why modulo numbers different?

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SherlockWante

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Why are all numbers different if you use a module.

Let’s say you use module 22 for 22 numbers.
Numbers 1,2,3...
So 23 becomes 1, 24 2...
If you multiple each number by 3, all numbers are still different.
1 becomes 3, 2 becomes 6,15 becomes 1.... no numbers are the same.

I don’t understand why this is the case, why are there not 2 1’s, which is the case if we take like the square.
 
I'm not sure I understand what you are asking, but I'll try. (There are several ways in which the concept of modular arithmetic can be introduced, so the way I say things may not match what you were taught.)

I think that you are saying that with modulus 22 (that is, working in mod 22 arithmetic), we can replace any number with its remainder after division by 22, so 22 is equivalent to 0, 23 is equivalent to 1, and so on. There are 22 distinct values in the mod 22 number system.

When we multiply each number by 3, we find that 3 times each of the 22 distinct numbers is a different number (mod 22).

But multiplying by 2 instead, we find that [MATH]2\times 1\equiv 2[/MATH] and [MATH]2\times 12\equiv 2[/MATH], and so on: pairs of results are the same. (I'm assuming you meant this rather than squaring.)

What is the difference? Did you notice that 2 is a factor of 22, but 3 is not? How might this be relevant?
 
Dr P knew what I was going to write, wrote it before me and did a better job. The real questions helpers on this forum should be working on is how does he do it?
 
Sorry.

But this is a good example of a question that may be best answered by two or more people saying things in slightly different ways, because the OP may need slightly different wording to understand, according to how they are being taught. So please don't hold back!
 
Thanks you very much!
I have noticed that 3 and 22 are coprime= no common divisors.
2 and 22 on the other side, have common divisors, because 2*11=22. So their common divisor is 2.

Because they have a common divisor, you can do 2*11*x. If you take x as 3, we come to 0. If we take x=2, we come to 0...

Can I formulate this on a more common way?
And what about x^2,^3?

Thank you again
 
SherlockWante said:
Because they have a common divisor, you can do 2*11*x. If you take x as 3, we come to 0. If we take x=2, we come to 0...
Click to expand...
I'm not quite sure what you are saying here; 2*11 times anything is congruent to 0, so 2 and 3 are not different in this respect.

I would say that for any x, 2x and 2(x+11) are congruent (since they differ by 22), so you get pairs of distinct numbers whose doubles are congruent.

This can't be said of 3x; on the contrary, if 3x and 3y are congruent, then 3(x-y) must be a multiple of 22, which can only happen if x-y is a multiple of 22 -- that is, if x and y are themselves congruent.

The difference is, as you say, that 3 is coprime with 22, while 2 is not.
 
Dr.Peterson said:
, which can only happen if x-y is a multiple of 22 -- that is, if x and y are themselves congruent.
Click to expand...
Thank you very much! I just didn't understand what you meant by this part.
And what if I take the ^4 of the numbers? Can I also prove it in this way?
 
If a product ab, in this case 3(x-y), is a multiple of another number n, here 22, then each prime factor of n must be a factor of at least one of the factors a and b. Since the prime factors of 22 are 2 and 11, and neither is a factor of 3, they must both be factors of x-y -- that is, x-y is a multiple of 22. I presume you know that two numbers that differ by the modulus are congruent.

I'm not sure what you mean by "take the ^4 of the numbers". Do you mean raising x and y to the fourth power? And what do you want to prove by doing that?
 
What do you find when you try it? What do you think there is to prove about it?
 
Let’s see.
Normal x:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

x^2:
0 1 4 9 16 3 14 27 20 15 12 11 12 ...

We see here that 10*10 mod 22=12*12 mod 22
Because 12*12=10*10 +2*22
So in each x*x=y*y+a*22 , x and y are congruent.
x^2-y^2=a*22

If we take a=1
we have x^2-y^2=22
where x must be >y and x=sqrt(22+y^2)
this has no solution

if we take a=2
we have x^2-y^2=44
solution: x=12 and y=10
...

Am I correct?
And can I formulate this in a more general/formulated way?

Thank you very much in advance.
 
