Why is $\frac{1}{\frac{1}{X}}=X$?

Can someone help me understand in basic terms why $$\frac{1}{\frac{1}{X}} = X$$

And my book says that "to simplify the reciprocal of a fraction, invert the fraction"...I don't get this because isn't reciprocal by definition the invert of the fraction?

• The reciprocal is by definition the inverse of the fraction --- and you calculate the inverse of the fraction by inverting the fraction. Jun 16 '13 at 3:53
• The reciprocal of $a/b$ is defined to be the unique $x$ such that $a/b\cdot x=1$, denoted by $(a/b)^{-1}$ or $1/(a/b)$. As it happens, $x=b/a$ is a formula for it, which can be easily verified, but there is a difference between a definition and an obvious formula.
– anon
Jun 16 '13 at 3:58
• @user82306 I always ask my students, how many half pizzas in a pizza? The answer, $1/(1/2)=2$. Jun 16 '13 at 4:10

$$1/x$$ is, by definition, the number that you multiply $$x$$ by to get $$1$$.

Similarly, $$1/\left(1/x\right)$$ is the number that you multiply $$1/x$$ by to get $$1$$.

But wait a sec: we just learned in the first sentence that that number is $$x$$.

• +1 for explaining why rather than just demonstrating how by algebraic manipulation. Jun 16 '13 at 8:18
• For this to be true, the inverse needs to be unique. And it is in the context of $\mathbb R^\times$, the group of non-zero real numbers under multiplication. Jun 16 '13 at 12:42
• (+1) By the magic of democracy, the super-voted answer is indeed the one to heed. Jun 16 '13 at 15:35

Maybe this will help you see why $\;\dfrac 1{\large \frac 1X}= X.\;$ We multiply numerator and denominator by $X$, which we can do because we can multiply any number by $\dfrac XX = 1$ without changing the actual value of the number:

$$\frac 1{\Large \frac 1X}\cdot \frac XX = \frac X1 = X$$ 

$$y=\frac1{\frac 1 x}$$ $$y'(x)=\left(\frac1{\frac 1 x}\right)' = -\left(\frac 1 {\left(\frac 1x\right)^2}\right)\left({\frac 1 x}\right)' = \frac 1 {\left(\frac 1x\right)^2} \cdot {\frac 1 {x^2}} = \frac {y^2(x)}{x^2}$$

So we have that $$x^2dy = y^2dx\\ \int \frac{dy}{y^2} = \int \frac{dx}{x^2}\\ -\frac{1}{y} = -\frac 1 x + C\\$$

Let's take a look at $y(1)$. $\frac 1 1 = 1$, this is already explained in a more common problem here: Why is $n$ divided by $n$ equal to $1$? So $y(1)=\frac{1}{\frac{1}{1}} =\frac 1 1 = 1$.

Note that I lost one possible solution, $y(x)=0$, by dividing by $y$. But since $y(1)=1$, it isn't really the solution.

Again: $y(1)=1$, so $~-\frac 1 1 = -\frac 1 1 + C ~~\Rightarrow~~ C=0$. Then $\frac {1} {y} = \frac {1}{x} \Rightarrow x=y$.

• I'm sure OP will find this very helpful. Jun 16 '13 at 11:35
• It seems people here love to make mathematical Rube-Goldberg machine.
– tia
Jun 16 '13 at 12:50
• Some of us (especially with cool names) still do appreciate. :-) Jun 16 '13 at 21:45
• I think that understanding the question via this method is actually a conceptual step backwards. Aug 16 '13 at 7:04
• I am pretty sure that in one of the derivatives evaluations you have already used that $\frac1{1/x} = x$, so your argument have a good chance of being cyclical.
– Ilya
Jun 13 '16 at 13:23

Does this work for you? Start with $${1\over1/x}$$ Multiply top and bottom by $x$: $${1\over1/x}{x\over x}={x\over(1/x)x}={x\over1}=x$$

Symbolically: $$\frac{1}{\frac{1}{X}}=1\div \frac{1}{X}=1\times\frac{X}{1}=X$$ OR $$\frac{1}{\frac{1}{X}}= \frac{1}{\frac{1}{X}} \times \frac{X}{X}=\frac{X}{1}=X$$ Intuitively:

We are asking how many 1/X pieces we can fit into a whole. Clearly there must be X of them!

