<p>can someone explain the answers to the questions below?
1) If n distinct planes intersect in a line, and another line L intersects one of these planes in a single point, what is the least number of these n planes that L could intersect?
a) n
b) n-1 answer
c) n-2
d) n/2
e) (n-1)/2</p>
<p>2) Which of the following has an element that is less than any other element in that set?</p>
<p>I. the set of positive rational numbers.
II. the set of positive rational numbers r such that r^2 is greater than or equal to 2.
III. the set of positive rational numbers r such that r^2 is greater than 4.</p>
<p>a) none ( answer)
b) I only
c) II only
d) III only
e) I and III</p>
<p>I know the answer for the first one. But I'm not native, so don't expect a clear explanation from me.
You know a plane is unlimitedly spreading as well as the line. So unless they're parallel, they must intersect. So line L can have only one plane, with which it won't intersect. So the answer is n-1. Guys, be judge, i'm not sure about my solution.
also i don't know even understand 2-rd problem and hope that someone would clarify it for me. perhaps i can do that!!! kk</p>
<p>okey, dokey. i think i figured out it now. you know, rational numbers aren't whole number, including decimals.
like 1.000001 is very small number, but it's not the least one, or even there isn't the least one. 1.0000000000000000000000001 or 1.0000000000000000000000000000000001 like this. the same is about choice B and C. though r^2 > 2 or r^2 > 4 it's still rational number.</p>
<p>Unless I'm making a stupid mistake by overlooking something, the answer to 2) should be a).</p>
<p>In a more mathematical fashion than how Tsenguun explained, I'll make an attempt at an interpretation. Let's try to find a smallest element of each set. How would we find a smallest member of set I)? Well, we find the smallest positive rational number. If we pick any positive rational number (for example, 1/1000), we can always choose a positive rational smaller than this (1/10000, with respect to the previous example). For set II), you're going to have to find the least rational that is greater than the square root of 2. The same method applies to that of I). The problem is trying to trick you by allowing you to assume the square root of 2 is rational. A whole lot more people fall for this than what you would expect. The same method as I) and II) applies to III) as well.</p>
<p>Slightly more mathematical language: if R0 = a/b is a rational, where a and b both positive integers, then R1 = a/(b+1) is also a positive rational integer and is less than R0. Thus the set of rationals contains no least element since for every positive rational you choose, you can always choose a smaller one.</p>
<p>It would be much easier explaining this in terms of least upper bounds, but I don't think many people studying for the SAT II know what these are.</p>
<p>Sorry about my bad explanation there. That's just because of my poor English, which is hopefully improving now!!!
Oh, then, what about the first one??</p>
<p>for the second question, just remember that there are an infinite amount of positive numbers(I), numbers greater than 2(II), and the same for greater than 4(III). you can always add one more decimal point, so you'll never find a single element that is the very least.</p>
<p>tsenguun, your explanation is exactly how i thought of it. if the planes intersect, none can be parallel. therefore, any line drawn can relate to the planes in one of two ways: it can be parallel to one plane or parallel to none. if it is parallel to one, it can be parallel to only one, because none of the planes are parallel to each other. if this is the case it'll intersect with n-1 planes. if the line is parallel to none, it'll intersect with all n planes. n-1<n, and therefore n-1 is the answer.</p>
<p>if you still are confused, simplify the problem by thinking of the planes as lines(take away the factor of depth). lets say n=4 lines. one line is y=x, one y=2x,y=3x, and y=4x. they all intersect at a point, which would be a line if it had depth. now, you draw another line, L, which we're going to say y=5x. it will intersect with all 4 (or all n) lines. however, change that equation L to y=2+x. it'll intersect w/ only three lines, because it is parallel to the line y=x. therefore the least amount of lines is n-1. if you still can see how it works, plug this into a graphing calculator. hope this helped!</p>