About an exercise appearing in TAOCP volume one's "Notes on the Exercises"

Question

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.

Orethaorferd

Out of context that is a calculation, it doesn't require any proof. – Habitancy

Is there a programming related question here? – Scabies

I'd guess given that the book in question is The Art of Computer Programming it is at least marginally related - but I think it is more a case of Knuth wanting to explicitly let other maths people know what was considered out of scope. – Plank

Answer 1

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).

Plank

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) / 2 and 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 = 2197 ? – Orethaorferd

Answer 2

I initially considered it as follows:
n³ = n * n * n
log_n(n³) = log_n(n*n*n)
log_n(n³) = log_n(n) + log_n(n) + log_n(n)
3 = 1 + 1 + 1
3 = 3

This seems fairly circular in its use of logarithmic identities, but given where I'm at in my algorithms research, it was oddly comforting.

Vasiliu

Answer 3

Got stuck at the same exercise and 'solved' it this way: a^b = mult(i=1 to b) a

After a bit of thinking I came to the conclusion that this is a prime factorization (both 13 and 3 are primes). Look up fermat's little theorem.

(I know, it's an old thread but maybe this'll help somebody who is also seeking an answer to this execise.)

Captivate