Mathematical proofs

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Feb 12th 2009, 07:03 PM
treetheta

Mathematical proofs

If a mod would like to move this to it's assigned section please do so, I don't know which one it belongs into.

sorry i'm not good with latex

Q1. Prove the following biconditional statement

1. For any a in Z(set of integers) a not logically equivalent to 0(mod 3) if and only if a^2 is logically equivalent with 1(mod 3)

2.Prove that if a is an odd integer, then for any x in Z(set of integers), x^2 - x - a does not equal 0

3. Use mathematical induction to prove that for any natural number n, 3 divides (2^(2n) - 1 ). Specify each step in the process clearly.

Thank you soo much (:
Feb 12th 2009, 07:47 PM
Jhevon

Quote:

Originally Posted by treetheta

If a mod would like to move this to it's assigned section please do so, I don't know which one it belongs into.

sorry i'm not good with latex

Q1. Prove the following biconditional statement

1. For any a in Z(set of integers) a not logically equivalent to 0(mod 3) if and only if a^2 is logically equivalent with 1(mod 3)

do you know how to change modular equations into algebraic ones? if so, i suggest you do that. for instance, $a \equiv 1 \mod{3}$ means $a - 1 = 3k$ for some $k \in \mathbb{Z}$ .

after rewriting the statements that way, see if you can figure out what to do.

Quote:

2.Prove that if a is an odd integer, then for any x in Z(set of integers), x^2 - x - a does not equal 0

consider two cases: (1) x is even, (2) x is odd. show that the statement holds in either case. good ol' fashion algebra should do the trick. recall how to define even and odd integers.

Quote:

3. Use mathematical induction to prove that for any natural number n, 3 divides (2^(2n) - 1 ). Specify each step in the process clearly.

Thank you soo much (:

do you know the method of mathematical induction? what have you tried?

Let $P(n):$ 3 divides $2^{2n} - 1$ for all $n \in \mathbb{N} = \{ 0,1,2,3, \cdots \}$

Clearly $P(0)$ is true. (check this!)

Assume $P(n)$ is true. Now we need to show that this implies $P(n + 1)$ is true.

Note that since $P(n)$ is true, we have

$2^{2n} - 1 = 3k$ for some $k \in \mathbb{Z}$ .

So, $P(n + 1) = 2^{2n + 2} - 1$

$= 4 \cdot 2^{2n} ~{\color{red} + 4 - 4}~ - 1$

$= \cdots$
Feb 12th 2009, 07:58 PM
treetheta

okay i think i got the #2

for the first 1 do i assume 3 cases for 0,1,2

and the third one how did u get

$ P(n + 1) = 2^{2n + 2} - 1 $

or did u multiply everything by 2^2 in which case u get

$ P(n + 1) = 2^{2n + 2} - 4 = 12p 2^(2n + 2) - 1 = 3(4n + 1) $

and does 2n + 2 = 2k

o.o i did these steps but i ended up getting 9/10 i don't know where i went wrong..
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Feb 12th 2009, 08:12 PM
Jhevon

Quote:

Originally Posted by treetheta

okay i think i got the #2

ok

Quote:

for the first 1 do i assume 3 cases for 0,1,2

for (=>) direction you need only 2 cases: 1 (mod 3) and 2 (mod 3)

for (<=) direction, use the contrapositive

Quote:

and the third one how did u get

$ P(n + 1) = 2^{2n + 2} - 1 $

replace n with (n + 1)

Quote:

or did u multiply everything by 2^2 in which case u get

$ P(n + 1) = 2^{2n + 2} - 4 = 12p 2^(2n + 2) - 1 = 3(4n + 1) $

and does 2n + 2 = 2k

that's not what i did

Quote:

o.o i did these steps but i ended up getting 9/10 i don't know where i went wrong..
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the answer here is not a number...
Feb 12th 2009, 08:17 PM
treetheta

Quote:

replace n with (n + 1)

i see how u got 2k + 2 now

Quote:

that's not what i did

and sorry the latex came out wrong
$P(n + 1) = 2^{2n + 2} - 4 = 12p$
$2^(2n + 2) - 1 = 3(4n + 1)$

so n = 2k

Quote:

the answer here is not a number...

and that wasn't the answer that is what i got on a marking scale
Feb 12th 2009, 08:23 PM
Jhevon

Quote:

Originally Posted by treetheta

i see how u got 2k + 2 now

and sorry the latex came out wrong
$P(n + 1) = 2^{2n + 2} - 4$

no, this is not P(n + 1)

Quote:

$= 12p$

how??

Quote:

$2^(2n + 2) - 1 = 3(4n + 1)$

so n = 2k

no, and you are not looking for n in the first place. are you sure you know what the method of mathematical induction is?

look up how to prove something via mathematical induction, then re-read my post and make sure you understand what is going on. then just pick up where i left off.
Feb 12th 2009, 08:36 PM
treetheta

okay, I've given up on what I did,
and now I'm working on what you did

Why did you multiply and add and subtract 4 on both sides

can you explain it to me or link me to something that will help me understand it
Feb 12th 2009, 09:45 PM
treetheta

Omg, thanks jhevon i gett it ! : D
Feb 12th 2009, 09:55 PM
Jhevon

Quote:

Originally Posted by treetheta

Omg, thanks jhevon i gett it ! : D

you got it? really? great! (Clapping)

that adding and subtracting 4 (hence adding zero) is a neat trick, right?
Feb 12th 2009, 10:09 PM
treetheta

yea it was hard for me to follow why u did it
but now it all makes sense : D