which is greater?

If you want a purely analytical method:

Note that $\sqrt{n}$ is a twice-differentiable function, yielding a second-derivative of $-.25n^{-1.5}$ which is negative for all positive n. Thus the first-derivative of the function decreases monotonically for positive n.

Apply the mean value theorem to the intervals [11,12] and [12,13]. You will get an interesting result which wil give you your answer.

BiP

 Quote by Mark44 I doubt that any instructor would accept a proof in which most of the symbols are ?.
You could make it an equation and use the same method... i.e. let root(13)-root(12)+x=root(12)-root(11). If x>0, then the left side is greater. If x<0, the right side is greater

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 Quote by piercebeatz You could make it an equation and use the same method... i.e. let root(13)-root(12)+x=root(12)-root(11). If x>0, then the left side is greater. If x<0, the right side is greater
If the left side is greater, then you can't write an = between the sides.

Am I understanding you wrong?

 Quote by micromass If the left side is greater, then you can't write an = between the sides. Am I understanding you wrong?
let x be the difference between the two. if, in the above scenario, x>0, then the left side must be less than the right, and vice versa.

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 Quote by piercebeatz You could make it an equation and use the same method... i.e. let root(13)-root(12)+x=root(12)-root(11). If x>0, then the left side is greater. If x<0, the right side is greater
And then the problem becomes determining the sign of x.

 Quote by Mark44 And then the problem becomes determining the sign of x.
Why's that?

Edit: Never-mind. I see why. It's because you square both sides, so when you solve for x, you get |x|=some number.

I would go with Bipolarity's method if you need a proof.

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