The Power of a Mathematical Picture

The Power of Mathematical Visualization

If a picture is worth a thousand words, in mathematics, a picture can spawn a thousand ideas. A picture can provide deep understanding, prompt an idea or process to suddenly make sense, and lead the way to finally understanding a tricky piece of mathematics.

The focus of this article is on sums. You will learn how to quickly add all the numbers up to 1000 and back down, learn about sums of odd numbers and of even numbers, and even establish Galileo’s results on ratios of sums of numbers—all through the use of a single picture.

Basic Arithmetic 

We can circle groups of dots in pictures to make sense of division. For example, the division problem 18 ÷ 3 is asking the following question: How many groups of 3 can you find in a picture of 18 dots? There are 6 of them, so 18 ÷ 3 = 6.

We can push this visual picture further and make sense of some complicated division problems. For example, what is 808 ÷ 98? We can see that the answer has to be 8 with a remainder of 24.

You can imagine looking for groups of 100, rather than 98. (The number is 98 is too difficult.) If we visualize this, we see that there will be 8 of these groups, with 8 dots left over.

But each group of 100 is itself off by 2 dots—we wanted groups of 98—so we have an extra 16 dots floating around. That makes for 8 groups of 98 and a remainder of 16 and 8, which is equal to 24 dots. Therefore, 808 ÷ 98 = 8 with a remainder of 24.

When asked to do 34 − 18, we can certainly do the traditional algorithm and get the answer, 16.

But can’t we just see in our minds that the answer has to be 2 + 10 + 4, which is 16? Line up a row of 34 blocks and a row of 18 blocks side by side.

Now we can see that the 2 rows differ by 2 and 10 and 4 blocks, so the difference is 16.

In the same way, 1012 − 797 has to be 3 and 200 and 12—which is 215. From 797 to 800 is 3, from 800 to 1000 is 200, and there is an extra 12, for a total of 215.

This flexibility of thought helps with subtraction in general. For example, consider 1005 − 387.


We have a lot of borrowing to do if we follow the traditional approach: 5 − 7, 0 − 8, and 0 − 3 all need borrows.
But we can make this work simpler.


We are looking for the difference between 1005 dots and 387 dots. Let’s make 1005 friendlier and turn it into 1000. Remove 5 from each and just compute the difference between 1000 and 382 instead. Now we can see the answer: 8 + 10 + 600, or 618.

But if we still want to do the traditional algorithm, then we can remove 1 more dot from each pile and make the problem 999 − 381.

Now we can do the algorithm without any borrows: 9 − 1, 9 − 8, and 9 − 3. This way, we’ve made the problem much easier to do, even if someone insists that we use the algorithm.

Isn’t multiplication really a geometry problem? Isn’t 24 × 13, for example, just asking for the area of a rectangle that is 24 units wide and 13 units high?

Then why not just chop up the rectangle into pieces that are manageable? For example, think of 24 as 20 and 4, and 13 as 10 and 3.

Then we see that 24 × 13 must be the areas of the individual pieces added together: 200 + 40 + 60 + 12 = 312.

Grids

In a 5-by-5 grid of squares, there are 25 small 1×1 squares within the grid. But we can count 2×2 squares as well. There are 16 of these in total. If we count the 3×3 squares, there turns out to be 9 of those. And there are 4 of the 4×4 squares. Finally, there is 1 large 5×5 square.

So, there are 25 1×1 squares, 16 2×2 squares, 9 3×3 squares, 4 4×4 squares, and 1 5×5 square. Each count of squares is itself a square number!

Why does counting squares on a square grid give square-number answers? Let’s focus on the lower-left corners of the squares we’re counting. For example, of the 2×2 squares, the following are some possible lower-left corners can be seen in figure 1.14.

Let’s draw all of the possible lower-left corners. Now we see that there is a square array of them, 4 × 4 of them, which is 16. Thus, there are 16 2×2 squares.

Let’s view the 5‑by‑5 grid as an array of dots as in figure 1.16. This is certainly a picture of 25 dots, but can you see in this picture the sum 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 +1?

Look at the diagonals: 1, 2, 3, 4, 5, 4, 3, 2, 1.

The sum we seek matches the diagonals of the square. There are 25 dots in all, so without doing any arithmetic, we can say that the value of the sum must be 25.

What is the sum of all the numbers 1 + 2 + 3 + … up to 10 and back down again?

This sum must come from the diagonals of a 10‑by‑10 array of dots. Again, without any arithmetic, the value of the sum must be 10 squared (102): 100.

What is the sum of all the numbers from 1 to 1000 and back down again? It must be 1000 squared, from a 1000-by-1000 array of dots. That’s 1 million.

If you were to compute this on a calculator—1 + 2 + 3 + …—it would take forever. But the answer is available to us quickly via this picture.

1 + 2 + 3 + … + 998 + 999 + 1000 + 999 + 998 + … + 3 + 2 + 1 = 1000 × 1000 = 1,000,000

The Sum of Numbers

There is a general formula for the sum of numbers.

1 + 2 + 3 + … + n = n2 + n ÷ 2

The sum of the first n numbers, 1 + 2 + 3 all the way up to some number n, is (n2 + n) ÷ 2. For example, the sum of the first 5 numbers, 1 + 2 + 3 + 4 + 5, is 52 + 5 = 25 + 5 = 30, and 30 ÷ 2 = 15. And we can check that 1 + 2 + 3 + 4 + 5 is indeed 15.

