mersenneforum.org Multiplication Tendency
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 2005-03-10, 03:20 #1 clowns789     Jun 2003 The Computer 3·7·19 Posts Multiplication Tendency This seems to work all the time. Square an integer. i.e. 2^2=4. Now, take the integer below it (1) and above it (3) and multiply it together. You get 3 and it is one less than the previous integers. So in other words... x^2= [(x-1)*(x+1)]-1. Is this already known? It seems quite simple, yet true.
 2005-03-10, 03:46 #2 amateur     Aug 2004 22×5 Posts Your are right. Here is why: Attached Thumbnails   Last fiddled with by amateur on 2005-03-10 at 03:49
 2005-03-10, 04:06 #3 axn     Jun 2003 3×11×157 Posts The basic identitiy (a+b)(a-b) = a^2-b^2 is at work here. Set a=x, b=1, and rearrange the terms, you'll get your result.
 2005-03-10, 04:57 #4 Mystwalker     Jul 2004 Potsdam, Germany 14778 Posts Personally, I like the formula a*b = ((a+b)/2)² - ((a-b)/2)² It looks complicated at first glance, but it helps me doing quick multiplication (mentally) in certain situations. Example: 21*19 = 20² - 1² = 399 22*18 = 20² - 2² = 396
2005-03-10, 20:32   #5
ewmayer
2ω=0

Sep 2002
República de California

1167410 Posts

Quote:
 Originally Posted by clowns789 x^2= [(x-1)*(x+1)]-1. Is this already known?
I think you mean x^2= [(x-1)*(x+1)]+1.

2005-03-11, 00:23   #6
clowns789

Jun 2003
The Computer

3×7×19 Posts

Quote:
 Originally Posted by ewmayer I think you mean x^2= [(x-1)*(x+1)]+1.
Yes, I do. I was making it up as I went along and thought about that this morning.

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