natural. even number. like this?

[jones1223]

Hi. m is a natural number. And I just want to show that m^2 + m is an even number, that
is "evenly divisible" by 2.

I tried like this,

m = 2m
m^2 + m
2m + 1

if m is 1; 1^2 + 1 = 2
if m is 2; 2^2 + 2 = 6
if m is 3; 3^2 + 3 = 12
if m is 4; 4^2 + 4 = 20

2,6,12 and 20 ... are all even numbers.

Is this a correct way to prove that m^2 + m is an even number??

If it isn't, or something is missing, then can you please tell me. I have really tried.

 

[Mrspi]

Hi. m is a natural number. And I just want to show that m^2 + m is an even number, that
is "evenly divisible" by 2.

I tried like this,

m = 2m
m^2 + m
2m + 1

if m is 1; 1^2 + 1 = 2
if m is 2; 2^2 + 2 = 6
if m is 3; 3^2 + 3 = 12
if m is 4; 4^2 + 4 = 20

2,6,12 and 20 ... are all even numbers.

Is this a correct way to prove that m^2 + m is an even number??

If it isn't, or something is missing, then can you please tell me. I have really tried.

You have showed that m[sup:1iu862ff]2[/sup:1iu862ff] + m is even for SOME natural numbers (1, 2, 3, and 4), but this does not PROVE that m[sup:1iu862ff]2[/sup:1iu862ff] + m is even for EVERY natural number.

Let m be a natural number.

m[sup:1iu862ff]2[/sup:1iu862ff] + m = m(m + 1)

Now, suppose m is EVEN. Then m = 2n for some natural number n, and
m(m + 1) = 2n(2n + 1)

Clearly this has a factor of 2, so m(m + 1) is even.

Suppose m is ODD. Then m = 2n + 1 for some natural number n, and
m(m + 1) = (2n + 1)[(2n + 1) + 1], or
m(m + 1) = (2n + 1)(2n + 2)
m(m + 1) = 2(n + 1)(2n + 1)
And the right side has a factor of 2, so m(m + 1) is even.

Since every natural number is either even or odd, and regardless of whether we started with an even number or an odd number for m, it turns out that m[sup:1iu862ff]2[/sup:1iu862ff] + m is an even number.

 

[Deleted member 4993]

Hi. m is a natural number. And I just want to show that m^2 + m is an even number, that
is "evenly divisible" by 2.

I tried like this,

m = 2m ? that is a false statement for all m <> 0
m^2 + m
2m + 1

if m is 1; 1^2 + 1 = 2
if m is 2; 2^2 + 2 = 6
if m is 3; 3^2 + 3 = 12
if m is 4; 4^2 + 4 = 20

2,6,12 and 20 ... are all even numbers.

Is this a correct way to prove that m^2 + m is an even number?? NO .... you have not proven your case for m= 97 or m= 2567

If it isn't, or something is missing, then can you please tell me. I have really tried.

Have you been taught proof by induction?

Another way to prove it.

If two consecutive natural numbers are chosen - one of those must be even.

and

The product of any natural number with an even number - is even.

So

(m) and (m+1) are two consecutive numbers.

One of those must be even.

m*(m+1) must be even.

 

[jones1223]

Thanks!! Great. But just one question; why there is also n, and not just m?


well, I have another. Can I ask another?

Let n > 3 be a natural number. If n^2 + 2 is a prime number, then n is dividable
with number 3 - I want to prove this.

Here's my solution;

let's try with numbers which are dividable with 3.
6,9,12,15,18,21 and so on.

6^2 + 2 = 38 . but 38 isn't a prime number.
9^2 + 2 = 83 . this is.

So, 83 is a prime number and 9 is dividable with 3.

12 doesn't produce a prime number.

also 15^2 + 2 = 227, is a prime number and 15 is dividable with 3.
18 doesn't produce a prime number.
21 does produce and it is dividable with 3.

I don't know why some of these ( 6,9,12,15,18,21 and so on. )
numbers + 2 produce a prime number but not all.

But still, I think this is quite good solution? ( hopefully! )

 

[jones1223]

Thanks!! Great. But just one question; why there is also n, and not just m? I'm not sure.
Sorry, I ask again. But I want to know, please.


quote;
You have showed that m[sup:1tuhibxu]2[/sup:1tuhibxu] + m is even for SOME natural numbers (1, 2, 3, and 4), but this does not PROVE that m[sup:1tuhibxu]2[/sup:1tuhibxu] + m is even for EVERY natural number.

Let m be a natural number.

m[sup:1tuhibxu]2[/sup:1tuhibxu] + m = m(m + 1)

Now, suppose m is EVEN. Then m = 2n for some natural number n, and
m(m + 1) = 2n(2n + 1)

Clearly this has a factor of 2, so m(m + 1) is even.

Suppose m is ODD. Then m = 2n + 1 for some natural number n, and
m(m + 1) = (2n + 1)[(2n + 1) + 1], or
m(m + 1) = (2n + 1)(2n + 2)
m(m + 1) = 2(n + 1)(2n + 1)
And the right side has a factor of 2, so m(m + 1) is even.

Since every natural number is either even or odd, and regardless of whether we started with an even number or an odd number for m, it turns out that m[sup:1tuhibxu]2[/sup:1tuhibxu] + m is an even number.[/quote]