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5 answers

"If n is a perfect square, then n is not the product of 2 and an odd number."

Remember that an odd number can be expressed in the form
n = 2k + 1 (for an integer k).

Let's try proving this by contradiction. Assume that n IS the product of 2 and an odd number. That is, assume

n = 2(2k + 1). Then

n = 4k + 2

Since n is a perfect square, it follows that sqrt(n) is an integer. Therefore, sqrt(4k + 2) is an integer. Factoring out a 4,

sqrt(4[k + 2/4])

Reducing

sqrt(4[k + 1/2])

Splitting up the square root into 2,

sqrt(4)sqrt(k + 1/2)

2sqrt(k + 1/2)

This means k + (1/2) is a perfect square. This is impossible (the square root of non-integers are non-integers), and this is a contradiction.

2007-03-14 18:51:14 · answer #1 · answered by Puggy 7 · 3 0

Let n be a perfect square. The square root of n, call it q, is either even or odd.

If q is odd, then n is odd and obviously cannot be a product of two and any integer.

If q is even, then it has 2 as a factor; since n is equal to 2 squared, it has 2 as a factor at least twice. (That is, you can divide it by 2 at least twice). Therefore, when you divide n by two, you get another even number, not an odd number.

In neither case can n=2x, with x being an odd number.

2007-03-15 02:25:19 · answer #2 · answered by William S 3 · 0 0

well, if n is a perfect square, it can be 0 1 4 9 16 ...
the product of 2 * odd number = x
if u want x to equal to n, then x has to also be a perfect square
since x is 2 times sumdin, the sumdin HAS to be 2 also, or else x wont be a perfect square, but 2 is not an odd number, thatz y n cant be product of 2 and an odd number

2007-03-15 01:45:10 · answer #3 · answered by wenzhengsf 3 · 0 0

There are two cases. Let n = z^2.
- If n is odd, then it obviously is not the product of 2 with anything.
- If n is even, then z is even: it is the product of 2 and some number k, and n is 4k*2. For n to be the product of 2 and an odd number, 2k*2 would have to be odd, which it clearly is not.

2007-03-15 02:43:07 · answer #4 · answered by Anonymous · 0 0

If 2 is a factor of a perfect square, then there must be a matching 2 amongst the other factors, thus prohibiting the product of the other factors from being an odd number.

2007-03-15 02:04:15 · answer #5 · answered by Helmut 7 · 1 0

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