Re^4: Closest factor less than sqrt(N) [proof]

Here's an informal proof of our conjecture "The factor of N closest to sqrt(N) is less than sqrt(N)".

Let M be the positive square root of N. (M*M = N, M>0)
The closest factor to M is bewteen 1 and 2*M. (i.e. 1 is closer to M than any number greater than twice M. (M-1 < 2*M-M))
For every pair of numbers whose product is N, one of the pair will be greater than M and the other less than M. (the product of two numbers greater than M is greater than N, and the product of two numbers less than M would be less than N)
Take a pair of numbers whose product is N. Call the smaller (M-X) and the larger (M+Y). (i.e. X>0, Y>0)
So, (M-X) * (M+Y) = N
Multiplying out, M^2 + M*(Y-X) - X*Y = N
But M^2 = N (by definition)
Substituting: N + M*(Y-X) - X*Y = N
Subtracting N: M*(Y-X) - X*Y = 0
Since X and Y are positive (by definition, step 4), the term -X*Y is negative
If -X*Y is negative, the only way for the entire sum to equal 0 is if the term M*(Y-X) is positive
But M is positive (step 1), so (Y-X) must also be positive, which means that Y is greater than X.
Finally, if Y is greater than X, the smaller factor, (M-X) is closer to M than (M+Y), so you can limit your factor search to numbers less than sqrt(N).

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Comment on Re^4: Closest factor less than sqrt(N) [proof]

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Re^5: Closest factor less than sqrt(N) [proof] by hv (Prior) on Feb 21, 2005 at 09:45 UTC
Here's an easier way: take `n = m^2`, and a factor greater than the square root `m + d`, then the other factor is: `m^2/(m + d) = m - dm/(m + d) = m - d + d^2/(m + d)` [download] Now `d^2/(m + d)` is greater than zero, so the cofactor is greater than `m - d` and hence closer to the square root. Hugo	[reply] [d/l] [select]
Re^5: Closest factor less than sqrt(N) [proof] by BrowserUk (Patriarch) on Feb 21, 2005 at 00:52 UTC
sleepingsquirrel++ Makes sense to me--but then my version did (to me) too :) Who'd've thunk it--even informality is relative :)	[reply]