Complex fraction question

• ChiralSuperfields
In summary, to simplify the fraction ##\frac{2}{\frac{5}{3}}##, you can use the method of multiplying by the reciprocal, which is equivalent to dividing by the fraction. This can be seen by rewriting the fraction as ##2 \cdot \left(5 \cdot 3^{-1}\right)^{-1}##, which simplifies to ##\frac{6}{5}##. This method is based on the concept of inverse elements and can be thought of as multiplying by 1 in a strategic way.

ChiralSuperfields

Homework Statement
I am trying to simplify ##\frac {2}{\frac{5}{3}}##.
Relevant Equations
My first method to simplify the fraction is to to I flip ##\frac{5}{3}## up I get ##2 \times \frac{3}{5} = \frac{6}{5}##

Method 2: if I flip 3 up I get ##\frac{2}{5} \times \frac{1}{3} = \frac{2}{15}##.

Method 3: I could use it multiply ##\frac{3}{3}## since this is the same as mutlipying by ##1##. ##\frac{2}{\frac{5}{3}} \times \frac{3}{3}## so the ##3## cancels giving ##\frac{6}{5}##

I know the method in bold gives the wrong answer. However, why dose it not work?

Many thanks!

if I flip 3 up I get ##\frac{2}{5} \times \frac{1}{3} = \frac{2}{15}##

The wording is horrifying: "Flipping up"

You multiply numerator and denominator by 3 to get ##\frac{2}{5} \times {3} = \frac{6}{5}##

##\ ##

Grelbr42 and ChiralSuperfields
BvU said:
if I flip 3 up I get ##\frac{2}{5} \times \frac{1}{3} = \frac{2}{15}##

The wording is horrifying: "Flipping up"

You multiply numerator and denominator by 3 to get ##\frac{2}{5} \times {3} = \frac{6}{5}##

##\ ##
Thank you for your help @BvU!

Many thanks!

There's this method I knew as the " Double C"
Sorry, will edit when I get on my pc.
##\frac {2}{\frac {5}{3}}##=##\frac {\frac {2}{1}}{\frac {5}{3}}##
Now do a" double C"

Top 2 with bottom 3, bottom 1 with top 5= ##\frac {2.3}{1.5}=\frac{6}{5}##

Last edited by a moderator:
ChiralSuperfields
I like to think of reducing fractions as just multiplying by 1, with a judicious choice of how that's represented to cancel or move the parts I want. Like this:
$$\frac{2}{\frac{5}{3}} = \frac{2}{\frac{5}{3}} ⋅ 1 = \frac{2}{\frac{5}{3}} ⋅ \frac{3}{3} = \frac{2⋅3}{\frac{5}{3}⋅3} = \frac{6}{5}$$
It's pretty simple in this context, but it's also really useful later in your studies with complex numbers, polynomial fractions, unit conversions, etc.

ChiralSuperfields
WWGD said:
There's this method I knew as the " Double C"
Sorry, will edit when I get on my pc.
##\frac {2}{\frac {5}{3}}##=##\frac {\frac {2}{1}}{\frac {5}{3}}##
Now do a" double C"

Top 2 with bottom 3, bottom 1 with top 5= ##\frac {2.3}{1.5}=\frac{6}{5}##

No need to edit it!

Many thanks!

Last edited by a moderator:
WWGD
DaveE said:
I like to think of reducing fractions as just multiplying by 1, with a judicious choice of how that's represented to cancel or move the parts I want. Like this:
$$\frac{2}{\frac{5}{3}} = \frac{2}{\frac{5}{3}} ⋅ 1 = \frac{2}{\frac{5}{3}} ⋅ \frac{3}{3} = \frac{2⋅3}{\frac{5}{3}⋅3} = \frac{6}{5}$$
It's pretty simple in this context, but it's also really useful later in your studies with complex numbers, polynomial fractions, unit conversions, etc.

That is a good way to think of it!

Many thanks!

Callumnc1 said:
Homework Statement:: I am trying to simplify ##\frac {2}{\frac{5}{3}}##.

My first method to simplify the fraction is to to I flip ##\frac{5}{3}## up I get ##2 \times \frac{3}{5} = \frac{6}{5}##
Division by a fraction is equivalent to multiplying by the reciprocal of that fraction. So ##\frac a {\frac b c} = a \cdot \frac c b##.

"Flipping" is something that one does to a hamburger patty. It is not a recognized mathematical operation.
BvU said:
The wording is horrifying: "Flipping up"
Amen!

fresh_42, SammyS and ChiralSuperfields
I like to think about the division that it does not exist! What we call division is basically a multiplication with inverse elements: ##\dfrac{x}{y}=x\cdot y^{-1}## where ##y^{-1}## is the unique element with ##y\cdot y^{-1}=1.##

What we do have is ##\dfrac{2}{\frac{5}{3}}=2\cdot \left(5\cdot 3^{-1}\right)^{-1}.## Inverse elements are defined. They are the solution to ##a \cdot x =1.## This means we really have
\begin{align*}
\dfrac{2}{\frac{5}{3}}&=2\cdot \left(5\cdot 3^{-1}\right)^{-1}=2\cdot \left(3^{-1}\right)^{-1} \cdot 5^{-1}=2\cdot 3 \cdot 5^{-1}=6\cdot 5^{-1}
\end{align*}
and in common phrasing ##=\dfrac{6}{5}.##

You do not have to follow this way of thinking about it, but it is the mathematical background. Multiplying with the inverted or flipped or whatever quotient is only the crutch. It is actually the multiplication with the multiplicative inverse element.

DaveE and ChiralSuperfields
fresh_42 said:
I like to think about the division that it does not exist! What we call division is basically a multiplication with inverse elements: ##\dfrac{x}{y}=x\cdot y^{-1}## where ##y^{-1}## is the unique element with ##y\cdot y^{-1}=1.##

What we do have is ##\dfrac{2}{\frac{5}{3}}=2\cdot \left(5\cdot 3^{-1}\right)^{-1}.## Inverse elements are defined. They are the solution to ##a \cdot x =1.## This means we really have
\begin{align*}
\dfrac{2}{\frac{5}{3}}&=2\cdot \left(5\cdot 3^{-1}\right)^{-1}=2\cdot \left(3^{-1}\right)^{-1} \cdot 5^{-1}=2\cdot 3 \cdot 5^{-1}=6\cdot 5^{-1}
\end{align*}
and in common phrasing ##=\dfrac{6}{5}.##

You do not have to follow this way of thinking about it, but it is the mathematical background. Multiplying with the inverted or flipped or whatever quotient is only the crutch. It is actually the multiplication with the multiplicative inverse element.
Thanks for sharing @fresh_42 ! That is so cool!