So I think we all agree that in a case like this, the radical is clearlyExample 2: √3000 is closest to which integer?

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.

*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:

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; itClaim: If x is closest to n^{2}, then√x is closest to n.

*isn't*true for arbitrary x ∈ ℝ.

Now, let's characterize the kinds of numbers for which the claim in questionCounter-example: Consider x = 12.4. It's closest to the perfect square 3^{2}(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).

*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}= n

^{2}+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 n

^{2}, while any x above the average is closer to (n+1)

^{2}. This average-of-the-squares is ((n)

^{2}+(n+1)

^{2})/2 = (n

^{2}+n

^{2}+2n+1)/2 = (2n

^{2}+2n+1)/2 = n

^{2}+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 n

^{2}+n+1/4 < x < n

^{2}+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 n

^{2}. 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) = n

^{2}+n by a value of between 1/4 and 1/2. Since n

^{2}+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.

Nice work!

ReplyDeleteI don't think my textbooks have made that claim. But this is a great question to ask students to ponder. I wonder if I might ask it of my pre-calc students at some point.

Sue, thanks for reading that and commenting. It does seem like an assumption that's easy to make (even for some book authors) although it's not true for all values of x.

DeleteAnd that's why I liked the idea of possibly walking through it with students.

ReplyDelete"Is this true always, sometimes, never?" Feels like always, but when we try to show it, we get to sometimes. It seems like a good item for helping students see why proof is necessary.

Sounds like a great exercise! Interested in how it will work in-class...

DeleteI'd be very uncomfortable too, and not just because of the factual error.

ReplyDeleteFor me the bigger concern is that a great by-product of math is that it promotes rigorous thinking -- that is, math should help one become sensitive to assumptions, and help one understand that even reasonable-sounding assumptions can be wrong. But this text is giving an implicit endorsement of a bad assumption.

It might be argued that rigorous thinking is too lofty of a goal to ask of remedial students. That might be right (though I think it's a more important skill than estimating square roots). But at the very least, I think a math text should not be endorsing such an assumption!

I totally agree, of course!

DeleteInteresting! I'll admit to using the rule of thumb your textbook suggests, though I didn't learn it anywhere, and since I never got past the first year of my math major, lo these many moons ago (another one lost to the humanities, alas), I lacked the mathematical acumen to notice the cases where it's false. I recently had call to do a lot of estimating of cube-roots (don't ask), so I'm curious if there's a similar problem with estimating cubes. Seems as though there'd have to be...

ReplyDeleteGood question and yes, in fact it's worse, this one fails for lots of integers: like any floor of the average of sequential cubes. Example: Let x = 4. It's closest to the cube 1^3 = 1 (4-1=3 but 8-4=4). But the cube root of 4 ~ 1.59, which is actually closer to 2.

DeleteAlgebraically comparing the cutoffs like I did above: cube-of-average (n+1/2)^3 = n^3 + 3/2*n^2 + 3/4*n + 1/8, while average-of-cubes (n^3+(n+1)^3)/2 = n^3 + 3/2*n^2 + 3/2*n + 1/2. So in this case there's a difference in the n term, indicating the gap between cutoffs gets wider and wider, trending arbitrarily large as n increases.

Which is maybe an even better illustration that while you can get a correct

interval(bracketing range) with this technique, saying that you're closer to one end or the other by looking at the power is not generally correct.