What's the question exactly? Do you mean: If p/q < 1, IS sq root(p/q) > p/q and sq root(p/q) < 1?

If p/q < 1, the two main scenarios to think about are p/q < 0 and 0 < p/q < 1. In the first, yes (p/q)^2 > p/q because the square will be positive (e.g. if p/q = -1/2, then (p/q)^2 = 1/4). However, if 0 < p/q < 1, then (p/q)^2 < p/q because you're multiplying fractions less than 1 (e.q. if p/q = 1/2, then (p/q)^2 = 1/4).

(p/q)^2 < 1 doesn't necessarily hold one way or another either. If p/q = -2, then (p/q)^2 > 1. if p/q = 1/2, then (p/q)^2 < 1.

Just tacking on here ...

As SoCan says, if 0 < x < 1, x^2 will be less than x, whereas the square root of x will be greater than x, and this interval is the only interval on the entire number line in which this is the case. So long as x is in that interval, it will indeed also be the case that both its square and its square root are less than 1 (e.g. √(1/4) = 1/2, and (1/4)^2 = 1/16.

One way to remember/conceptualize this is to think about these graphs.



Notice that the only "abnormal" region is the space between x=0 and x=1, and that it is abnormal in the sense that the graph of y = x^2 dips BELOW the graph of y = x. This unique dipping-below is precisely due to the fact that that's the only interval in which x^2 is less than x. Or, equivalently, the only interval in which x is greater than x^2.

As SoCan points out, squaring any negative number will yield a greater number (since it will yield a positive, and all positives are greater than all negatives). Furthermore, if we tried to take the square root of any negative number, we'd break math (at least within the realm of real numbers, which is all we talk about for GMAT purposes). So the original

If p/q < 1, then sq root(p/q) > p/q and sq root(p/q) < 1

would be best amended to

If 0 < x < 1, then sq root x > x and sq root x < 1

Best,

Thanks ya'll, it's not a question from a book. Just a random thought I had. Yes, I was assuming p/q >0 and p/q is a positive decimal. Thanks for clearing up guys.