Don't Borrow Tens Ever Again in a Subtraction Problem!

When the traditional method of subtraction is taught in school, it can often be quite confusing, especially when the number you are subtracting from has lots of zeros. It is a pain to go borrow the tens from other numbers, especially when they are ones or zeros. But, there is an easier way to approach this that isn't taught in class.

In geometry, the word "complimentary" is used to describe the relationship between two angles that sum to 90°. For instance, 36 and 54 are complimentary angles. Similarly, the compliment of 36 would be 54 when talking about geometry.

A geometric application of complimentary angles (and a funny picture)

With arithmetic on the other hand, a number's compliment is its difference from the power of ten above it. So, the compliment of 36 would be 64 (100 - 36 = 64), or the compliment of 473 would be 527 (1000 - 473 = 527). This is more practical for solving arithmetic problems than the distance from 90.

One-hundred or one-thousand are very difficult numbers to subtract from. These are ones where you need to keep turning zeros into nines until you reach that first digit, and then figure out the problem from there. This can be quite the pain. But, there is a shortcut for these examples.

All you have to do is subtract the last digit of the number from ten. This is the last digit of the compliment. Then, subtract all of the rest of the digits from nine, and place them in their corresponding place value. For instance, let's find the compliment of 36.

10 - 6 = 4
9 - 3 = 6

100 - 36 = 64

Pretty easy, right? Let's try it with 473.

10 - 3 = 7
9 - 7 = 2
9 - 4 = 5

1000 - 473 = 527

This can easily be extended to numbers in the thousands, millions, billions, and more as long as you can keep track of the digits. For example, 5647823 would have a compliment of:

10 - 3 = 7
9 - 2 = 7
9 - 8 = 1
9 - 7 = 2
9 - 4 = 5
9 - 6 = 3
9 - 5 = 4

10000000 - 5647823 = 4352177

Couldn't be easier! This method is basically just remembering the fact that no matter what the situation, you will have borrowed a ten from each zero, making them all nines except for the very last one which will remain a ten. Try a few subtraction problems yourself and you will see why it works.

This can be applied to geometry in a way as well. To subtract quickly from ninety, you still subtract the last digit from ten, and the first one from eight (one less than the nine). So, the compliment of 29° is:

10 - 9 = 1
8 - 2 = 6

90° - 29° = 61°

For supplements of angles (the angle's difference from 180), you can do a similar technique as well. Subtract the last digit from ten and the first from seventeen. For instance, the supplement of 48° is:

10 - 8 = 2
17 - 4 = 13

180° - 48° = 132°

In fact, any number ending in zero(s) can be subtracted from just by altering this method. I have found this extremely helpful when performing mental math (three-digit and four-digit squaring requires you to quickly identify how far you are from the nearest hundred/thousand, which often needs compliments). It is also very practical. When you give the cashier a hundred dollar bill, they are usually impressed when you tell them the change before they have time to punch the bill into the register. I'd recommend practicing this technique because it is useful and also quite a bit of fun.

Noam Elkies: Summing the Fourth Powers

A few weekends ago, I went to the MOVES (Mathematics of Various Entertaining Subjects) conference held by the Museum of Mathematics in New York City. I gave a workshop on the Dynamic World of Mathematics, and got to see some very interesting presentations and people.

Meeting Noam Elkies at MOVES

One of the people who I met was Noam Elkies. Noam Elkies is currently a professor at Harvard, and is the youngest full tenured professor in the history of the school.

In 1772, Leonhard Euler conjectured that there was no way to solve the following equation with whole numbers:

a⁴ = b⁴ + c⁴ + d⁴

For over 200 years, mathematicians were trying to prove this conjecture, but to no avail. Finally, in 1986, a twenty-year-old Noam Elkies offered a different type of answer.

In most posts, I will explain how someone proved a conjecture true. But in this instance, Noam Elkies proved the conjecture false. He found values of a, b, c, and d that made the equation true, and thereby showing that the equation is possible.

a = 20,615,673
b = 18,796,760
c = 15,365,649
d = 2,682,440

In addition, he proved that there are an infinite number of solutions to this equation. That completely went against what Euler had thought! This shows that you can never be 100% sure if something is true until you have proof. That philosophy applies to math, science, critical thinking in general, and it is also a great story to show the importance of mathematical proofs.

Ironically, my TEDx talk from this last June just went up a week and a half ago, and I told that exact story in my talk. So, I thought this would be an appropriate post to share the video in.

Time to Double Your Money!

Last year, I did a post about a number called e. This number is around 2.71828, and has many applications to calculus and finance. In the other post, I discussed its application to finding compound interest. Click here to see that post.

You might remember that the formula for finding the amount of money you have when your interest gets compounded continuously is:

Pe^rt

P = amount of money originally deposited
r = interest rate
t = time (years)

This is a prettier formula, but a more practical formula, which also has variable n for the number of times the money was compounded in the year is:

P(1 + r/n)^nt

For instance, if you deposited 1000 dollars in the bank, and your money got compounded every year with a 3% interest rate, after 30 years, you would have:

P = 1000
r = 3% = .03
t = 30
n = 1

P(1 + r/n)^nt
1000(1 + .03/1)¹⁽³⁰⁾
1000(1 + .03)³⁰
1000(1.03)³⁰
1000(2.42726)
2427.26

So, after thirty years, you would have about $2427.26, which means you were able to more than double your money! This might be a little surprising that it is possible for your money to double if you leave it alone for long enough.

