Number of handshakes in a room

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Homework Help Overview

The problem involves determining the total number of handshakes that occur in a meeting as each new person arrives and shakes hands with everyone already present. The goal is to prove by induction that the number of handshakes is given by the formula \(\frac{n(n-1)}{2}\).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the inductive reasoning behind the handshake formula, with attempts to clarify the steps involved in the induction process. Questions arise regarding the assumptions made in the inductive step and the basis case.

Discussion Status

There is ongoing exploration of the inductive proof, with some participants questioning the clarity of the explanations and the assumptions made. Guidance has been offered regarding the correct interpretation of the inductive step and the basis case, but no consensus has been reached on the best approach to articulate the proof.

Contextual Notes

Participants express concern over the choice of the basis case for the induction and the implications of starting with zero participants. There is also mention of alternative methods to solve the problem without induction.

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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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