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## #1 2005-08-26 09:43:27

mikau
Super Member

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### factor theoroms

I made a small oberservation which is really obvious but I never quite realized before and it has helped me to simplify quicker once I've solved for the values of some variables.

if xy = 1, then x and y are reciprocals of eachother, so if you know x = 2 you know y = 1/2. Of course solving for x or y does the same thing but its slightly faster to just take the reciprocal.

Likewise, if xy = 5, then x = 5 times the reciprocal of y and y = 5 times the reciprocal of x.  Of course you can just solve for x or y but I find it faster to simply take the reciprocal and multiply times 5.

So obviousy if xy= n then x = n/y and y = n/x      but if I think of it as x = 1/y * n, or y = 1/x*n its slighty faster to do in my head, for me at least.

Likewise, if xyz = 1, then the product of 2 of the variavles equals the reciprocal of the other, and I think if xyz = n, then the product of 2 of the variables will equal n times the reciprocal of the other.

I find its just good to be aware of this, it tends to speed things up.

A logarithm is just a misspelled algorithm.

## #2 2005-08-26 10:45:31

kylekatarn
Power Member

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### Re: factor theoroms

that's a pretty obvious, but interesting observation

## #3 2005-08-27 02:08:20

mikau
Super Member

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### Re: factor theoroms

Precisly. But the obvious observations are the ones that are most often overlooked. For me at least. lol.

A logarithm is just a misspelled algorithm.

## #4 2005-08-27 08:15:25

MathsIsFun

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### Re: factor theoroms

Inverses in Mathematics:

Multiply/Divide
Powers/Roots
Integrate/Differentiate

If you can do something in maths, then someone is bound to say "can you go backwards?"

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman