Simplifying terms

[Probability]

I've got a little stuck here in my understanding. I think its a procedural problem that I just have not quite got to grips with at the moment. Your guidance would be greatly appreciated.

I want to simplify this expression.

[MATH]\frac{1}{3}{r}\times{3s}[/MATH]
I have no idea how to correctly word this or lay it out correctly mathematically, so if I present this wrong, please show me the correct method.

[MATH]\frac{1}{3}{r}\times{3s}[/MATH] Multiply across by 3rs

[MATH]\frac{1}{3}{r}\times{9}{s}=\frac{9}{3}{rs}[/MATH]
[MATH]\frac{9}{3}={3}rs\div{3}={1}rs[/MATH]

 

[topsquark]

Multiplication is commutative. That means ab = ba. (I'm also throwing in the associative law, (ab)c = a(bc).)

So
[math]\dfrac{1}{3}r \times 3 s = \dfrac{1}{3} \times r ~ \times 3 \times s = r \times \dfrac{1}{3} \times 3 \times s = r \times 1 \times s = rs[/math].

-Dan

 

[Otis]

… I want to simplify this expression.

[MATH]\frac{1}{3}{r}\times{3s}[/MATH]

Hello Probability. Here's another way to view the simplification: Think in terms of multiplying two fractions.

Let's recall the rule for multiplying two fractions. It is [numerator × numerator] over [denominator × denominator], like this:

\(\displaystyle \frac{A}{B} × \frac{C}{D} = \frac{A×C}{B×D}\)

We may view \(\frac{1}{3}{r} × {3s}\) as a product of two algebraic fractions:

\(\displaystyle \frac{r}{3} × \frac{3s}{1} = \frac{r3s}{3}\)

The common factors \(\frac{3}{3}\) cancel.

?

 

[lookagain]

We may view \(\frac{1}{3}{r} × {3s}\) as a product of two algebraic fractions:

\(\displaystyle \frac{r}{3} × \frac{3s}{1} = \frac{r3s}{3}\)

A variable with a constant immediately to the right of it is not universally taken
to be a product. For example, WolframAlpha reads "r3" as \(\displaystyle \ "r^3" \ \) and "r3s" as \(\displaystyle \ "r^3s." \ \) Grouping symbols around either the variable or the constant here
would be a fix.

But, in this context, I would place them as such:

\(\displaystyle \dfrac{r(3)s}{3} \ = \ r(\tfrac{3}{3})s \ = \ rs\)


Or, I might eliminate the middle step.

 

[Otis]

… WolframAlpha reads "r3" as r³ …

I'm not using WolframAlpha in this thread, but I'd type asterisks (when using a CAS) with expressions like r*3*s.

\(\;\)