What Does a ≡ b (mod m) Mean in Mathematics?

[flyingpig]

OKay, what is a "mod"...?

Homework Statement

Verify the proposition
[PLAIN]http://img442.imageshack.us/img442/7808/unledah.png

The Attempt at a Solution

I don't know how to start, all I see is that 0/m belongs to Z...

 

[Mark44]

Surely wikipedia can offer some help, right? Don't waste our time on trivial questions that you can easily answer for yourself.

 

[DaveC426913]

Surely wikipedia can offer some help, right? Don't waste our time on trivial questions that you can easily answer for yourself.

I'm not sure that wiki will help him solve this. Yes, he might learn what mod means, but he said he doesn't know where to start. Surely he could use a shove in the right direction.

 

[SammyS]

If (b-a)/m ∊ ℤ, that means b-a is an integer multiply of m, i.e. (b-a)/m has a remainder of zero.

 

[Mark44]

I'm not sure that wiki will help him solve this. Yes, he might learn what mod means, but he said he doesn't know where to start. Surely he could use a shove in the right direction.

It would probably help him answer the question in his thread title.

 

[SammyS]

I'm not sure that wiki will help him solve this. Yes, he might learn what mod means, but he said he doesn't know where to start. Surely he could use a shove in the right direction.

Yes, what Mark said sounded rather harsh, but Mark & I & several others (even micromass worked with him though a multi-page thread recently) have a fairly extensive history of working with fp (flyingpig), so Mark knows him fairly well, and had his reasons for that post.

I'm sure Mark didn't intend for that remark to be mean-spirited.

 

[DaveC426913]

It would probably help him answer the question in his thread title.

Yes, but you didn't finish reading his post before spewing venom.

 

[flyingpig]

Okay all that

"congruent module" (big words) mean is that some number m beside the "mod" means whatever comes in front is a divisor, nothing more than that.

Jesus these unnecessary abstract words are killing me

Sorry for causing the ruckus.What exactly is there to verify? As Sammy said there is no reminder (0? )

I got to satisfy the three rules of relations

1) reflexivity (again what's with all these words?)

aRa

(a - a)/m = 0/m = 0 $$\in Z$$

2) symmetry

I am going to use M for some other integer and m the original integer

So there exists a b such that

b - a = Mm so that a - b = -Mm = M(-m)

BUt -m is also in Z, so that holds

3) Same reasoning as (2)

There exists some c

such that

a - b = Mm

b - c = Nm

a + c = Mm + Nm = m(N + M)

But N + M is also in Z, so that holds and thus completes the proof

aRb = bRa

 

[uart]

The question gives you the definition of "modulo equals" and then asks you to (use that definition to) verify that it's an equivalence relation.

For example, Show that if "a=b mod m" and "b=c mod m" that "a=c mod m" and so on.

 

[Mark44]

Yes, what Mark said sounded rather harsh, but Mark & I & several others (even micromass worked with him though a multi-page thread recently) have a fairly extensive history of working with fp (flyingpig), so Mark knows him fairly well, and had his reasons for that post.

I'm sure Mark didn't intend for that remark to be mean-spirited.

Right, I didn't. "Peeved" is closer to what I was feeling. In retrospect I should have toned what I said down a notch or so.

Yes, but you didn't finish reading his post before spewing venom.

No, Dave, I did read his entire post, which also includes, "I don't know how to start..." My intent was to get flyingpig to do due diligence by looking up some basic terminology.

 

[flyingpig]

COuld someone please critique my work?

 

[flyingpig]

I just want to correct one small thing in the problem. It actually asks me to PROVE and not VERIFY, though I don't think there is a difference in this case.

 

[Mark44]

Your work in post #8 looks OK at a quick glance, but your book's definition of modulo requires me to do mental translation, and your use of m and M makes it more difficult to follow than it needs to be.

a $\equiv$ b (mod m) is usually defined to mean that a - b = km for some integer k.

Modulo arithmetic divides the integers into equivalence classes based on their remainders when divided by m.

For example, in modulo 7, the only possible remainders are 0 (multiples of 7), 1, 2, 3, 4, 5, and 6.

13 $\equiv$ 6 (mod 7), because 13 - 6 = 7 = 1*7.

 

[flyingpig]

Your work in post #8 looks OK at a quick glance, but your book's definition of modulo requires me to do mental translation, and your use of m and M makes it more difficult to follow than it needs to be.

So it's just my poor choice of letters that's really wrong because I was worried about transitivity part ? It's not my book it's my notes our prof sent to us lol

 

[SammyS]

It's not too bad.

See red inserts.

...

these unnecessary abstract words are killing me     (They're not unnecessary.)

What exactly is there to verify? As Sammy said there is no reminder (0? )

Originally Posted by Flyingpig's Work

"

I got to satisfy the three rules of relations

1) reflexivity (again what's with all these words?)

aRa    Here you are showing that a ≡ a (mod m)

(a - a)/m = 0/m = 0
∈Z

Write what you've just proved !

∴ congruence (mod m) is reflexive.

2) symmetry       Here you are showing that if a ≡ b (mod m), then b ≡ a (mod m)   Let the reader know (that you know) this.

I am going to use M for some other integer and m the original integer

So there exists a b such that

b - a = Mm so that a - b = -Mm = M(-m)

But -m is also in Z, so that holds

∴ congruence (mod m) is symmetric.

3) Transitivity    Here you are showing that if a ≡ b (mod m), and b ≡ c (mod m) then a ≡ c (mod m)

[STRIKE] Same reasoning as (2)[/STRIKE]

There exists some c

such that

a - b = Mm

b - c = Nm

a + c = Mm + Nm = m(N + M)

But N + M is also in Z, so that holds

∴ congruence (mod m) is transitive.

... and thus completes the proof

​

"

aRb = bRa

And finally, ... Why did you put in a quote what wasn't in a post anywhere? Makes it difficult for others to include that stuff in a quote.

Had to CUT & PASTE it.

 

[icystrike]

Let me give you a loose defn of "mod"

If a=b+km
s.t k is an integer,

a==b (mod m)

 

[flyingpig]

And finally, ... Why did you put in a quote what wasn't in a post anywhere? Makes it difficult for others to include that stuff in a quote.

Had to CUT & PASTE it.

LOl, I actually did it because I thought it would look neat. I had no idea it would be the other way around. Reminds me of that time I had so many spoilers no one (HallsofIvy, you, and someone else, I think it was vela) knew what on Earth I was doing

(They're not unnecessary.)

What's wrong with saying "$$a \equiv b$$ is divisible by m"?

 

[SammyS]

LOl, I actually did it because I thought it would look neat. I had no idea it would be the other way around. Reminds me of that time I had so many spoilers no one (HallsofIvy, you, and someone else, I think it was vela) knew what on Earth I was doing

Well, I just had to give you some crap about that "QUOTE" thing. I did make you post look nice, and the cutting & pasting wasn't really that difficult.

 

[diligence]

LOl, I actually did it because I thought it would look neat. I had no idea it would be the other way around. Reminds me of that time I had so many spoilers no one (HallsofIvy, you, and someone else, I think it was vela) knew what on Earth I was doing



What's wrong with saying "$$a \equiv b$$ is divisible by m"?

I sympathize with struggling with terminology, but not every abstract concept has an explanation as succinct as this. If math was structured the way you propose, with no terminology beyond the basic stuff, most advanced concepts would require a tangled mess of implications to explain, and textbooks and research papers would require A TON more paper. It would always have to go back to the basic axioms, which would obviously be completely ridiculous.

As Sammy said, the are certainly NOT unnecessary