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133

answers:

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Hi,

There is a question in TAOCP vol 1, in "Notes on Exercises" section, which goes something like:

"Prove that 13^3 = 2197. Generalize your answer. (This is a horrible kind of problem that the author has tried to avoid)."

Questions:

  1. How would you actually go about proving this ? (Direct multiplication is one way, another way could be using formula of (a+b)^3). Does the solution requires using some method that will allow us to make some kind of generalization ?

  2. What is the generalization here ?

  3. Why is this a horrible kind of problem ?

  4. What are some other kind of similar horrible problems that you are aware of ?

Appreciate any answers.

P.S. I apologize if the statement of problem above makes it look like a homework problem, but its not. Request people to not tag this as a homework problem, so that more people can give answers.

+2  A: 

I'd guess that he's alluding to perhaps proving it starting from just the Peano axioms. Then constructing the integers, and going on to formally show that 13^3 = 2197 is a natural, logical conclusion that flows from the definition of exponentiation.

We could generalize to show that given an a and b, there exists some integer c, that is a^b.

This is a horrible kind of a problem because most people find it uninteresting.

Similar sorts of problems can be found in a course on analysis (along with some greatly more interesting).

garethm
Hi garethm, I doubt it. If the above problem required using Peano axioms, it would have the rating of at least M30 or HM30, where as i think that this particular question has the rating of less than 15.Is it possible that, the expectation is something like this (for e.g.):Prove that 1 + 2 + 3 + ... + 10 = 55. Generalize your answer.And the answer would be something like:(1+10) + (2+9) + ... + (5+6) = 5 x 11 = (10 x 11) / 2and the generalization would obviously be (at least to Gauss :-)1 + 2 + 3 + ... + n = (n x (n+1))/2.If so, what such identity is hidden in 13^3 = 1397 ?
vshenoy