Simple Algebra Division question

[DeepGround]

Hello,

I am not grasping how to deal with multiple divisions properly.

If I have a/1 divided by b/1 divided by c/1

How do I know if the compressed form is ac/b or a/bc?

 

[JukkaVayrynen]

It's simple: you don't.

 

[coverband]

a/bc

 

[dodo]

What poster #2 is saying, is that (a/b)/c is not equal to a/(b/c). So talking about a/b/c, without properly using parenthesis to tell which of the two cases is meant, is simply ambiguous.

 

[mathwonk]

i would have chosen a/(bc) as what was meant, but i see the problem.

 

[DeepGround]

Is it ever possible to be working on a problem and end up with a\b\c?

 

[dodo]

a/b/c are just written-down symbols that stand for an idea on your mind; if you work on a problem and get that result, in your mind you'll know what you mean (if you're not insane). Now, other people won't understand you unless you use parenthesis, or write something more graphical like \frac {a/b}{c} or \frac {a}{b/c}.

 

[coverband]

I think what this problem is lacking is parentheses! ( )

 

[HallsofIvy]

Hello,

I am not grasping how to deal with multiple divisions properly.

If I have a/1 divided by b/1 divided by c/1

How do I know if the compressed form is ac/b or a/bc?

Is a/1= a, b/1= b, c/1= c? If so why write it that way?

Is it ever possible to be working on a problem and end up with a\b\c?

What does "\" mean here?

 

[DeepGround]

Is a/1= a, b/1= b, c/1= c? If so why write it that way?


What does "\" mean here?

To specify that all variables are already a fraction. Some math texts show a/b/1 is a/b and a/1/b is ab/1

I meant / by "\"

 

[HallsofIvy]

a/b/1 can be interpreted as (a/b)/1= a/b or a/(b/1)= a/b so that's not a problem. a/1/b could be interpreted as (a/1)/b= a/b or a/(1/b)= ab. That's a problem.

It really doesn't matter whether a or b are "already" fractions.

 

[DeepGround]

To specify that all variables are already a fraction. Some math texts show a/b/1 is a/b and a/1/b is ab/1

I meant / by "\"

Oh wow, I just figured out where I went wrong, now I see how it does not matter where the main division is located because the if all denominators are 1 then it doesn't matter if you multiply the 1 by the numberator or the denominator.

 

[Diffy]

I thought the general rule for an ambiguous case was to work from left to right and which would be ((a/1)/(b/1))/(c/1) = a/bc.