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Old 2005-03-10, 03:20   #1
clowns789
 
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Default 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.
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Old 2005-03-10, 03:46   #2
amateur
 
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Your are right. Here is why:
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Old 2005-03-10, 04:06   #3
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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.
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Old 2005-03-10, 04:57   #4
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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
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Old 2005-03-10, 20:32   #5
ewmayer
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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.
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Old 2005-03-11, 00:23   #6
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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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