I see it a little differently.

3X-1=mod6

there is no number you can put in X to get integer multiples of 6 on the left.

One example. 3(7/3)=1 mod6 => 7=1 mod6 ; this would be correct, but it is not a solution since 2.3333333333333333333333333333 is not an integer.

Correct me.

Correct you? I don't get it.

I just meant that in the case that I did something wrong.

The way I see it, we need 3*(some integer) to be equal to (some integer)*6 with a remainder of 1.

Is that right?

Try re-writing the equation. The ~ equals the modulus in my equation

For 3x ~ 1 mod 6 to be true, 3x -1 = k6 must be true, where x and k are integers (with me so far?)

Then solve for x to see it can be an integer.

3x - 1 = k6

x - 1/3 = k2

x = k2 + 1/3

Therefore, for every case where k is an integer, x cannot be an integer so 3x ~ 1 mod 6 cannot be true.

Your solution in post 15 is also correct, you just didn't draw the conclusion. x and k cannot simultaneously both be integers.

"Since a  b mod m if and only if b = a+mk for some k 2 Z, adjusting an integer modulo
m is the same as adding (or subtracting) multiples of m to it." It's more legible here: http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/modarith.pdf

The different text may also help with the other problems.

ditto

so I was correct but I didn't conclude.

So what would my conclusion look like.. math symbols or just saying "x cannot be an integer because for every 2k+1/3, there is no x such that x belongs to Z"

A proof won't be considered "correct" simply because your conclusion appears the same. You need to define the criteria the equation needs, then show your work on how the equation meets that criteria, then conclude.

The level of documentation really depends on your teacher. Showing your work is important. It's been a long time since I've done proofs, but something like the following may work.

"Therefore, for 3x ~ 1 mod 6; x {insert symbol for not a member of} Z" or

"Therefore, 3x {insert symbol for not congruent to} 1 mod 6"

Either of these conclusions are true and prove the original is not correct, but it depends on which is most satisfactory to your teacher

I was not trying to check in to this one... because I can not...
but just wanted them to ' label ' the subject...
I took ' discrete ' to mean ' confidential ' since it was not capitalized ... and could not figure out why math needed to be confidential...
LOL

http://i3.photobucket.com/albums/y74...rx24/page2.jpg

how about #5?

the triple bar has me confused.

% is mod

so

a % n = r

5 % 11 = 5

but


11 % 5 = 1

so it does not work out, correct?


As well as #1 parts A-D

A) F(10) = 0
F(2) = 2
F(55) = 0

B) F(10) = F(0)

C and D I am lost on

0 bar = n bar (a bar over the 0 and n)... I need to prove it but I cannot prove it

n > 1 is a fixed natural number.