Well, I think this is a matter of what is multiplication and what is division. First, we denote that $$\frac{1}{x}=y$$ which means $$xy=1\qquad(\mbox{assuming x\ne0 in fundamental mathematics where there isn't Infinity(\infty)})$$ Now, $$\frac{1}{\frac{1}{x}}=\frac{1}{y}$$ by using the first equation. Here, by checking the second equation ($xy=1$), it is obvious that $$\frac{1}{y}=x$$ thus $$\frac{1}{\frac{1}{x}}=x$$ Q.E.D.

Naturally, if we inverse some inverted object, we will have the object itself! This matter take place for numbers and their operations: $$-(-x)=x$$ and for any non-zero $x$ $$\frac{1}{\frac{1}{x}}=x$$ (if we know "our limits" this fact is true for all objects in mathematics also for functions $(f^{-1})^{-1}=f$, directions(vectors, matrices, ...) and so on.)

Let, $\frac{1}{\frac{1}{x}} = y$, where $y \neq x$. Now,

$\frac{1}{\frac{1}{x}} = y \\\Rightarrow 1 = \frac{y}{x}\\\Rightarrow x = y$

A contradiction! So, $\frac{1}{\frac{1}{x}} = y$, where $y \neq x$ is false. So, $\frac{1}{\frac{1}{x}} = x$

• Wouldn't this work better as a direct proof? Jun 21 '13 at 20:06
• Probably, but to some people 'indirect' way is more understandable than 'direct' way. Jun 22 '13 at 6:39

I look at it this way. How do you check division? With multiplication.$\frac{15}{5}=3$ because $3\cdot 5=15$. Similarly, $\frac{1}{\frac{1}{x}}=x$ because $x\cdot\frac{1}{x}=\frac{x}{1}\cdot\frac{1}{x}=\frac{x\cdot 1}{1\cdot x}=\frac{x}{x}=1$

• Here's another way I look at it: say $x=2$. $1\div\frac{1}{2}$ asks How many one halves fit inside one?" Well, two one halves fit inside one. Dec 5 '13 at 7:48

Note that: $1/x=x^{-1}$ and also that: $(x^a)^b=x^{ab}$

Therefore $1/(1/x)=(x^{-1})^{-1}=x^1=x$

• $(x^a)^b=x^{ab}$ is not always true but $1/1/x = x$ is always true $x\ne =0$.
– user312097
Feb 22 '17 at 6:39

All the other answers use algebra, but I prefer the intuitive explanation based on this comment.

Suppose that you need to divide a large pizza for infants, who cannot safely eat an entire pizza. Instead, each infant can eat only a slice. Each slice is smaller than the original pizza; so the size of each slice must necessarily be a fraction $< 1$.
In the picture below, each slice is $1/6$ of the original pizza. Then $\dfrac{\color{red}{1}}{\color{forestgreen}{\dfrac{1}{6}}} = \color{red}{1}$ pizza divided by slices of size $\color{forestgreen}{\dfrac{1}{6}}$.

Well, what occurs if you divide $\color{red}{1}$ pizza totally into slices of size $\color{forestgreen}{\dfrac{1}{6}}$?
You end with 6 slices of pizza, after you cut the pizza per the 3 black straight lines as pictured above!

Simply because:

$$\frac{1}{\frac{1}{X}}=\frac{1}{(\frac{X}{1})^{-1}}=\frac{1}{1}\frac{X}{1}=\frac{1X}{1}=X$$

And yes, the reciprocal of a fraction, is the inverted fraction...