Number of handshakes in a room

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Homework Statement



as each new person arrives to a meeting, they shake hands of all who are present.

prove the number [by induction] of occurred handshakes are \frac{n(n-1)}{2}




The Attempt at a Solution



if no one shows up then clearly no handshakes are exchanged

\frac{0(0-1)}{2} = 0

assuming k people show up, there will be \frac{k(k-1)}{2} handshakes

now if one more person shows up, that will make a total of k+1 people and because one does not shake their own hand there will be k +1 -1 more shakes than when there where k people .

there will be

\frac{(k+1)(k+1-1)}{2} handshakes in total

=\frac{k^2+k-k+k+1-1}{2}
=\frac{k^2-K}{2} + \frac{k+k+1-1}{2}
=\frac{k(k-1)}{2} + k


is this explained properly?
 
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jonroberts74 said:

Homework Statement



as each new person arrives to a meeting, they shake hands of all who are present.

prove the number [by induction] of occurred handshakes are \frac{n(n-1)}{2}




The Attempt at a Solution



if no one shows up then clearly no handshakes are exchanged

\frac{0(0-1)}{2} = 0

assuming k people show up, there will be \frac{k(k-1)}{2} handshakes

now if one more person shows up, that will make a total of k+1 people and because one does not shake their own hand there will be k +1 -1 more shakes than when there where k people .

there will be

\frac{(k+1)(k+1-1)}{2} handshakes in total

I don't follow that step. Aren't you assuming what you are trying to prove?

=\frac{k^2+k-k+k+1-1}{2}
=\frac{k^2-K}{2} + \frac{k+k+1-1}{2}
=\frac{k(k-1)}{2} + k


is this explained properly?
 
I felt like I was assuming but wasn't for sure.

I feel like there's a point where I need to state the new number of handshakes with each incoming person is the new total of people minus 1, so I don't know if I said that at the right time.
 
If there are ##k## people in the room your induction hypotheses says there have been ##\frac{(k-1)(k)}{2}## handshakes. If you now walk into the room, how many handshakes will you add (easy question)?
 
I'll add k handshakes

so start here

\frac{k(k-1)}{2} + k
 
Yes. Simplify that and see if you get the right formula for ##k+1##. You worked it backwards originally.
 
In addition to above, I'm concerned about the basis case, ##n = 0##. In the context of the problem, I'm wondering why you did not chose ##n = 1## or ##n = 2##.
 
More explicitly: make certain that the inductive step works when ##k = 0## in the steps LCKurtz mentioned
 
jonroberts74 said:

Homework Statement



as each new person arrives to a meeting, they shake hands of all who are present.

prove the number [by induction] of occurred handshakes are \frac{n(n-1)}{2}




The Attempt at a Solution



if no one shows up then clearly no handshakes are exchanged

\frac{0(0-1)}{2} = 0

assuming k people show up, there will be \frac{k(k-1)}{2} handshakes

now if one more person shows up, that will make a total of k+1 people and because one does not shake their own hand there will be k +1 -1 more shakes than when there where k people .
I would have stated this a little differently. If one more person arrives he must shake hands witheach of the k people who were already there so the number of handshakes is increased by k:
\frac{k(k-1)}{2}+ k

there will be

\frac{(k+1)(k+1-1)}{2} handshakes in total

=\frac{k^2+k-k+k+1-1}{2}
=\frac{k^2-K}{2} + \frac{k+k+1-1}{2}
=\frac{k(k-1)}{2} + k


is this explained properly?
 
  • #10
HallsofIvy said:
I would have stated this a little differently. If one more person arrives he must shake hands witheach of the k people who were already there so the number of handshakes is increased by k:
\frac{k(k-1)}{2}+ k

Apparently you didn't read posts 4, 5, and 6?
 
  • #11
I didn't notice them, rereading all now

nvm, not to me.

so many long days, haha
 
Last edited:
  • #12
Of course this can be done more simply without induction. Each of the n people must shake hands with each of the n-1 others which would give n(n-1) people shaking hands. But each hand shake involves 2 people so that is n(n-1)/2 hand shakes.
 

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