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(64m-3n)*7n*3m = 1344m2n - 63mn2

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Q: How do you multiply out 64m-3n 7n 3m?
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What is 9m plus 6n-3m plus 7n?

This question as written can be taken two ways: 9m + 6n - 3m + 7n = 6m + 13n, or (9m + 6n) - (3m + 7n) = 9m + 6n - 3m - 7n = 6m - n


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Is the root of 7 irrational?

Yes, and here's the proof: Let's start out with the basic inequality 4 < 7 < 9. Now, we'll take the square root of this inequality: 2 < √7 < 3. If you subtract all numbers by 2, you get: 0 < √7 - 2 < 1. If √7 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √7. Therefore, √7n must be an integer, and n must be the smallest multiple of √7 to make this true. If you don't understand this part, read it again, because this is the heart of the proof. Now, we're going to multiply √7n by (√7 - 2). This gives 7n - 2√7n. Well, 7n is an integer, and, as we explained above, √7n is also an integer; therefore, 7n - 2√7n is an integer as well. We're going to rearrange this expression to (√7n - 2n)√7, and then set the term (√7n - 2n) equal to p, for simplicity. This gives us the expression √7p, which is equal to 7n - 2√7n, and is an integer. Remember, from above, that 0 < √7 - 2 < 1. If we multiply this inequality by n, we get 0 < √7n - 2n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √7p < √7n. We've already determined that both √7p and √7n are integers, but recall that we said n was the smallest multiple of √7 to yield an integer value. Thus, √7p < √7n is a contradiction; therefore √7 can't be rational, and so must be irrational. Q.E.D.


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