so this is confusing, with mod arithmetic i see that you only need the remainder as an answer but i don't get how it is solved.

like

2 (equivalence sign) 4(mod 3)

(3+4)(mod 5)

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- March 25th 2013, 04:02 PMzhengcl86Modular Arithmetic help!
so this is confusing, with mod arithmetic i see that you only need the remainder as an answer but i don't get how it is solved.

like

2 (equivalence sign) 4(mod 3)

(3+4)(mod 5) - March 25th 2013, 04:10 PMAntRe: Modular Arithmetic help!
- March 25th 2013, 04:14 PMzhengcl86Re: Modular Arithmetic help!
2 = 4(mod3) is a true/false question and the = sign after 2 is the one with the 3 bars the equivalence sign. since i don't know the code to put that in.

like if i get something like (2-4)(mod 7) and (2+9)(mod 10) how would i go about solving those? - March 25th 2013, 04:28 PMAntRe: Modular Arithmetic help!
2 = 4 (mod 3) is false.

Modulo arithmetic really isn't too bad once you get used to the idea. In fact, we often use modulo arithmetic to tell the time; For example, 13 o'clock = 1 o'clock. And, 20:00 = 8pm. This is all modulo 12.

Two number are equal modulo n if they differ by a multiple of n.

Perhaps the best way to write a number modulo n, is to write it as a number less than n but greater than 0.

If you number is lager than n, subtract multiples of n until you reach a number less than n. If you have a negative number, add multiplies of n until you arrive at a number greater than 0 but less than n.

so what is 4 modulo 3? Well, 4 -3 = 1. So 4 = 1 mod 3.

What is -10 modulo 3? Keep adding multiples of 3 until you reach a number greater than 0 but less than 3. -10 = -7 = -4 = -1 = 2 mod (3)

Addition and subtraction are easy, just work out the answer as usual, then add or subtract multiples of 3 until you arrive at a number greater than 0 but less than n. - March 25th 2013, 04:40 PMzhengcl86Re: Modular Arithmetic help!
this kinda of helped a bit, even tho i have my phone's time set on 24 hour clock, but it wasn't clicking in my head. thank you

- March 25th 2013, 04:58 PMzhengcl86Re: Modular Arithmetic help!
so how would congruence equations come into play?

i have 2+n=3(mod7)

so would the answer be n = 6 (mod 7) ?? - March 26th 2013, 03:06 PMAntRe: Modular Arithmetic help!

Treat like a normal equation, and subtract 2 from both sides...

And you're done.

Of course, this isn't the unique way of writing it, in fact:

- March 26th 2013, 03:50 PMPlatoRe: Modular Arithmetic help!
- March 28th 2013, 06:53 AMzhengcl86Re: Modular Arithmetic help!
hmm that seems understandable,

and makes alot more sense now

how about fractions? n/2 = 4(mod7)

so n = 2(mod7)

n=2(mod7) n=9(mod7) n=16(mod7) n=23(mod 7)

would that be the solution?

then it comes regular equations like

3=(n-4)(mod7)

so 3n-12(mod7)

n=-3(mod7) n=5(mod7) n=12(mod7) n=19(mod7)

is my math correct? lol thanks in advance! - March 28th 2013, 08:56 AMSelindeRe: Modular Arithmetic help!
For the first one, you get the idea, but you made a small mistake.

If you have

then, as you would do usually,*multiply*(not divide!) both sides by 2, and you get

Now, as Ant explained, this is not the only answer. In fact, the "best" or most "standard" answer is the one where n is equivalent to some number between 0 and 7; in this case, that would be .

For the second equation, note that

is the same equation as

.

So, we again can just solve it like we would any regular equation, by adding 4 to both sides:

So our answer is (or better: ).