How to make sure no numbers are the same?

While your adding numbers to the array, just check if the array already has the value, if so, then try again. This seems to do the trick.

#include <iostream>

using namespace std;

int main()
{
    int i, j, k, randomNumber[10];

    for(i = 0; i < 10; i++)
    {

            randomNumber[i] = rand() % 10 + 1;
            for (int q = 0; q < i; ++q)
            {
                    if (randomNumber[q] == randomNumber[i]) // Array already has value
                    {
                            --i; // Retry
                            break;
                    }
            }
    }

    for(k = 0; k < 10; k++)
    {
            cout << randomNumber[k] << ", ";
    }

    cin.get();
}

Hope this helps.

C or C++ program without functions is a nonsense.
The 1st step on the right way: try to select code fragments to define usefull functions, for example:

int isHere(const int array[], int nelem, int number)
{
    for (int i = 0; i < nelem; ++i)
        if (array[i] == number)
            return true;
    return false;
}

const int N = 10;

int main(int argc, char* argv[])
{
    int randomNumber[N];
    int r;

    for (int n = 0; n < N; ++n)
    {
        do {
            r = rand() % N + 1;
        } while (isHere(randomNumber,n,r));
        randomNumber[n] = r;
    }
    for (int i = 0; i < N; ++i)
        cout << i << ".\t" 
        << randomNumber[i] << "\n";
    cout << endl;

    return 0;
}

Fill the array with unique values - a loop can do this.

Then shuffle it (think deck of cards, a finite non-repeating set in a random order).
http://www.sgi.com/tech/stl/random_shuffle.html

The suffling is a nice solution.

However let's come back to the assignment. Why r = rand() % N + 1; (or randomNumber[i] = rand() % 10 + 1 as in original code)? Why all numbers in the array must be in 1..N range?

If we have another range, a shuffle method is no good.