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[Vector]

Well, you should probably get the 6th Edition, which is the one that I used (the one I linked you to was earlier, whoooops).

But, you know, for this sort of standard text they're usually pretty easy to pick up cheap if you work one or two editions behind the present iteration.

[Mego]

6th edition is currently cheaper then 4th, $23.95 vs $26.52.

[ed boy]

What the actual fuck? A math textbook for under $50? Vector, what kind of demonic ritual did you have to perform for this to exist?

You can loads of textbooks dirt cheap if you look in the right places. I've bought a fair few math textbooks from here before, and although the shipping prices are high, the book prices are so low I can quite often get them for £20 or less.

[Bauglir]

6th edition is currently cheaper then 4th, $23.95 vs $26.52.

Got a used one of the sixth for $14, actually.

[Karlito]

Plus it's the beginning of summer of textbook prices drop as students try to sell their uneeded books. They'll rise again once it's closer to the start of the term.

[Tarqiup Inua]

http://www.amazon.com/Invitation-Discrete-Mathematics-Jiri-Matousek/dp/0198570422

This one is a good book for discrete mathematics... it's easy to digest, so to say, and explains the subject well :-)

EDIT: At least in original language... seems like most of the favourable reviews are from those who had Czech version in hands, so who knows...

[Skyrunner]

This is a Korean math problem!

The graph plot might not be correct. >_>

There's a caveat: you cannot use a calculator in any form or function. o_O
This one stumped me for a while an hour ago, so I'm curious if other people can solve it well.

[Sergarr]

Well, the first one is obviously false, and the second one... y4-y4? Wat.

[Another]

1) x2 is obviously less then x1.
2) x3 is not equal to x4 and it is zero on the right side...
3) both x3 < x4 and y4 < y3, so this one statement is correct.

The plot seems correct.

[MagmaMcFry]

It kinda seems like all these problems are just a simple typo away from some harder questions.

Here's some harder questions:

1) Is x2 < 1/9 < x1?
2) Is (x3-x4)/(x2-x1) = (y4-y3)/(y2-y1)?
3) Is x4y4 < x3y3?

Spoiler: Solutions (click to show/hide)

All three are true.
1) Take the log3 of the left inequality and the log2 of the right, and substitute the intersection equations. You'll get x2 > 0 and x1 < log2(9)-2, which are both true.
2) Rewrite to -(y4-y3)/(x4-x3) = (y2-y1)/(x2-x1), which is easy to see since both sides are exactly 1.
3) Substitute y = 2-x, then negate both sides, add 1, and take the square root, you'll get |x4-1| > |x3-1|, which is true.

[Sergarr]

So it's basically "remove all wrong answers to get the right one" sort of math problem? No wonder calculators are banned.

EDIT: Disregard that, I'm stupid.

[MagmaMcFry]

So it's basically "remove all wrong answers to get the right one" sort of math problem? No wonder calculators are banned.

It's the sort of problem that tests for actual ability to think mathematically, not just the ability to use calculators. I approve of these problems.

[Skyrunner]

It kinda seems like all these problems are just a simple typo away from some harder questions.

Here's some harder questions:

1) Is x2 < 1/9 < x1?
2) Is (x3-x4)/(x2-x1) = (y4-y3)/(y2-y1)?
3) Is x4y4 < x3y3?

Spoiler: Solutions (click to show/hide)

All three are true.
1) Take the log of the equation and substitute the intersection equations. You'll get x2 > 0 and x1 < log2(9)-2, which are both true.
2) Rewrite to -(y4-y3)/(x4-x3) = (y2-y1)/(x2-x1), which is easy to see since both sides are exactly 1.
3) Substitute y = 2-x, then negate both sides, add 1, and take the square root, you'll get |x4-1| > |x3-1|, which is true.

Number 2 there is the correct version of that problem
Anyhow, apparently that was easier than expected <3 Mind if I throw another one at you people?

Also, seeing different ways to solve problems is strangely amusing... For example, Another's answer for no3 wasn't something I would have thought of. I divided both sides by y4y3 and saw that it was basically the average slope to those points.

Also, it might have been easier than expected, because the original problem didn't have the plot.

[MagmaMcFry]

Anyhow, apparently that was easier than expected <3 Mind if I throw another one at you people?

Go for it!

[Another]

A form of nerd-sniping: present a simple problem that can be trivially made much harder.