. From basic math, we know that this can be visually represented as a two-by-two square.

Now suppose we are multiplying 11 by 2. From basic math, we get the two and times it by the ones digit, and we get a two for the products one digit. We get the two again and times it by the tens digit, and we get another two for the product’s tens digit. Thus, 22. However, this is visually represented by having two 2-by-1 rectangles, where one rectangle represents the tens digit and the other one represents the ones  digit. This is pretty much the basic concept of Japanese multiplication.

The second example in the video- 123 x 321- is pretty much the same thing, except this time you group two or more “intersections” or in the terms of my explanations, two or more “squares” and count that as one digit. This could be visually representative of the fact that when you multiply a multiple-digit number by a multiple-digit number, you will have to add. (Right? Hope you still remember your basic multiplication.)

Anyhow, this was something I found interesting. If you want me to clarify my explanation a little bit more, than just contact me. For tomorrow, I will focus on a political cartoon. (I love those!)

A Mathematical Beauty of Hyperbolas

This morning in my Algebra 2 Honors class, my teacher decided to show my class and I some extra tid-bits. These tid-bits were perhaps one of the most beautiful mathematical concepts I ever saw. We were learning about analytic geometry and hyperbolas at that time. For those of you who don’t know what a hyperbola is, check this graph below:

To sum it up in general terms, a hyperbola is pretty much any graph that looks like the one above- the red curves only-(vertical or horizontal). However, there is one property of the hyperbolas that I will focus on for this post: notice that besides red curves, there are black lines also. These black lines are asymptotes, in which the graph of the hyperbola- the red curves- approach them but never ever touch them.

Visually, yes, we understand this property. However, what my math teacher showed me this morning was how to explain this in mathematical terms, specifically in limits. Isn’t that almost magical? The fact that you can explain a visual quality in numbers and equations?

Anyway, let us first take take the equation of this hyperbola in the graph above.

 x²/16 – y²/9 = 1

Notice that in order to illustrate this relationship in which the graph approaches but never touches the asymptotes, y has to be near positive/negative infinity. So let us get the y-coordinates of this equation. To do so, simply solve y in terms of x using algebra. This will result in-

                                         y=±¾ √(x²-16)

   [which equals by factoring out an x² in the radical..] y=±¾ √[x²(1 –  16/x²)]

   [which equals by taking the x^2 out of the radical..] y=±¾x √(1 –  16/x²)

 Notice again that for this relationship to occur not only does y have to near pos/neg infinity, so does x. So suppose x is approaching negative or positive infinity. Or in math, x→±∞. Let us using this value plug x into the equation that we have now. This will cause 16/x² to be approximately equal to 0, because 16, a small number, out of infinity, a super large number, is a really small fraction that has to be very very close to zero. In math terms, 16/x²→0.

Now, let us get this value for 16/x^2 and plug it back into the equation. This will cause the terms under the radical to go like this:

Since 16/x²→0, (1 –  16/x²)→ (1-0) → 1

Substitute this (1 –  16/x²)→ 1 into the equation y=±¾x √(1 –  16/x²) and get-

y→±¾x √(1 –  16/x²)→ ±¾x √(1)→ ±¾x

given x→±∞ ; y→±¾x

Tada! Let’s look at this limit equation we got. First of all, refer back to the graph. Notice that the ±¾x  is in fact the equations of the asymptote lines. So what this limit equation is basically saying is that as x approaches negative/positive infinity, y or the function will approach  ±¾x aka the asymptotes. In other words, we have just mathematically illustrated the asymptotic relationship in which the hyperbola will near but never touch the asymptotes. Not just that, we in a sense derived it from a hyperbolic equation; thus we proved that in any hyperbola, the graph will always have this relationship. Isn’t that beautiful?

Perhaps the way I’m showing it to you on the computer is not as beautiful as it really is. (It’s actually pretty hard to write these equations on a computer. I couldn’t even use Words for this.) I would agree that showing this on paper and actually writing this out makes this derivation seem so much more beautiful.

But what makes this beautiful, computer or not? I dunno really, but I think it is the fact that it is so simple and so elegant. Just a small number of steps and you can explain a visual property in mathematical terms. I don’t know about you, but whenever I read through this process, there is a thrill down my spine. It is just so mathematically beautiful.

Proving (-1)(-1) = 1

Hello guys. For today I have decided to do a little bit of math. We all know that a negative times a negative equals a positive. Yet few of us rarely know why this is so.

When I was in seventh grade, I didn’t know either. I just knew this so I could get good math grades. Until one day, one of my genius friends named Will Chen showed me how. Below is a proof of (-1)(-1) = 1, which is essentially the same as showing that a neg x neg = positive.

-1=-1    reflexive property

-1(0) =-1(0) multiplication property

-1(0) = -1(1-1)  substitution of 1-1 for 0

0= -1(1) + (-1)(-1) substitution and distributive prop

 1(1) = (-1)(-1) addition property

1 = (-1)(-1) simplification

Well, that’s all for today! Hope your winter vacations were good!