1.for any 2 non-zero integers x and y there exist a unique integer q and r called the quotient and remainder respectively such that x=yq+r where <=r|y|

You might have gotten help by now if you had simply put "Math" in the School Subject box. Please follow directiions.

m|xy => m|x or m|y

not and

2|6*7

the first one has a typo, aNd google will provide many proofs of the division algorithm, and euclidean division

Sure! Let's go through each question step by step:

1. For any two non-zero integers x and y, there exists a unique integer q and r called the quotient and remainder respectively, such that x = yq + r, where -|y| ≤ r < |y|.
- To prove this, we use the Division Algorithm. First, divide x by y: x = q1 * y + r1, where q1 is the quotient and r1 is the remainder.
- Now, if r1 = 0, then the quotient q1 is the unique integer q and r = 0 is the remainder, so the statement holds true.
- Otherwise, consider the case where r1 ≠ 0. We can rewrite the equation as x - q1 * y = r1.
- Now, we can divide r1 by y (since y ≠ 0) and get: r1 = q2 * y + r, where q2 is the quotient and r is the remainder.
- Substituting r1 in the equation: x - q1 * y = q2 * y + r. Rearranging this gives: x = (q1 + q2) * y + r.
- Here, (q1 + q2) is the new quotient, let's call it q. So we have x = q * y + r, where q is an integer and r is the remainder.
- Now, we need to prove the uniqueness of q and r. Suppose there are two sets of values (q1, r1) and (q2, r2) satisfying the condition. Then, we have:
x = yq1 + r1 and x = yq2 + r2.
- Subtracting the two equations, we get: 0 = y(q1 - q2) + (r1 - r2). Now, since x and y are given constants, this equation implies that q1 - q2 = 0 and r1 - r2 = 0.
- Therefore, q1 = q2 and r1 = r2, showing the uniqueness of the quotient and remainder.

2. If m is a prime integer and m is divisible by xy, then m is also divisible by x and y separately.
- To prove this, we assume m/xy, which means that m is divisible by xy, i.e., there exists an integer k such that m = k * xy.
- Now, we need to show that m is divisible by x and y individually.
- Assume m does not divide x or y, i.e., there is no integer p such that m = p * x or m = p * y.
- Since m = k * xy and x does not divide m, x must divide y. Let's write y = xq, where q is an integer.
- Substituting y = xq in the equation m = k * xy, we get: m = k * x * (xq). Simplifying this further gives: m = (k * x^2) * q.
- Here, (k * x^2) is an integer, let's call it r. So we have m = r * q. Thus, m is divisible by q.
- But since q = y/x, we can rewrite it as q = p * y, where p is an integer. Therefore, we have m = r * (p * y).
- This means m is divisible by y also, which contradicts our assumption that m does not divide y.
- Hence, our assumption was incorrect, and it is proved that if m is divisible by xy, it must also be divisible by x and y individually.

I hope that helps! Let me know if you need any further clarification.