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Archy [21]
2 years ago
9

N is a positive integer

Mathematics
1 answer:
Murrr4er [49]2 years ago
6 0

Part (1)

n is some positive integer. Let's say for now that n is even. So n = 2k, for some integer k

This means n-1 = 2k-1 is odd since subtracting 1 from an even number leads to an odd number.

Now multiply n with n-1 to get

n(n-1) = 2k(2k-1) = 2m

where m = k(2k-1) is an integer

The result 2m is even showing that n(n-1) is even

------------

Let's say that n is odd this time. That means n = 2k+1 for some integer k

And also n-1 = 2k+1-1 = 2k showing n-1 is even

Now multiply n and n-1

n(n-1) = (2k+1)(2k) = 2k(2k+1) = 2m

where m = k(2k+1) is an integer

We've shown that n(n-1) is even here as well.

------------

So overall, n(n-1) is even regardless if n is even or if n is odd.

Either n or n-1 will be even. If you multiply an even number with any number, the result will be even.

=======================================================

Part (2)

n is some positive integer

2n is always even since 2 is a factor of 2n

2n+1 is always odd because we're adding 1 to an even number. The sequence of integers goes even,odd,even,odd, etc and it does this forever.

-----------

Another way to see how 2n+1 is odd is to divide 2n+1 over 2 and you'll find that we get (2n+1)/2 = 2n/2+1/2 = n+0.5

The 0.5 at the end is not an integer, so there's no way that (2n+1)/2 is an integer; therefore 2n+1 is odd.

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