simple exponent distributive law question

nadroj · 07-24-2007, 08:52 PM

im studying for a midterm for a course which is basically about proving statements. im working on a practice question and am stuck because i forget basic distributive laws for exponents.

what it boils down to, is that i have to expand the following

Code:

2^a(2b + 1)

i know (ab)^c = a^c * b^c but cant work out with confidence what the above would be when expanded.

is it something like: (4^a)(2b^a) + 2^a?
or maybe just (4^a)b + 2^a?

thanks for any input

wjevans_7d1@yahoo.co · 07-24-2007, 09:03 PM

I don't know that I can help, but it would help if you clarify.

Which of these do you mean?

Code:

2^(a*(2b + 1))
(2^a)*(2b+1)

nadroj · 07-24-2007, 09:04 PM

sorry, it always looks poor with simple characters.

i meant the second one you listed:
(2^a)(2b + 1)

reverse · 07-25-2007, 12:03 AM

Basically you have something of the form: x*(y+1) which equals x*y + x.

"2^a * (2*b + 1)" is enough from a parenthesis point of view and it "expands" to: 2^a * 2*b + 2^a which equals: 2*2^a*b + 2^a = 2^(a+1) * b + 2^a.

You might want to give us the original exercise, to avoid any surprises.

qanopus · 07-25-2007, 06:58 AM

2^a*(2b + 1)=2^a*2b+2^a=2^(a+1)*b+2^a

wjevans_7d1@yahoo.co · 07-25-2007, 07:34 AM

Assuming for the moment that nadroj has his answer, I'd like to hijack this thread temporarily.

Quoth the highly esteemed schatoor:

Quote:

la illah ha illalah.
Muhammad ur rasul ullah
Ameer-ul-Momineen Ali-Ullah

I can guess what the first two lines mean.

What does the third line mean?

qanopus · 07-25-2007, 09:08 AM

Dude not the place to discuss here. But to shortly answer your question:
the first two lines is something that all Muslims hold as true; the unity of the Creator and the prophethood of Muhammad (peace be on him)
The third line is something that Shia Nizari Ismailis add and is an variation of "Ali un Wali Allah" that other Shias add. Basically it testifies that Ali was designated by Muhammad as his spiritual successor, something that is denied by the majority of the Muslims, the Sunnis.

sundialsvcs · 07-25-2007, 07:49 PM

Getting back to the math, I find it very helpful to grab a sheet of paper and "expand the sucker all the way out." The expression, I mean.

nadroj · 07-25-2007, 10:49 PM

sundialsvcs: i did, believe me! i just forgot these rules--did some searching on them before posting but couldnt quite get an example of the form above. i tried it again after seeing the above answers and could get it though.. makes perfect sense.

reverse and schatoor: thanks guys, i get it now.
so what you guys have above is that it expands to: 2^(a+1) * b + 2^a. if you factored out a 2 would it look like: 2[ 2^a * b + 2^(a-1) ] (which of course isnt much help).

reverse, since you asked, ill try and put here the original exercise. again, this is a course based on proofs.

as above, we are given a function f that takes two natural numbers (integers > 0) and sends it to a third natural number. it is defined as: f(a,b) = 2^a * (2b + 1)
what i had to prove for the particular question, was that this function is injective. that is, take 4 natural numbers (a,b) and (c,d). assume that f(a,b) = f(c,d) is a true statement. what you have to prove for it to be injective, is that, (a,b) = (c,d) (that is, a = c and b = d).

so i came here looking on how to expand it to see if i could factor out variables, etc, because as you can see it will be pretty messy. i did end up being able to prove it earlier today (due to a different frame of mind) which i will now show, for those interested.

take a, b, c, d from N (again, set of natural numbers). _assume_ that f(a,b) = f(c,d). you have to now prove that a = c and b = d. lets proceed..

Code:

f(a,b) = f(c,d) ("is true")
2^a * (2b + 1) = 2^c * (2d + 1) [now add 2^(-a) to each side, keeping the equality true]
2^a * (2b + 1) + 2^-a = 2^c * (2d + 1) + 2^-a  [add the exponents with same base, gives you:]
2^(a - a) * (2b + 1) = 2^(c - a) * (2d + 1)
2b + 1 = 2^(c - a) * (2d + 1)

after doing alot of these proofs in class, one will notice that the 2b + 1 is always odd, and 2^(any natural number) * (2d + 1) is always even, _unless_ 2 is raised to exponent 0. since the equality states that the left hand side is odd, the right hand side must be odd, and the only way that can happen is if 2^(c-a) = 1 = 2^(0). this means that c = a. now back to the equation:

Code:

2b + 1 = 2^0 * 2d + 1
2b + 1 = 1 * 2d + 1
2b + 1 = 2d + 1
2b = 2d
b = d

the proof is now complete.

thanks again to those interested, and sorry for the mess of a post.

reverse · 07-26-2007, 12:29 AM

Hold on a bit.

Quote:

2^a * (2b + 1) + 2^-a = 2^c * (2d + 1) + 2^-a [add the exponents with same base, gives you:]
2^(a - a) * (2b + 1) = 2^(c - a) * (2d + 1)

That's not how it works. 2^a + 2^(-a) does not equal 2^(a-a). A clear example: 2^1 + 2^(-1) = 2 + 1/2 = 5/2 and not 1 (which = 2^(1+(-1))).

Instead: 2^a * 2^(-a) = 2^(a - a). So instead of ADDING 2^(-a) you actually multiply both sides with it.

P.S.: naturally, this doesn't hold true only for 2 as a base.

qanopus · 07-26-2007, 05:20 AM

reverse is quite right... these are the rules:

a^b*a^c = a^(b+c)
a^(b*c) = (a^b)^c

and

a^b+a^c is not equal to a^(b+c)

nadroj · 07-26-2007, 08:29 AM

sorry, you are correct reverse. i used addition rather than multiplication, which is i think what i meant (since i was using that rule but with the addition operator). i guess what else i could have done is divide each side by 2^a, which would have the same effect i think.

thanks guys