This reminds me of something I discovered, believe it or not, when I took the SAT (or one of those tests) in high school! In my free time after finishing, I tried graphing something like x^2 mod 22 (probably a smaller modulus) and saw some interesting patterns.

Do you see a specific reason why 10^2 and 12^2 might be congruent mod 22, related to the fact that 10 + 12 = 22?
 
Wow, that’s cool!

In each x*x=y*y+a*22 , x and y are congruent.
x^2-y^2=a*22

If (x+y)*(x-y) is a multiple of 22, then x and y are congruent.
Each prime factor of 22 must be a factor of at least one of the factors (x+y) and (x-y). The prime factors of 22 are 2 and 11.
In the case with 10 and 12, both 2 and 11 are factors of (x+y) and 2 is a factor of (x-y)?
Am I correct?
Thank you.
 
No, 10 and 12 are not congruent. But they do have a similar relationship. If they were congruent, their difference would be a multiple of 22. What is true instead?

Now think about the congruence [MATH]x^2\equiv y^2[/MATH]. What would you expect to be the solution, in terms of ordinary algebra?

Your last paragraph is correct; this is a useful technique for finding integer solutions of quadratic equations. For example, when you pointed out that x^2-y^2=22 has no (integer) solutions, I observed that if there were a solution, either x+y would be 22 and x-y = 1 (which is impossible because the sum and difference of two integers are either both odd or both even), or x+y=11 and x-y=2 (also impossible). You should consider negative numbers, too, but that adds nothing.
 
Dr.Peterson said:
If a product ab, in this case 3(x-y), is a multiple of another number n, here 22, then each prime factor of n must be a factor of at least one of the factors a and b. Since the prime factors of 22 are 2 and 11, and neither is a factor of 3, they must both be factors of x-y -- that is, x-y is a multiple of 22. I presume you know that two numbers that differ by the modulus are congruent.

I'm not sure what you mean by "take the ^4 of the numbers". Do you mean raising x and y to the fourth power? And what do you want to prove by doing that?
Click to expand...
And can I also formulate this in a more general way?
So like a*x ≡ a*y <=> ...? And is there maybe a proof of this?
 
SherlockWante said:
Thank you.
This is only true if x=y or x=-y?
Click to expand...
Yes. And so the reason [MATH]10^2\equiv 12^2[/MATH] is that [MATH]12\equiv -10[/MATH] -- that is, their sum is a multiple of 22.
SherlockWante said:
And can I also formulate this in a more general way?
So like a*x ≡ a*y <=> ...? And is there maybe a proof of this?
Click to expand...
This is a bigger topic to explore -- how to solve equivalences.

I don't think you've told us what you have learned about this topic, or what your current goal is. We could continue forever, now that you have opened up bigger questions; you're asking for a course in number theory, or at least a chapter! If you are taking such a course, the answer to this question should come soon. If you are not, find a book on the subject! My goal has been to whet your appetite for exploration, rather than to teach you the whole thing.
 
Thank you!
Since I don't know much about modulus, I don't want the proof anymore :D.
The second power seemed easy, but what if we go till the power of 5? Can I just say that x^5≡ y^5 <=> x≡ y
And if I say that, how can I proof that my statement is right? Thank you.
 
SherlockWante said:
Thank you!
Since I don't know much about modulus, I don't want the proof anymore :D.
The second power seemed easy, but what if we go till the power of 5? Can I just say that x^5≡ y^5 <=> x≡ y
And if I say that, how can I proof that my statement is right? Thank you.
Click to expand...
Try defining modulus carefully and going from there

a mod b, given a and b are positive integers, means

[MATH]a - \left ( b * \left \lfloor \dfrac{a}{b} \right \rfloor \right ).[/MATH]
Now start deducing theorems.
 
JeffM said:
Try defining modulus carefully and going from there

a mod b, given a and b are positive integers, means

[MATH]a - \left ( b * \left \lfloor \dfrac{a}{b} \right \rfloor \right ).[/MATH]
Now start deducing theorems.
Click to expand...
Thank you. Can you deduce this using like x^3 so that I can do it for x^5. Or can you give me tips how to start because I have no idea how to use your given info for a deduction.
 
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