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Mathematical Proof To....something


W_L

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Weird blog entry, I was just trying to figure something in my head and a sequence of equations came to my mind.

 

1 x 2 +3
3 x 2 - 1

 

Both would equal 5 right.

 

So I thought, pretty cool coincidence, why not test it out on other sequential numbers:

 

2 x 3 +4
4 x 3 -2

 

Equal 10

 

3 x 4 + 5
5 x 4 - 3

 

Equals 17

 

4 x 5 +6
6 x 5 - 4

 

Equals 26

 

5 x 6 + 7
7 x 6 -5

 

Equals 37

 

6 x 7 + 8
8 x 7 -6

 

Equals 50

 

....

 

It keeps working out

 

I know this is a mathematics proof of some kind, but what did I just stumble across

 

set "a" multiplied by following set "b" plus following set "c" = set "c" multiplied by preceding set "b" minus preceding set "a"

 

a x b + c = c x b - a as long as it is a whole set sequence

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I had never thought about this... :2thumbs:

 

P.S. Are you trying to hit on Drew using a maths theory? :gikkle:

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What you have is two equation:

 

n x (n+1) + (n+2)

 

(n+2) x (n+1) - n

 

Both of these simplify to:

 

n x n + 2 x n + 2

 

So, yes, they'll always be equal. Basic mathematics :)

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a*b + c = c*b - a

 

A lovely pattern W_L and quite easy to prove.

In fact you were almost there.

 

All you have to realise is that in your pattern b is the same as a+1 and c is the same as a+2.

Rewrite the left side of your equation as :

a*(a+1) +(a+2)

Expand and you have a*a +a + (a+2) which is a*a +2a +2

 

Now do the same with the right side:

(a+2)*(a+1) - a

Expand and you have a(a+1) +2(a+1) - a  which is a*a + a + 2a +2 -a which is a*a +2a + 2

 

Lo and behold , we've just shown that the left side is exactly the same as the right side.

In other words, as long as the three numbers are consecutive your pattern is always true.

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Well it's a proof to the following equation: x2+1

 

Observe:

a= x-1

b= x

c= x+1

 

So let's substitute the variables in your equation, a*b+c=c*b-a with variables in terms of x.

 

That gives us: (x-1)(x)+(x+1)=(x+1)(x)-(x-1)

 

When we simplify: x2+1=x2+1

 

 

That was fun! :D

 

Edit: Didn't see Palantir's or Graeme's explanations before I posted, which are a far better explanations than my own in my opinion. :)

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Well it's a proof to the following equation: x2+1

 

Observe:

a= x-1

b= x

c= x+1

 

So let's substitute the variables in your equation, a*b+c=c*b-a with variables in terms of x.

 

That gives us: (x-1)(x)+(x+1)=(x+1)(x)-(x-1)

 

When we simplify: x2+1=x2+1

 

 

That was fun! :D

 

Edit: Didn't see Palantir's or Graeme's explanations before I posted, which are a far better explanations than my own in my opinion. :)

 

 

So I've proven that Left and Right can equal out in perfect harmony, at least in this instance

 

"Wink" Drew "Wink"

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What you have is two equation:

 

n x (n+1) + (n+2)

 

(n+2) x (n+1) - n

 

Both of these simplify to:

 

n x n + 2 x n + 2

 

So, yes, they'll always be equal. Basic mathematics :)

Put  in the familiar form used in US text books, a^2 + 2ab + b^2  and I forget the name of that equation

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Put in the familiar form used in US text books, a^2 + 2ab + b^2 and I forget the name of that equation

That's ( a + b )^2 but that's not the case here because n^2+2n+2 isn't a perfect square.

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Let me just finish the proof:

 

a x b + c = c x b -a

 

b= a +1

c= a + 2

 

Expressed as a

 

a(a + 1) + a + 2 = (a + 2)(a + 1) - a

a^2 + a + a + 2 = a*2 + a + 2a + 2 -a

a^2 + 2a + 2= a^2 +2a +2

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You can also get the answer like this:

 

The equations as

 

a×b+c and

c×b-a

 

As the three are consecutive numbers, their solutions are equal to either

 

c^2 - 2b or

a^2 + 2b

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