Monochromatic life savers
Life savers come in a variety of colors...
But Ted decides to prove that they only come in one color,
To be more specific, he wants to prove that in any group of n lifesavers, they are all the same color.
So he offers the following inductive argument:
1) Consider n = 1 : Clearly one life saver will have the same color as itself.
2) Now assume that that the statement is true for n life savers and prove that it is true for n + 1 lifesavers.
3) Take a group S of n + 1 lifesavers, { L 1 , L 2 , … , L n , L n + 1 } .
4) Now { L 1 , L 2 , … , L n } is a subgroup A of n lifesavers, and thus by the induction hypothesis are all of the same color.
5) As { L 2 , L 3 , … , L n + 1 } is also a subgroup B of n lifesavers, again by the induction hypothesis these lifesavers are all of the same color.
6) Since subgroups A and B have at least one element in common, the color of the lifesavers in subgroup A must be the same as the color of the lifesavers in subgroup B , and thus the n + 1 lifesavers in group S are all of the same color.... voilà!
7) Since its:
- True for n = 1
- True for n + 1 if its true for n
then its true for all n !
Where does Ted's argument breakdown. i.e. Which step is flawed?
Image credit: http://www.candywarehouse.com
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1 solution
Do steps 4 and 5 not violate the spirit of POI somehow? They seem to be the set-up for the punchline of step 6, so they are all in on the conspiracy. I was going to go with step 5 but I then chose not to answer, (my aversion to MC questions kicking in, I guess). I'll be curious to see what others have to say. Fun problem, regardless, and it definitely made me crave some Life Savers. :)
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Yeah, I see your point about 4 and 5... :0) Lemme think about if there is a better way to word the question...
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Your wording is probably fine, but just as a thought, one possible rewording is as follows:
1) and 2) stay as is.
3) Take a group S of n + 1 lifesavers, { L 1 , L 2 , . . . . . . , L n , L n + 1 } .
4) Now { L 1 , L 2 , . . . . . , L n } is a subgroup A of n lifesavers, and thus by the induction hypothesis are all of the same color.
5) As { L 2 , L 3 , . . . . , L n + 1 } is also a subgroup B of n lifesavers, again by the induction hypothesis these lifesavers are all of the same color.
6) Since subgroups A and B have at least one element in common, the color of the lifesavers in subgroup A must be the same as the color of the lifesavers in subgroup B , and thus the n + 1 lifesavers in group S are all of the same color .... Wallah!
7) stays as is.
I'm more comfortable with steps 4) and 5) now, and step 6) breaks down in the case of n = 1 , (even though it is legit for n ≥ 2 ), since in this one case A and B will not have an element in common. This then preserves the intent of your problem and the integrity of your solution.
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@Brian Charlesworth – Good idea... I've updated the problem accordingly... Thanks!
Step 6 doesn't work when you go from n = 1 to n = 2 life savers.