Can anyone solve question 13? I'll give you a million dollars for the proof?

>Point 7:

What is x=2, y=3, z=4?

The questions are true/false, baka! That's not the hard part, the hard part is the proof using mathematic logic. Hence that chapter title: logic.

Oh, just glanced at one at random. Going over in detail, they're all trivial except 13 obviously. Why would they put that in? Just to fuck with people?

Here's what I have so far OP:
01 False; the line would start at -1,1 and dip down to 2,-1, then rise to 3,0, which is impossible.

02 True; for x positive, x^2 = x + x + x + ... + x (x times) >= x. For x negative, x^2 is always positive so x^2 >= x. For x = 0, we have x^2 = 0 = x.

03 False; consider x = -2: x^3 = -8 < -2 = x.

04 False; consider x = 2: x^3 = 8 > 2 = x.

05 True; use x = 0: x^3 = 0 = x.

06 True; suppose sqrt(2) was rational. Then sqrt(2) = p/q where p/q is a rational number in simplest terms. Then 2 = p^2/q^2 implying 2q^2 = p^2. Then p^2 is even, so p must be even; say p = 2m. Then 2q^2 = 4m^2 implying q^2 = 2m^2. Similarly, q^2 and q are even. Following, p/q cannot be in smallest terms, which contradicts our assumption.

07 False; consider x = z = 1 and y = 2. Then x +y = y + z = 3, but x + z = 2.

08 True; if x is even then x = 2n and x^2 = 4n^2 = 2(2n^2) which is also even.

09 True; note that 1 = 2^0. Suppose n = 2^m_1 + 2^m_2 + ... + 2^m_k, where m_i < m_j when i < j. Then n+1 = 2^0 + 2^m_1 + 2^m_2 + ... + 2^m_k. If m_1 > 0, we are done. If m_1 = 0, then 2^0 + 2^m_1 = 2^1. If m_1 > 1, we are done. If m_1 = 1, then 2^1 + 2^m_1 = 2^2. We repeat this process noting that it must eventually terminate since k is finite, and see that n+1 is a sum of distinct powers of 2. By induction, we are done.

10 False; consider the positive integer 2.

11 True; every integer can be written as a sum of 1s, i.e. x = 1 + 1 + ... + 1 (x times). If we can pair the 1s together, say into n pairs, then we have x = 2n. If there is one left over, then x = 2n + 1 where n counts the number of pairs we could make.

12 True; suppose that we had x = 2m = 2n + 1. Then 2(m-n) = 1. But 2(m-n) is even, which is a contradiction.

13 Hm, I have been working for a while on this one, I'll get back to you.

>doing grade school arithmetic in that much detail
>copying out the sqrt(2) irrationality proof

whew lad

Get back once you've solved 13

>m.
kill yourself

Just Google Goldbach's conjecture you fucking retard.

It's easy to say you're smart, which is what you're doing when you just wave your hands and call things "trivial" or "grade school arithmetic" and proofs you couldn't figure out on your own "copied."
However, I've actually shown work, which speaks for itself, so people KNOW I'm smart.

The book is for a class on mathematical proofs. It's an intro to the subject, just a 200 level class, and the first assignment was specifically noted as not being for a grade and not to spend too long on it. The expectation isn't that we'll solve goldbach's conjecture, it's that we'll realize there isn't necessarily an accepted proof for any question.

>so people KNOW I'm smart.

Damn you really showed them you're smart by handling complex questions like if three points lie on a straight line

4

yeah, whether the statements themselves are true or false is obvious, but proving them isn't so simple.

Every prime number is odd (other than 2) or it could be divided by two.

Two odds added together is even every time.

Since your intervals between primes is only 1 at the beginning you can safely say every number above can be covered.

Not a proof, but might help...

>Not a proof, bcoz can't into proof
gtfo fgt pls
Lrn2goldbach

>but proving them isn't so simple
Kek. Please be in high school

True. It is enough to prove it is true for 10^200

All it takes is one number after that not following it to disprove the entire conjecture.

Such a number does not exist

14 is false. Is that worth anything?

are u srs? questions like 3 points on a line have been asked since pyramids the proof without 'hurr durr there they r' is not for hs kids

Well... WILL you give me a million dollars for the proof? You framed it as a question, so I have to ask.

charter zero is an okay book.

book of proof is better

en.wikipedia.org/wiki/Goldbach's_conjecture

>Can anyone solve question 13?
>On Veeky Forums
kek

If you can prove goldbach's conjecture, there is literally a $1,000,000 USD prize.

This fags underage for sure.

?
i would bet he/she is probably 18/19 based on the book in the pic.

it's a math text titled Chapter Zero