P = V * ( V / R ) = V2 / R

[LarryFine]

Certainly some law must govern this?

Yes: inside quotes first. Still, try it multiple ways and see whether you get the same results.

 

[wwhitney]

Kind of. I always assumed that you multiply both numbers in the brackets by the one outside the brackets.

No, you may be thinking of this:

a * (b + c) = (a * b) + (a * c ) (distributive law of multiplication)

The parentheses on the right hand side are usually omitted, as it is understood that you group multiplication operations together before doing addition operations.

With multiplications only, the order doesn't matter:

a * b = b * a (commutative law)
a * (b * c) = (a * b) * c (associative law)

And so again, when multiplying more than two numbers, the parentheses are usually omitted, because both options give the same result.

Division throws a wrench into things, order starts to matter. I think of division as "take the reciprocal and then multiply". So if you make that explicit (using Recip(a) to mean 1/a, the reciprocal of a) and keep track of which number get the reciprocal, it all works out:

a * (b / c ) = a * (b * Recip(c)) = (a * b) * Recip(c) = (a * b) / c

Because of the above, it's safe to write a * b / c, there's no ambiguity. But something like a * b / c * d should be avoided, as it is not clear if you mean a * (b / c) * d or a * b / (c * d), which are different.

Lastly, Recip(b * d) = Recip(b) * Recip(d), so you have the general rule for multiplying two fractions of "multiply the numerators and multiply the denominators":

(a / b ) * (c / d) = a * Recip(b) * c * Recip(d) = a * c * Recip(b) * Recip (d) = a * c * Recip(b * d) = (a * c) / (b * d)

Wayne

 

[oldsparky52]

Kind of. I always assumed that you multiply both numbers in the brackets by the one outside the brackets.

I understand. I have always had a hard time with factoring and distributing. V/R is not two (so no both) numbers, it is one number that is obtained by dividing V by R. Same for V standing alone, it is a number that is obtained by dividing V by 1. If the inside of the parentheses was V+R instead of V/R then I believe you are correct that you multiply the outside V to both numbers, so it would be V(V+R) would be V²+VR (I think, ) Of course that's not the equation we are working with.

 

[Jerramundi]

@wwhitney hitting us with the actual mathematical rules and their names. Well done sir.
Pythagorean what?! It's the 3-4-5 rule!

I once corrected an elderly sparky I was working for, letting him know it's called the Pythagorean Theorem. He looked at me with so much hate in his eyes, lmfao

 

[mbrooke]

No, you may be thinking of this:

a * (b + c) = (a * b) + (a * c ) (distributive law of multiplication)

The parentheses on the right hand side are usually omitted, as it is understood that you group multiplication operations together before doing addition operations.

With multiplications only, the order doesn't matter:

a * b = b * a (commutative law)
a * (b * c) = (a * b) * c (associative law)

And so again, when multiplying more than two numbers, the parentheses are usually omitted, because both options give the same result.

Division throws a wrench into things, order starts to matter. I think of division as "take the reciprocal and then multiply". So if you make that explicit (using Recip(a) to mean 1/a, the reciprocal of a) and keep track of which number get the reciprocal, it all works out:

a * (b / c ) = a * (b * Recip(c)) = (a * b) * Recip(c) = (a * b) / c

Because of the above, it's safe to write a * b / c, there's no ambiguity. But something like a * b / c * d should be avoided, as it is not clear if you mean a * (b / c) * d or a * b / (c * d), which are different.

Lastly, Recip(b * d) = Recip(b) * Recip(d), so you have the general rule for multiplying two fractions of "multiply the numerators and multiply the denominators":

(a / b ) * (c / d) = a * Recip(b) * c * Recip(d) = a * c * Recip(b) * Recip (d) = a * c * Recip(b * d) = (a * c) / (b * d)

Wayne

I'm going to re-read this a few times. Stellar reply! A whole math textbook summed in a good way.

 

[mbrooke]

Yes: inside quotes first. Still, try it multiple ways and see whether you get the same results.

You're right. The reply below might hint at why... lemme figure all this out...

 

[Dennis Alwon]

This might work only on old-fashioned keyboards with number pads.

That is correct... I work from a desktop so I have the keypad on top and on the side

 

[Dennis Alwon]

You're right. The reply below might hint at why... lemme figure all this out...

Try learning some algebra free at the Khan Academy

Khan Academy

www.khanacademy.org

 

[don_resqcapt19]

windows character map also has lots

½
Ø

alt#0216 = Ø

 

[wwhitney]

I'm going to re-read this a few times.

