mathwonk said:
your answer makes no sense to me. a/b can be either greater or smaller than sqrt(2). one can prove that a/b < sqrt(2), if and only if (a+2b)/(a+b) > sqrt(2), however. And there really is no "trick", just obvious rearrangements of fractions (and squaring)
Either I didn't correctly express myself or I really did something wrong.
Here's how I worked it out:
If [tex]\frac ab < \sqrt2 < \frac{a+2b}{a+b}[/tex]
Then,
[tex]\frac ab < \frac {a+2b}{a+b}[/tex]
So,
[tex]a(a+b) < b(a+2b)[/tex]
Thus,
[tex]a^2 < 2b^2[/tex]
Therefore,
[tex]\frac ab < \sqrt2[/tex]
---
That was how I figured this out a couple of hours ago. It felt like a great deal to me and I was smiling to myself while half walking, half jumping around the room*. My over excitement was quite short-lived as I couldn't figure out how to proove the second part, i.e, [tex]\sqrt2 < \frac {a+2b} {a +b}[/tex].
After looking at the book, I understood that I might have approached the question the wrong way. (this is actually an "example question" w/answer from the book) At the top of page 14 of
this book is the solution to the question, which (obviously) is a good way to approach this. Anyway, I hope that I've explained myself clearly enough this time.
I'll ask the question again, if I have no formal algebra knowledge and mainly learned via doing, could I still pull this off? :-)
*Consider that before having done "Example 1 & 2" on that book, the hardest mathematics I had ever been in contact with was that "monkey math exam" I linked to in the previous page. And yes, you're correct, in that this test is essentially an AP equivalent although it covers more material than AP Calculus and Stats. Even then, I'm not sure what good this does, if any at all...