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Avenyr

Basic algebra question

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I can't believe I don't remember this but I got confused over some very basic algebra stuff. I have 1/(x+1)/(x+2) <--- no brackets this is vertically put I taught it gave 1/(x+1)*(x+2) but the math book's solution says it's (x+2)/(x+1) and I don't remember why it is like that and I feel ashamed of doing calculus when I don't even know such a basic operation This is only the part of the problem and this is not in it's entirety help me out please T_T

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Guest Anonymous Poster
The answer depends on the order in which the divisions are performed. If it is grouped as 1 / ((x+1)/(x+2)) then the book's answer is correct.

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from the text, I have to replace all the X in F(x) with G(x) meaning

F(g(x)) is what I'm looking for and it turns out F(x) is 1/x and G(x) is (X+1)/(X+2)

so then it becomes 1/(x+1)/(x+2) but I don't know if and why I should put parenthesis to make it become like this 1/[(x+1)/(x+2)]

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Basically I'd like to know the common rules in those type of situation with the fact that I know that dividing something means multiplying with the inverse of the divisor.

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you realy should put some paranthesis...

what you thought about is:
(1/(x+1)) / (x+2) = 1 / ((x+1)*(x+2))

what the book meant is:
1 / ((x+1)/(x+2)) = (x+2) / (x+1)


why is this so? the philosophical abstract algebra behind it is deep but the for calculations you need only remember the method which is best explain by an example:
when you devide by 3 (=3/1), its the same as multipling by 1/3 (third)
same way when you devide by 2/3, its the same as multipling by 3/2

so when you do 1 devided by (x+1)/(x+2) its the same as multipling 1 by (x+2)/(x+1).


the same logic worked for your understanding of the problem
deviding (1/(x+1)) by (x+2) is the same as multpling by 1/(x+2)
so the result is (1*1) / ((x+1)*(x+2))

I hope I was clear...
if not ask again and ill try to rephrase

Iftah.

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So if I follow what you said Iftah,
my reasoning is good for the technique but I got confused because I forgot to make proper use of parenthesis because the problem was 1/x and replacing that x byt (x+1)/(x+2) making it in reality something like

1/[(x+1)/(x+2)]

them meaning that this would be the same as (1/1)*[(x+2)/(x+1)] which in term would become

(x+2)/(x+1) < final answer

is that right ?

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just read the posts that were posted while i wrote my first reply (hate when that happens)

what does G(F(X)) means? its a new function, different from F and from G.
the new function first activates F on X, then activates G on the result.
so you see if you want to write a formula for G(F(X)) you need to put F(X) in place of Y, but you must surround it with paranthesis to make sure it is calculated first.

*always* when you replace X in F(X) think of it as surrounded by paranthesis.
as you saw, without it the results are often a mistake

for example:
F(X) = X+1
G(Y) = 2*Y

G(F(X)) = 2*(X+1) = 2*X + 2

without paranthesis you would get
G(F(X)) = 2*X+1 <--- wrong!!!


Iftah.

[Edited by - Iftah on September 12, 2005 8:49:49 AM]

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Ok I get it now

so the only problem I had was considering the parenthesis by forgetting them or ignoring them for that matter since I thought it didn't make a difference by I see it does

alright I understood

so something like X / Y / Z if there are no parenthesis means

X * 1/Y * 1/Z which would give X/Y*Z

but it would have been otherwise if there would have been parenthesis like in my problem

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well, there is no meaning to A/B/C since you cant tell if its (A/B)/C or A/(B/C)

and these two are different:

(A/B)/C = A/(B*C)
A/(B/C) = (A*C)/B

without the paranthesis you guessed one option and did the calculations correct, but you guessed the wrong one.

I feel I wasnt clear on the explaination of G(F(X))...

anyways, if you are not sure if you got the right formula for G(F(X)) you can always check yourself -
choose X (say 2) then activate F on it (put 2 in F's formula) then activate G on the result (put the result in G's formula).
then try putting 2 in the formula you got for G(F(X)). it should come out the same.
(ofcourse getting the same doesnt prove you got it right...)


Iftah.

[Edited by - Iftah on September 11, 2005 10:00:25 PM]

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