how to know if my ideas didnt tought before?

[R.D. Silverman]

Originally Posted by blob100

I'll try proving again (I wish less confusing):
m, d(m) and a are defined.
To prove a is a real unity (1 or -1) we will prove it may not be deviding m or d(m).
If a|m, d(m) must be devided by a too, which does not fit it's definition.
If a|d(m), m must be devided by a too, which does not fit d(m)'s definition.
We base it on (d(m),m)=1. If these both (d(m),m) have a devisor a, it does not fit d(m),m)=1.
This gives a must be a real unity.

Try the following.

Let the units be x1, x2, ....xk where k = phi(m).

Consider the subset of these units such that each element has an
inverse that is different from itself. PAIR (all of) THESE.

Multiply them together. Their product is trivially equal to 1. i.e.
each unit times its inverse equals 1 by definition, so the product of
all the pairs must be 1.

[BTW, this result, when m is prime gives Wilson's Theorem; look it up]

Now consider the elements whose inverse is not different from themselves.
Thus, a = a^-1 mod m. What can you say about the product of any
two of these elements?

[R.D. Silverman]

Originally Posted by philmoore

Nice catch! At least someone is paying attention (not me...)

Ah. Total brain damage on my part. I had some examples in a file
and I cut/pasted together some incorrect parts of the file. Very careless.

Try: Units of Z/10Z are 1,3,7,9. 3*7 = 1 mod 10, so what is left is
1*9 = 9 mod 10.

[blob100]

Originally Posted by R.D. Silverman

Now consider the elements whose inverse is not different from themselves.
Thus, a = a^-1 mod m. What can you say about the product of any
two of these elements?

These numbers a which sutisfy the following are real unities right?
The product of 1 and -1 is -1.

[jyb]

Originally Posted by R.D. Silverman

Ah. Total brain damage on my part. I had some examples in a file
and I cut/pasted together some incorrect parts of the file. Very careless.

Try: Units of Z/10Z are 1,3,7,9. 3*7 = 1 mod 10, so what is left is
1*9 = 9 mod 10.

Sure, you gave that latter example in post #464 too. My point was that this is not total brain damage. It's just a mistake. All of us make them sometimes, whether students or advanced practitioners. We shouldn't be berated so mercilessly when it happens.

[R.D. Silverman]

Originally Posted by blob100

These numbers a which sutisfy the following are real unities right?
The product of 1 and -1 is -1.

Don't invent terminology. What is a "real unity"?

What do you mean by "satisfy the following" when you write 1* -1 = -1??

[R.D. Silverman]

Originally Posted by jyb

Sure, you gave that latter example in post #464 too. My point was that this is not total brain damage. It's just a mistake. All of us make them sometimes, whether students or advanced practitioners. We shouldn't be berated so mercilessly when it happens.

I disagree. It was total brain damage on my part because I failed
to check the data that I pasted into my message. This is such an
oversight that "brain damage" is appropriate.

[flouran]

Originally Posted by R.D. Silverman

I disagree. It was total brain damage on my part because I failed
to check the data that I pasted into my message. This is such an
oversight that "brain damage" is appropriate.

Unless you are actually brain damaged, then I disagree that brain damage is appropriate (not to mention inaccurate).

[blob100]

Originally Posted by R.D. Silverman

Don't invent terminology. What is a "real unity"?

What do you mean by "satisfy the following" when you write 1* -1 = -1??

When I said "sutisfy the following" I meant:
(1) Sutisfiy: Now consider the elements whose inverse is not different from themselves.
Thus, a = a^-1 mod m.
*You asked what is the product of these numbers.
(2)As I know, a unity is i=(-1)^(0.5) or 1 or -1.
When I said "real unity" I meant 1 or -1.
So, back to your question:
Numbers that agree the following (1) are the numbers: 1,-1.
Thier product is -1.

[R.D. Silverman]

Originally Posted by blob100

When I said "sutisfy the following" I meant:
(1) Sutisfiy: Now consider the elements whose inverse is not different from themselves.
Thus, a = a^-1 mod m.
*You asked what is the product of these numbers.
(2)As I know, a unity is i=(-1)^(0.5) or 1 or -1.

Not 'unity', as in 'root of unity'. The word is 'unit'. I already defined this
for you. And Shanks' book discusses them. A unit is an element
of a ring that has a multiplicative inverse.