One thing to know is that when I write those lines of the form A = B = C = D that's short hand for "A = B by some rule (unspecified, for the reader to figure out or hopefully it's clear); then B = C by some other rule ; and C = D by some other rule. So the result is that A = D, what we wanted to prove."

Cheers, Wayne

 

[Dennis Alwon]

alt#0216 = Ø

Yep here are a bunch more

You must use the numLock keys for these Codes

Alt + 1 = ☺
Alt + 2 = ☻
Alt + 0176 = °
Alt + 251 = √
Alt + 0178 = ²
Alt + 0179 = ³
Alt + 252 = ⁿ
Alt + 227 = π
Alt + 241 = ±
Alt + 0188 = ¼
Alt + 0189 = ½
Alt + 0190 = ¾
Alt + 247 = ≈
Alt + 242 = ≥
Alt + 243 = ≤
Alt + 236 = ∞
Alt + 234 = Ω
Alt + 0216 = Ø
Alt + 0174 = ®

 

[Jerramundi]

One thing to know is that when I write those lines of the form A = B = C = D that's short hand for "A = B by some rule (unspecified, for the reader to figure out or hopefully it's clear); then B = C by some other rule ; and C = D by some other rule. So the result is that A = D, what we wanted to prove."

Cheers, Wayne

Transitive law of logic and mathematics.

A=B, B=C, C=D are the "axioms," or "starting points." And through substitution, you can prove A=D.
The proof offered by REAL mathematicians behind this concept are complex and debatable, but it's a widely accepted theorem.

 

[mbrooke]

Try learning some algebra free at the Khan Academy

Khan Academy

www.khanacademy.org

Thanks. A bit over my head, but I will give it a try.

 

[mbrooke]

Transitive law of logic and mathematics.

A=B, B=C, C=D are the "axioms," or "starting points." And through substitution, you can prove A=D.
The proof offered by REAL mathematicians behind this concept are complex and debatable, but it's a widely accepted theorem.

A knew there was more into this.

 

[GoldDigger]

Kind of. I always assumed that you multiply both numbers in the brackets by the one outside the brackets.

That is only when you are adding terms inside the brackets, not when multiplying. a (b + c) = ab + ac

a/a = 1, therefore
1 * (b/c) = a/a * b/c, thus ab/ac = b/c.

 

[GoldDigger]

Yep here are a bunch more

You must use the numLock keys for these Codes

Alt + 1 = ☺
Alt + 2 = ☻
Alt + 0176 = °
Alt + 251 = √
Alt + 0178 = ²
Alt + 0179 = ³
Alt + 252 = ⁿ
Alt + 227 = π
Alt + 241 = ±
Alt + 0188 = ¼
Alt + 0189 = ½
Alt + 0190 = ¾
Alt + 247 = ≈
Alt + 242 = ≥
Alt + 243 = ≤
Alt + 236 = ∞
Alt + 234 = Ω
Alt + 0216 = Ø
Alt + 0174 = ®

Definitely does not work on laptops with no numeric keypad. Really impossible on phones, although most phones have other ways of getting extended character sets.

 

[wwhitney]

Definitely does not work on laptops with no numeric keypad.

Some laptops have a "NumLock" mode where part of the keyboard (a slanty array of letter keys) is remapped to provide the numeric keypad. In that mode it should work.

Cheers, Wayne

 

[mbrooke]

That is only when you are adding terms inside the brackets, not when multiplying. a (b + c) = ab + ac

a/a = 1, therefore
1 * (b/c) = a/a * b/c, thus ab/ac = b/c.

What determines these rules? For me its always been difficult to memorize them since I don't know the how or why behind it.

 

[GoldDigger]

What determines these rules? For me its always been difficult to memorize them since I don't know the how or why behind it.

What determines these rules is a process of logical proof as laid out by wwhitney. We start out knowing some things about the number system by the way we describe them and the two fundamental operations (addition and multiplication) that apply to them. From there we derive everything else.
As for remembering them, they are either something for which you follow the mathematical development and remember it that way or which you learn by rote in the old school method. One thing that helps is to substitute actual numbers (using a different integer or preferably a different prime number for each operand.) Using prime numbers reduces the odds of a wrong equation accidentally working for your test case.
As a math lover, I feel that they are just obvious, but that will not work for everyone.

Using concrete examples may both help you to understand how a rule works and to check whether you have remembered it incorrectly,

 

[mbrooke]

Guess it just works, you get the same numbers when you run em through like Larry Fine pointed out.

And more praise to Whitney, that post should be a sticky in a math forum.