Math Topics
Karl Freidrich Gauss
Karl Friedrich Gauss was born in Brunswick, Germany, in 1777. Some say he was the greatest German mathematician of the nineteenth century. In mathematics he made contributions in the fields of differential geometry, number theory, and statistics. Besides mathematics he studied physics and astronomy. He also correctly predicted the orbit of the asteroid Ceres.

Portrait of Gauss by S. Bendixen (1828)
Gauss was very brilliant mathematically from an early age. There are various stories that are often told about Karl's early life that demonstrate his early mathematical ability.
One of the most common stories told about Karl's early life concerns a day in elementary school. The teacher asked the students to add together all the numbers from one to one hundred. Why did the teacher ask the students to do this assignment? Some say to keep the students occupied. Nevertheless Karl quickly found the solution to the problem while the rest of the students struggled through the arithmetic.
How did he figure out the problem?
He noticed the following 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101. This pattern continues all the way to 50+51 = 101. So there are 50 pairs of numbers that each add up to 101. Thus 50 times 101 = 5050, which is the answer to the teachers problem.
Another way to see Gauss's shortcut is to write the numbers from 1 to 100 in both ascending order and in descending order. Then add column by column. Again, every sum adds up to 101.
1 |
2 |
3 |
... |
98 |
99 |
100 |
+ 100 |
+ 99 |
+ 98 |
... |
+ 3 |
+ 2 |
+ 1 |
= 101 |
= 101 |
= 101 |
... |
= 101 |
= 101 |
= 101 |
Doing the problem this way, one will end up with 100 copies of the number 101 rather than 50. So if one does the problem this way the answer would be:
(100)(101)/2 = 5050.
Whew ... I hope you are still with me. The reason we took this extra step to explain is because now we can generalize this specific case given to Gauss by his teacher to work for any number n. So if we want to know the sum of numbers from 1 to any number we can use the following formula:
1 + 2 + 3 + ... + n = n(n + 1)/2
For example say we wanted to add the numbers from 1 to 5. The answer from the left hand side of the above identity would be: 1 + 2 + 3 + 4 + 5 = 15. From the right hand side of the above identity the answer would be (5)(6)/2 = 15. So doing it either way will give you 15 for the answer!
Are we having fun yet? I'm getting hungry for some cereal. Didn't we have a problem about cereal?
Looking back at the table we put together for the cereal problem and the formula above: n(n+1)/2 do you see how we can calculate the numbers in the table on the left hand side to get the numbers in the table on the right hand side?
Before we get completely back to that cereal problem there is another way to look at this problem that relates to all that we have been working through so far.
In the year 500 BC the Greeks studied the patterns of dots that were in the shape of geometric figures. For example, if one arranged dots in the shape of equilateral triangles - one would get what are called triangular numbers. To learn more about triangular numbers click below on the words "Triangular Numbers"
[Cereal Box] [Gauss] [Triangular Numbers] [Pascal's Triangle] |