# What's the algebraic property where you can flip the fractions in an equation?

Earlier in algebra, we spent over 20 minutes trying to figure out

$$\frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_e} \,\,\,\, \text{solve for }R_2$$

when the teacher said "What you start out with is the same as what you learned in pre-algebra

$$\frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_e}$$

subtract $\frac{1}{R_1}$ from both sides:

$$\frac{1}{R_2} = \frac{1}{R_e} - \frac{1}{R_1}$$

and then the math gods said 'you may flip as long as all are flipped'"

$$R_2 = R_e - R_1$$

What is the name of this algebraic property?

(Sorry, I couldn't find any good tags for use here.)

• I'm afraid it's not true. That's probably why you couldn't remember it – davidlowryduda Dec 3 '13 at 23:10
• Whatever math God told you that is wrong =] – LASV Dec 3 '13 at 23:10
• @David Mitra Be careful, sometimes students misinterpret the instructors =P – LASV Dec 3 '13 at 23:11
• The property is named "error". Sometimes called a variation of the freshman's dream. In general $\frac{1}{\frac{1}{R_1}-\frac{1}{R_2}} = \frac{R_1-R_2}{R_1 R_2}.$ – deinst Dec 3 '13 at 23:12
• @Nick - I think the "generalized freshman's dream" would be: All operations are linear. This certainly fits into that. – Simon Rose Sep 23 '14 at 12:59

You can flip if you flip correctly. Flipping both sides of

$$\frac{1}{R_2} = \frac{1}{R_e} - \frac{1}{R_1}$$

gives you

$$R_2 = \frac{1}{\frac{1}{R_e} - \frac{1}{R_1}}$$

Well, that's not quite right: more pedantically, flipping both sides gives

$$\frac{1}{\frac{1}{R_2}} = \frac{1}{\frac{1}{R_e} - \frac{1}{R_1}}$$

but we know that the left hand side of this is the same thing as $R_2$. (at least in the current setting, where $R_2$ is known to be nonzero)

This is not an algebraic property, because it is not true.

For example, let $R_2=R_1=2$ and $R_e =1$. Then, $$\frac{1}{2} = \frac{1}{1} - \frac{1}{2}$$ but $$2 \neq 1 - 2$$

Yes, you can “flip”, so long as the two sides are single fractions: from $$\frac{1}{R_2}=\frac{R_1-R_e}{R_1R_e}$$ you can rightly deduce $$R_2=\frac{R_1R_e}{R_1-R_e}$$

Note that, in general, $$\frac{R_1R_e}{R_1-R_e}\ne R_e-R_1$$ Indeed the equality would imply $$R_1R_e=-R_e^2+2R_1R_e-R_1^2$$ or $$R_1^2-R_1R_e+R_e^2=0$$ Since your numbers are by hypothesis non zero, this would imply $$\left(\frac{R_1}{R_e}\right)^2-\frac{R_1}{R_e}+1=0$$ or, setting $t=R_1/R_e$, $t^2-t+1=0$. This equality is not true for every real $t$. So your conclusion is really wrong.

You can't just flip all willy-nilly. If what you said was true, then the resistance of electrical circuits in series and parallel connections would be the same! That just isn't true.

$$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}\mathbf{\neq} \frac{1}{R_1 + R_2 + R_3 + \dots + R_n}$$ Fastest way to see this is by setting $R_1 = R_ 2 = \dots = R_n = R$ (say)

Does it feel like $\frac{n}{R}$ and $\frac{1}{nR}$ are the same?
Now, you see. You must either change your math gods or listen to them more carefully.

The flipping property is called the Invertendo: $$\frac{a}{b} = \frac{c}{d} \iff \frac{b}{a} = \frac{d}{c}$$

There are many ways you can realize it and I leave it as an exercise to you to try.

The simplest way is by raising both sides of the equation to $-1$ :

$$\frac{a}{b} = \frac{c}{d} \implies \left(\frac{a}{b}\right)^{-1} = \left(\frac{c}{d}\right)^{-1} \implies \frac{a^{-1}}{b^{-1}} = \frac{c^{-1}}{d^{-1}} \implies \frac{1}{a} \cdot \frac{b}{1} = \frac{1}{c} \cdot \frac{d}{1} \implies \frac{b}{a} = \frac{d}{c}$$

The following is the best approach to your question:

$$\frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_e}\\ \implies \frac{1}{R_2} = \frac{1}{R_e} - \frac{1}{R_1} = \overbrace{\frac{R_1 - R_e}{R_1 R_e}}^{\text{Taking LCM}} \\ \implies \underbrace{R_2 = \frac{R_1 R_e}{R_1 - R_e}}_{\text{Applying Invertendo}}$$