(a-b)(a-b+c) = (b-a)(a-b+c)
(b-a)(c-b)(d-c) (a-b)(b-c)(c-d)
Can this be proven?
My opinion is that the part in bold is incorrect.
Please verify.
Thank-you!
(a-b)(a-b+c) = (b-a)(a-b+c)
(b-a)(c-b)(d-c) (a-b)(b-c)(c-d)
Can this be proven?
My opinion is that the part in bold is incorrect.
Please verify.
Thank-you!
It is quite simple.
Use the following fact
x - y = -1 * (y - x)
The result is as follows
(a-b)(a-b+c) -1*(b-a)(a-b+c)
----------------- = ------------------------------------
(b-a)(c-b)(d-c) [-1*(a-b)][-1*(b-c)][-1*(c-d)]
-1 * (b-a)(a-b+c) (b-a)(a-b+c)
= -------------------------- = ------------------
(-1)^{3} * (a-b)(b-c)(c-d) (a-b)(b-c)(c-d)
The last step works because -1 / (-1)^{3} = 1.
Comments
That's pretty much it.
(-1)^{n} = 1 if n is even and -1 if n is odd.
Using this you get -1 / (-1)^{3} = -1 / -1 = 1.
Comment