When I said "real unity" I meant 1 or -1.
So, back to your question:
Numbers that agree the following (1) are the numbers: 1,-1.
Thier product is -1.

No. We are NOT discussing 'unity' (or whatever you think 'unity' might
be). I have said this multiple times: The units of the integers taken mod m,
where m is an integer greater than 1 are those integers that have a multiplicative inverse mod m. The question is then: What is the product
of all of the units taken mod m???

Even you gave examples showing that you know what units are. So why
are you confused now? The product of two units does not have to be
1 or -1. Consider the integers mod 10. The units are 1,3,7,9. Note
that 3*9 is 7 mod 10. This clearly isn't 1 or -1. On the other hand,
3 and 7 are inverses.

[blob100]

Originally Posted by R.D. Silverman

Not 'unity', as in 'root of unity'. The word is 'unit'. I already defined this
for you. And Shanks' book discusses them. A unit is an element
of a ring that has a multiplicative inverse.




No. We are NOT discussing 'unity' (or whatever you think 'unity' might
be). I have said this multiple times: The units of the integers taken mod m,
where m is an integer greater than 1 are those integers that have a multiplicative inverse mod m. The question is then: What is the product
of all of the units taken mod m???

Even you gave examples showing that you know what units are. So why
are you confused now? The product of two units does not have to be
1 or -1. Consider the integers mod 10. The units are 1,3,7,9. Note
that 3*9 is 7 mod 10. This clearly isn't 1 or -1. On the other hand,
3 and 7 are inverses.

I know the difference between unit and unity.
Unity- i, 1, -1.
Unit of a ring- For any given natural number n, g is a unit of n if and only if g is natural <n, (n,g)=1.
I'm deffinietly not confused (maybe just because of not understanding what question am I needed to answer now),
Your last question was:
What is the product of natural numbers a which satisfy:
a=a^(-1)(mod m).
We easily mention that these numbers are equal to thier inverse.
These may be 1,-1 (I called these "real unities" becuase these are real and unities, I'm sorry for calling these like that, becuase of wasting too much time for it).

Can you please give me a hint how to explain why is d(m) modulo m is allways 1 or -1?

I want again to show my work:
If "a" which is d(m) modulo m (the residue) isn't 1 or -1, it may be devide d(m) or m.
If it devide any of them, it must be deviding the other one too, which is out of the definition of d(m).

My confusion come by not understanding the definition of inverse here:
"On the other hand,
3 and 7 are inverses. "
Inverses of what? Do you mean, a number b^(-1) for any real b?
Is there another meaning here by saying inverse?

[R.D. Silverman]

Originally Posted by blob100

<snip>

These may be 1,-1 (I called these "real unities" becuase these are real and unities, I'm sorry for calling these like that, becuase of wasting too much time for it).

Stop inventing terminology.
And your statement: "(I called these "real unities" becuase these are real and unities, ) is a misuse of existing terminolgy. 1 and -1 are units of Z.
In fact, they are the only units. Stop inserting extraneous words into
(i.e. the word 'real', as in 'real unities') your discussion because they
make it appear that you are just confused.


Elements of a ring that are equal to their
own inverse do not have to be 1 or -1. Why do you think that they are?
Consider m = (say) 24. 5 is its own inverse, yet 5 certainly is not 1 or -1.

Can you please give me a hint how to explain why is d(m) modulo m is allways 1 or -1?

For about the (3rd? 4th? 5th?) time: PAIR THE UNITS WITH THEIR
INVERSES.

I want again to show my work:
If "a" which is d(m) modulo m (the residue) isn't 1 or -1, it may be devide d(m) or m.
If it devide any of them, it must be deviding the other one too, which is out of the definition of d(m).

My confusion come by not understanding the definition of inverse here:
"On the other hand,
3 and 7 are inverses. "
Inverses of what? Do you mean, a number b^(-1) for any real b?
Is there another meaning here by saying inverse?

Stop adding words!!!!! (as in 'real' b). Do not use the word real
in a mathematical context unless you are in fact discussing the REAL
NUMBERS. We are discussing the units of the integers taken mod m.
These are the ONLY numbers under discussion. Why do you keep
babbling about "real" numbers? In R, every number except 0 is a unit.

I made it clear at the very beginning of this discussion what an
inverse is. It is a MULTIPLICATIVE INVERSE. How can this possibly be
unclear? The definition of a multiplicative inverse is basic, pre-algebra
arithmetic.