Why Does Flipping the Denominator in Complex Fractions Give the Wrong Answer?

[ChiralSuperfields]

Homework Statement

I am trying to simplify $\frac {2}{\frac{5}{3}}$.

Relevant Equations

Please see below

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!

 

[BvU]

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 $\frac{2}{5} \times {3} = \frac{6}{5}$

$\$

 

[ChiralSuperfields]

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!

I agree about the wording.

Many thanks!

 

[WWGD]

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= $\frac {2.3}{1.5}=\frac{6}{5}$

 

[DaveE]

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]

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= $\frac {2.3}{1.5}=\frac{6}{5}$

Thank you for your reply @WWGD !

No need to edit it!

Many thanks!

 

[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.

Thank you for your reply @DaveE!

That is a good way to think of it!

Many thanks!

 

[Mark44]

Homework Statement:: I am trying to simplify $\frac {2}{\frac{5}{3}}$.
Relevant Equations:: Please see below

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.

The wording is horrifying: "Flipping up"

Amen!

 

[fresh_42]

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.

 

[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.

Thanks for sharing @fresh_42 ! That is so cool!