Where does this formula come from, and why is it true?

Our 5×5 array of dots gave us something akin to this result. We have that 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25. Can we get from this answer to just 1 + 2 + 3 + 4 + 5?

If we look at what we have, we see that the sum we want, 1 + 2 + 3 + 4 + 5, is the left half of the equation.

1 + 2 + 3 + … + n = n2 + n ÷ 2

1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25

Actually, half is not quite right. The right portion of the equation is missing a 5. It’s just the sum 1 + 2 + 3 + 4. We want to see an additional 5, so let’s add a 5 on the left—and to keep things balanced, we need to add a 5 to the right as well.

1 + 2 + 3 + 4 + 5 + 5 + 4 + 3 + 2 + 1 = 25 + 5

Now we see 2 copies of what we want. Twice the sum we seek is 25 + 5. So, this means that the sum itself is half of this. 1 + 2 + 3 + 4 + 5 is indeed (52 + 5) ÷ 2. And this matches the general formula. There is nothing special about the number 5. The same ideas show that the sum of the first n counting numbers must be half of n2 + n.

2 × (1 + 2 + 3 + 4 + 5) = 25 + 5

1 + 2 + 3 + 4 + 5 = 25 + 5  ÷ 2

The Sum of Odd Number

Look at the 5-by-5 grid of dots again. Do you see the sum 1 + 3 + 5 + 7 + 9, the sum of the first 5 odd numbers?

We can certainly circle these groups randomly and make them fit.

But such a random picture isn’t enlightening. We want to see a picture that isn’t locked into this particular example of 25 dots. We want a picture that speaks to a higher truth and clearly holds for all possible square arrays. Mathematicians are always on the lookout for this sort of thing, and symmetry is often a pointer to higher truths.

Do you see 1 + 3 + 5 + 7 + 9 in the 5-by-5 array of dots in a way that speaks to a higher truth? Think L shapes.

The sum of the first 5 odd numbers is hidden in the 5-by-5 array as Ls. The sum 1 + 3 + 5 + 7 + 9 must be 52, or 25.

In the same way, the sum of the first 10 odd numbers, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19, sit in a 10-by-10 array of dots and therefore must have an answer of 100, the count of dots in that array.

In general, the sum of the first n odd numbers must be n2.

Galileo lived at the turn of the 16th century and is revered today for his work in science and mathematics, thought to make fractions out of the odd numbers. For example, take the first 5 odd numbers and use their sum for the numerator of a fraction and the sum of the next 5 odd numbers for its denominator. This gives a fraction that simplifies to 1/3.

Do the same for the first 2 odd numbers, followed by the next 2. You get 1/3 again.

Do it again for the first 10 odd numbers, and the next 10. It’s 1/3 again!

Galileo observed that all the fractions made out of the odd numbers this way are equal. They all equal 1/3. These fractions are today called the Galilean ratios. There is a connection between the ratios and the L shapes in squares. Figure 1.23 is purely visual proof of the Galilean ratios.

The first 5 L shapes, the sum of the first 5 odd numbers, makes 1 block of 25 dots. The next 5 L shapes for the next 5 odd numbers makes 3 blocks of 25 dots. So, the first 5 odd numbers make for 1/3 of the next 5 odd numbers.

The Sum of Even Numbers

Are there results about sums of even numbers, too? For example, we have a picture for the sum of the first 5 odd numbers. Can we get from this a picture of the first 5 even numbers, 2 + 4 + 6 + 8 + 10?

Just add a dot to each L shape!

This has turned the 5-by-5 square into a rectangle. The sum of the first 5 even numbers must be the 5×5 we had before plus 5 more, 52 + 5, which is 30.

In general, the sum of the first n even numbers must come from the picture of n2 dots plus an extra n dots: n2 + n.

We’re coming full circle, because we have seen the expression n2 + n before.

Take the sum of the first 5 even numbers. It equals 52 + 5.

Now divide everything by 2: 2 ÷ 2, 4 ÷ 2, 6 ÷ 2, 8 ÷ 2, 10 ÷ 2, and (52 + 5) ÷ 2.

And we’re back to the formula 1 + 2 + 3 + 4 + 5 = (52 + 5) ÷ 2.

We have come full circle. We’re back to the general formula for the sum of numbers.

Problems

  1. a. What is the sum of the first 1000 counting numbers?

b. What is the sum of the first 1000 odd numbers? (What is the thousandth odd number?)

c. What is the sum of the first 1000 even numbers?

2 Draw a picture to show that the sum of the first 3 odd numbers must be 1/8 the sum of the next 6 odd numbers.

Solutions

  1. a.

b. The one-thousandth odd number is 1999 and the sum of the first 1000 odd numbers: 1 + 3 + 5 + … + 1999, is 10002 = 1,000,000.

c. The sum of the first 1000 even numbers: 2 + 4 + 6 + … + 2000, is 10002 + 1000 = 1,001,000. (Divide this by 2 and get back to the sum of the first 1000 counting numbers!)

  1.  (See FIGURE 1.27.)   In general, we have:

 From the lecture series The Power of Mathematical Visualization.
Taught by Professor James Tanton, Ph.D.