Triangular NumbersDate: 07/07/98 at 08:09:35 From: frances Subject: Triangular numbers Dear Dr. Maths, How do you know a number is a triangular number or not, and how is n/2(n+1) derived - i.e. how did they get it? Frances Date: 07/07/98 at 09:41:36 From: Doctor Rob Subject: Re: Triangular numbers A number N is a triangular number if 8*N + 1 is a perfect square. Example: 45 is a triangular number because 8*45 + 1 = 361 = 19^2 is a perfect square. 88 is not a triangular number because 8*88 + 1 = 705 = 3*5*47 is not a perfect square. Draw a triangle of size n. Draw another next to it, upside down. Together they form a rectangle n by n+1. Here is an example of a size 6 triangle, two of which form a 6-by-7 rectangle: o oooooo oo ooooo ooo oooo oooo ooo ooooo oo oooooo o - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 07/08/98 at 03:41:18 From: Frances Subject: Use of triangular numbers Dear Dr Maths, I am presently writing an oral report on triangular numbers. I need to know the purpose of these numbers - i.e. what are they used for? Also, where does the 8 in '8*N + 1' when finding whether N is a triangular number come from? Date: 07/08/98 at 10:47:51 From: Doctor Rob Subject: Re: Use of triangular numbers If a triangular number has the form N = n*(n+1)/2, then 2*N = n^2 + n. Now I "complete the square" on the righthand side by adding 1/4: 2*N + 1/4 = n^2 + n + 1/4 = (n+1/2)^2. Now to clear fractions and get everything in terms of whole numbers, I multiply by 4: 4*(2*N+1/4) = 8*N + 1 = 4*(n+1/2)^2 = (2*n+1)^2. Another way to look at this is to take the equation N = n*(n+1)/2, and use the Quadratic Formula to solve it for n: (1/2)*n^2 + (1/2)*n - N = 0, n = (-1/2 +- sqrt[1/4 - 4*(1/2)*(-N)])/(2*[1/2]), = -1/2 +- sqrt[1/4 + 2*N], = (-1 +- sqrt[8*N + 1])/2. In order for n to be a rational number, 8*N + 1 must be a perfect square, and if it is a perfect square, it is the square of an odd whole number, so n is a whole number, too. One of the uses of triangular numbers is to count things. How many ways can you pick two objects out of a set of n objects, if order doesn't matter? The answer is the (n-1)st triangular number. Example: From {a,b,c,d,e} we can pick ab, ac, ad, ae, bc, bd, be, cd, ce, or de. There are 4 of these starting with a, 3 starting with b, 2 starting with c, and 1 starting with d. The total number is 10 = 4 + 3 + 2 + 1. There are 10 ways of choosing 2 objects from a set of 5, and 10 = 4*5/2, the 4th triangular number. By the way, this is how you count the number of edges and diagonals of a pentagon. There are 10 of them, 5 edges and 5 diagonals (pick two of the vertex points in all 10 ways possible, and connect them with line segments). - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/