Example 2: √3000 is closest to which integer?So I think we all agree that in a case like this, the radical is clearly between the two integers indicated (since the radical function is monotonic). But the additional step of saying which of the two it's closer to is not done in all textbooks. Here's another example (not our school's textbook). Let's clearly state the claim being made here:
Solution:... [after some preliminary estimates] Try between 50 and 60, (55)2 = 3025, still a bit too high. Try (54)2 = 2916, now a little too low. Thus √3000 is between 54 and 55, but closer to 55 since 3025 is closer to 3000 than is 2916.
Claim: If x is closest to n2, then √x is closest to n.Above, "closest" means the minimum distance from x to any n ∈ ℕ. This claim gave me a squirrelly feeling for some time, and with good reason; it isn't true for arbitrary x ∈ ℝ.
Counter-example: Consider x = 12.4. It's closest to the perfect square 32 (distance 3.4 from 9, versus 3.6 from 16). But the square root is actually closest to the integer 4 (√12.4 ≈ 3.52).Now, let's characterize the kinds of numbers for which the claim in question won't work. For some integer n, take the cutoff between it and its successor, n+1/2 (i.e., the average of n and n+1). Any √x below this value is closer to n, while any √x above it is closer to n+1. Under the squaring operation, this cutoff gets mapped to the square-of-the-average (n+1/2)2 = n2+n+1/4.
On the other hand, consider the cutoff between the squares of the integers in question. Any x below their average is closer to n2, while any x above the average is closer to (n+1)2. This average-of-the-squares is ((n)2+(n+1)2)/2 = (n2+n2+2n+1)/2 = (2n2+2n+1)/2 = n2+n+1/2.
So you can see that there's a gap between these two cutoffs, and in fact it's exactly 1/4 in all cases, no matter what the value of n. If you pick x in the range n2+n+1/4 < x < n2+n+1/2, then √x will be closer to its ceiling of n+1, but x itself will be closer to its floor-square of n2. Specifically, the problem cases for x are anything a bit more than the product of two consecutive integers (also called a pronic or oblong number), exceeding n(n+1) = n2+n by a value of between 1/4 and 1/2. Since n2+n is itself an integer (ℕ closed under add/multiply), we see that any x in violation of the claim must be strictly between two consecutive integers, and thus cannot itself be in ℕ.
In conclusion: While the claim in question is not true for all real numbers, it is a trick that does happen to work for all whole-numbered values of x. How important is that? Personally, I'm pretty uncomfortable with giving our students an unverified procedure which can leave them thinking that it works for any number under a radical, when in fact that's not the case at all.