Maths proofs Watch

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blag dahlia112
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#1
Report Thread starter 15 years ago
#1
are these correct?
1.)
Prove that 2 consecutives numbers multiplied are even...
(2n) (2n+1) = 4n^2 + 2n

and since 2n + 2n proves that 2 evens added must be even and both the number in the above answer are even (both divide by 2) so will be even.

2.)
Prove that any 3 consecutive numbers are divisible by 3?
(x)(x+1)(x+2) = 3(x+1)

if the second one is correct then how do you get 3(x+1) from factorising the brackets?
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Quimbly
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#2
Report 15 years ago
#2
First of all, I assume by "numbers", you mean integers... ?

-------------------------------------------------------------
1.)
Prove that 2 consecutives numbers multiplied are even...
(2n) (2n+1) = 4n^2 + 2n
-------------------------------------------------------------

This proves that any number multiplied by 2 and multiplied by the consecutive number to that is even. Big deal!

If you can't come up with a nice, simple, elegant solution, go the brute force approach:

Take any integer n
Case 1:
Assume n is odd (i.e. there is no integer i such that n = 2i)
It follows then that (n+1) is even (i.e. there is an integer j such that (n+1) = 2j)

So, write it out now: (n)(n+1) = (n)(2j) = 2(nj)

Case 2: Assume n is even (and therefore n+1 is odd), and do the same thing as in case 1.

Bam; You're done!
To simplify it, you could just assert that for any two consecutive integers, one of them must be even. And follow it up with proving that an even number multiplied by an odd number is always an even number.

-------------------------------------------------------------
Prove that any 3 consecutive numbers are divisible by 3?
(x)(x+1)(x+2) = 3(x+1)
-------------------------------------------------------------

Yes, you want to get it of the form 3(n).

This question is similar to the first. For any three cosecutive integers, one of them must be a multiple of 3. I'll leave it up to you, if you need to prove it.

Case 1:
Assume x is a multiple of 3; x = 3i
Then...
(x)(x+1)(x+2) = 3i(x+1)(x+2) = (3ix + 3)(x+2) = 3ix^2 + 6ix + 3x + 6
= 3(ix^2 + 2ix + x +2)

Case 2:
Assume (x+1) is a multiple of 3; (x+1) = 3j
Then...
(x)(x+1)(x+2) = (x)(3j)(x+2) = (3jx)(x+2) = 3jx^2 + 6jx
= 3(jx^2 + 2jx)

Case 3:
Assume (x+2) is a multple of 3; (x+2) = 3k
Then...
(x)(x+1)(x+2) = (x)(x+1)(3k) = (x^2 + x)(3k) = 3kx^2 + 3kx
= 3(kx^2 + kx)

Voila! You're done.

So, you can see where this is headed.... How about multiplying 4 consecutive integers? Would it be a multiple of 4? Try and solve it using a similar method.
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It'sPhil...
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#3
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#3
(Original post by Quimbly)
First of all, I assume by "numbers", you mean integers... ?

-------------------------------------------------------------
1.)
Prove that 2 consecutives numbers multiplied are even...
(2n) (2n+1) = 4n^2 + 2n
-------------------------------------------------------------

This proves that any number multiplied by 2 and multiplied by the consecutive number to that is even. Big deal!
Actually I would say this approach is sufficient as clearly two consequtive numbers multiplied together will take the form even x odd. Your method is long winded and not neccesary. Also pretty obviously if you do it mod 2 any even number = 0 (mod 2) hence rhs=lhs=0 hence 2 consecutives numbers multiplied are even
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meepmeep
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#4
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#4
(Original post by It'sPhil...)
Actually I would say this approach is sufficient as clearly two consequtive numbers multiplied together will take the form even x odd. Your method is long winded and not neccesary. Also pretty obviously if you do it mod 2 any even number = 0 (mod 2) hence rhs=lhs=0 hence 2 consecutives numbers multiplied are even
To further this method, if you have n consecutive numbers and write them all to mod n, then you will get 0,1,2,3,4.....n-1. If you multiply them together, you get 0(mod n) as any number multiplied by 0 is 0. Hence the product is 0(mod n) and so is divisible by n.
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Quimbly
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#5
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#5
Like I said, if you can't come up with a simple solution, or using an advanced method (such as moduli), then "brute force" is the way to go. I was just trying to help the guy out.

As for
(2n) (2n+1) = 4n^2 + 2n ...

Yes, I have seen this form before, but it isn't really accurate, as not every integer n can be expressed in the form (2i). It's trivial, but true. For it to work, you have to assert that the first integer in your sequence is even. And then we're not dealing with ANY sequence of two integer, only sequences that start with even numbers.

Nit-pick, nit-pick!
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It'sPhil...
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#6
Report 15 years ago
#6
(Original post by Quimbly)
Like I said, if you can't come up with a simple solution, or using an advanced method (such as moduli), then "brute force" is the way to go. I was just trying to help the guy out.

As for
(2n) (2n+1) = 4n^2 + 2n ...

Yes, I have seen this form before, but it isn't really accurate, as not every integer n can be expressed in the form (2i). It's trivial, but true.
For it to be true, you have to assert that your first integer in your sequence is even. And then we're not dealing with ANY sequence of two integer, only sequences that start with even numbers.

Nit-pick, nit-pick!
But any ordered pair of integers (a,b) can be expressed in the form (2n, 2n+1) or (2m-1, 2m) hence one is always a multiple of 2
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meepmeep
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#7
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#7
(Original post by Quimbly)
Like I said, if you can't come up with a simple solution, or using an advanced method (such as moduli), then "brute force" is the way to go. I was just trying to help the guy out.

As for
(2n) (2n+1) = 4n^2 + 2n ...

Yes, I have seen this form before, but it isn't really accurate, as not every integer n can be expressed in the form (2i). It's trivial, but true. For it to work, you have to assert that the first integer in your sequence is even. And then we're not dealing with ANY sequence of two integer, only sequences that start with even numbers.

Nit-pick, nit-pick!
Hey, nothing you said was wrong. Though it may seem long, all the relevant stuff was there. I'm not trying to compete with you. Just offering an alternative and how it extends into a more general situation.
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Quimbly
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#8
Report 15 years ago
#8
(Original post by It'sPhil...)
But any ordered pair of integers (a,b) can be expressed in the form (2n, 2n+1) or (2m-1, 2m) hence one is always a multiple of 2
Yes, you missed my point.. Oh well..
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Quimbly
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#9
Report 15 years ago
#9
(Original post by meepmeep)
Hey, nothing you said was wrong. Though it may seem long, all the relevant stuff was there. I'm not trying to compete with you. Just offering an alternative and how it extends into a more general situation.
Thanks meep,
Your answer was accurate, relevant and extensible. No complaints from me.
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