# Modulus of negative numbers

1. Jan 23, 2012

### EngWiPy

One of the discrete Fourier transform (DFT) properties (symmetry property) is that:

$$x^*[-n]_N \stackrel{DFT}{\leftrightarrow}X^*[k]$$

where * means conjugate, and $$[.]_N$$ means modulus N. What is the meaning of modulus of negative numbers?

2. Jan 24, 2012

### chiro

Hey S_David.

When you say modulus do you mean the standard definition found in whole number arithmetic?

If this is the case, then the normal definition applies. Basically it is the lowest remainder given dimension of some number by a whole number N.

For negative numbers, this means that you have to start from a multiple that is less than or equal to a multiple of N.

So lets say we have a number -8 and our modulus argument is 9, then the answer is going to be 1 since -9 is a factor of 9 and one more than -9 is -8. Basically we apply the same decomposition theorem of n = pq + r where r is the modulo result, n is our input to decompose and q is your 'N' in this case and p is a whole number which in the negative case is a negative number.

Is this what you are wondering about? I get a feeling it may not be since you are asking about things in the context of fourier transforms.

3. Jan 24, 2012

### EngWiPy

Yeah, I asked for standard definition. May be there is a physical meaning in DFT, but I wanted to know about the negative numbers in general.

Thanks

4. Jan 25, 2012

### HallsofIvy

A negative number, x, mod n, is defined in exactly the same way as for a positive number: write x= mn+ r where $0\le r< n$. Then the modulus is r.

For example, to find "-18 mod 5", I note that 3(5)= 15< 18< 20= 4(5). That is -18= -20+ 2 so "-18 mod 5" is 2.

Notice that to find "18 mod 5", I would start the same but write 18= 15+ 3 so "18 mod 5" is 3. It is the fact that r must be non-negative that is key.

Of course. 2+ 3= 5= 0 mod 5. Since -18 and 18 are "additive inverses", so must they be "mod 5".

So another way to find "-x mod n" is to find "x mod n" and subtract that from n.

5. Jan 25, 2012

### EngWiPy

Thanks HallsofIvy, that helped a lot.