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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?

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The reciprocal is by definition the inverse of the fraction --- and you calculate the inverse of the fraction by inverting the fraction. –  Gerry Myerson Jun 16 '13 at 3:53
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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
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@user82306 I always ask my students, how many half pizzas in a pizza? The answer, $1/(1/2)=2$. –  James S. Cook Jun 16 '13 at 4:10

12 Answers 12

up vote 30 down vote accepted

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:

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

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$1/x$ is, by definition, the number that you multiply $x$ by to get $1$ (assuming that $x\ne 0$).

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 $x$.

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+1 for explaining why rather than just demonstrating how by algebraic manipulation. –  200_success Jun 16 '13 at 8:18
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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. –  Ayman Hourieh Jun 16 '13 at 12:42
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(+1) By the magic of democracy, the super-voted answer is indeed the one to heed. –  John Bentin Jun 16 '13 at 15:35

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

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

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I'm sure OP will find this very helpful. –  Gerry Myerson Jun 16 '13 at 11:35
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It seems people here love to make mathematical Rube-Goldberg machine. –  tia Jun 16 '13 at 12:50
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I tried to explain it in more detail –  Harold Jun 16 '13 at 19:58
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Some of us (especially with cool names) still do appreciate. :-) –  Harold Cavendish Jun 16 '13 at 21:45
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I think that understanding the question via this method is actually a conceptual step backwards. –  Ryan Reich Aug 16 '13 at 7:04

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.

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

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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!

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Hey @john, I know you didn't mean anything by it, but can you try not to use the word "mathematically" when you mean "symbolically"? Mathematics is more than equations and symbols, as you know. A small part of how we can make the general public understand this is by being careful about how we as mathematicians use that word, because it conveys what we think about the nature of the subject. This was part of my vocabulary too, and I've been trying to change it myself-- I know it just slips out and is totally innocent. Consider this a friendly reminder :) –  Eric Stucky Jun 16 '13 at 5:49
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I agree completely Eric! Mathematics is indeed so much more than simple arithmetic and "solving for x" like many believe. As you said, this sort of vocabulary just slips out every now and then. –  john Jun 16 '13 at 11:30

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, $\frac{1}{\frac{1}{x}} = x$

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

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.)

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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 $x\cdot\frac{1}{x}=\frac{x}{1}\cdot\frac{1}{x}=\frac{x\cdot 1}{1\cdot x}=\frac{x}{x}=1$

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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. –  Alison Dec 5 '13 at 7:48

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

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

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As you learned in elementary school. $$\frac{1}{\frac{1}{x}}= \frac{\frac{1}{1}}{\frac{1}{x}} = \frac{1}{1} \div \frac{1}{x} = \frac{1}{1} \times \frac{x}{1} = \frac{x}{1} = x$$

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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...

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