You might be wondering how long it will take to double. How many years does it have to sit there? In other words, what value of t makes that equation equal to 2P? We deposited P, so to make it double, we must get 2P.

2P = P(1 + r/n)^nt

In our scenario, we deposited $1000 with an annually compounding interest rate of 3%. So, plug all of this in and we get:

2P = P(1 + r/n)^nt
2(1000) = 1000(1 + .03/1)^1t
2000 = 1000(1.03)^t
2000/1000 = (1000(1.03)^t)/1000
2 = (1.03)^t
ln(2) = ln((1.03)^t)
ln(2) = t • ln(1.03)
ln(2)/ln(1.03) = t
24 ≈ t

Note: this computation required something called logarithms. They look weird, but are very easy to understand. I explained them in my post on Benford's Law. Click here to read it.

So, it will take 24 years to double. So, let's set a rule to this - figure out a simple formula where you can figure out how long it will take for your interest to double, assuming it is compounded annually.

2P = P(1 + r/1)^1t
2P/P = (P(1 + r)^t)/P
2 = (1 + r)^t
ln(2) = ln((1 + r)^t)
ln(2) = t • ln(1 + r)
ln(2)/ln(1 + r) = t

This is as simplified as the equation will get without using calculus. And this doesn't look very simple anyways. But, let's look at a graph of it. It is easier to see as a picture than a messy jumble of logarithms.

This is a logarithmic function. However, it looks very much like a rational function, or a function that is the quotient of two polynomials (for example, 1/x is a rational function). So, let's try to find a rational function that fits this blue curve.

You can play around with it on your graphing calculator if you want, but I will just tell you that the function that fits it best is 72/(100r). Here are the two graphs:

As you can see, the two graphs are practically touching. In fact, all the way up through 0.5, they are very close together, as you can see here:

Since no bank on this planet offers 50% interest rates (if anyone has heard otherwise, please contact me), the 72/(100r) should be a good approximation for any of our purposes. Even to see the difference in the ones and tens, I had to set the graph below two. So, they are very close together.

This 72/(100r) equation can look even better. You might remember that r is currently in decimal form. To turn it into a percentage, you must move the decimal over twice, or multiply it by 100. So, dividing the interest percentage into 72 will give the same approximation.

Let's try it out on the original example. We said that it was a 3% interest rate. 72 ÷ 3 = 24, and we did conclude that it would take about 24 years. For a 6% interest rate, it would take about 72 ÷ 6 = 12 years. For a 4.5% interest rate, it would take about 72 ÷ 4.5 = 16 years. Couldn't be easier!

This rule is normally called the Law of 72. I find it very cool because you don't even need to know the amount of money you deposited to figure out this time. It doesn't matter. You don't even need to convert the interest rate into a decimal. I think this is a really intriguing and practical formula.

Math in the News: Playing with Pentagonal Tilings

This last weekend, I attended MathFest 2013, which is held by the Mathematical Association of America in Hartford, Connecticut. I gave a talk on magic squares, and got to see many other fascinating speakers. Many of my future posts will be content I learned at this convention.

One of the speakers was Frank Morgan, who talked about tiling the plane with different polygons, primarily pentagons. During the presentation, he mentioned that he wrote a news article about it in the Huffington Post. Since the presentation was so interesting, I thought this article would be a perfect "Math in the News" post.

http://www.huffingtonpost.com/frank-morgan/bees-honeycomb-mathematics_b_1318258.html

Multiplication of Fractions Proof

Let me show you a pretty cool proof. We will prove that:

1/2 • 3/4 • 5/6 • 7/8 • 9/10 • ... • 99/100 < 1/10

With most proofs, it is easy to see if the conjecture is correct on face, but the actual proof takes more work. In this example, it is extremely difficult to tell if 1/10 would be less or greater than the product.

Let's start by denoting the left-hand side with A.

A = 1/2 • 3/4 • 5/6 • 7/8 • 9/10 • ... • 99/100

We will also denote the sort of opposite of A with B.

B = 2/3 • 4/5 • 6/7 • 8/9 • 10/11 • ... • 98/99

You will notice that every term in the B product has a corresponding term in the A product. 1/2 and 2/3 are both the first term, 3/4 and 4/5 are both the second term, and so on. You will also notice that every term in the B fraction is greater than every term in the A fraction. So, we can conclude that B is greater than A, or A is less than B.

A < B

Let's multiply both sides of that inequality by A.

A • A < A • B

A² < AB

The product of A and B is something that we can figure out pretty easily. Since the numerators of A are the denominators of B and vice versa, everything will cancel out. All that will be left is 1/100, since 1 and 100 were not in the B product. So, AB = 1/100. Substitute 1/100 in for AB to get:

A² < AB
A² < 1/100

If we solve this inequality for A, we will find that A < 1/10.

A² < 1/100
√(A²) < √(1/100)
A < 1/10

But what is A equal to? At the beginning, we denoted it to be equal to the left-hand side of the hypothesis inequality. Substitute that in for A, and we get:

1/2 • 3/4 • 5/6 • 7/8 • 9/10 • ... • 99/100 < 1/10

And there is our proof. These types of proofs don't really innovate mathematics, but I do think they can be fun.