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Triangular Numbers

Date: 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?


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. 


45 is a triangular number because 8*45 + 1 = 361 = 19^2 is a perfect 

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   

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.  


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 

- Doctor Rob, The Math Forum   
Associated Topics:
High School Number Theory
Middle School Number Sense/About Numbers

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