Reply to comment

Maths in a minute: Shake to solve

handshake

Suppose you have $n+1$ people in a room and each person shakes hands with each other person once. How many handshakes do you get in total? The first person shakes hands with $n$ other people, the second shakes hands with the $n-1$ remaining people, the second shakes hands with $n-2$ remaining people, etc, giving a total of

$n+(n-1)+(n-2)+...+2+1$ handshakes.

But we can also look at this in another way: each person shakes hands with $n$ others and there are $n+1$ people, giving $n \times (n+1)$ handshakes. But this counts every handshake twice, so we need to divide by 2, giving a total of

  \[ \frac{n \times (n+1)}{2} \]    

handshakes.

Putting these two arguments together, we have just come up with the formula for summing the first $n$ integers and we’ve proved that it is correct:

  \[ n+(n-1)+(n-2)+...+2+1 = \frac{n \times (n+1)}{2}. \]    

Maths can be so easy!

Reply

  • Web page addresses and e-mail addresses turn into links automatically.
  • Allowed HTML tags: <a> <em> <strong> <cite> <code> <ul> <ol> <li> <dl> <dt> <dd>
  • Lines and paragraphs break automatically.

More information about formatting options

By submitting this form, you accept the Mollom privacy policy.