Intuition: $\frac{1}{p}+\frac {1}{q} = 1$ equivalent to $(p-1)(q-1)=1$.

Question

It is easy to show mathematically that if $p,q \ne 0$ then $$\frac{1}{p}+\frac {1}{q} = 1 \qquad \Leftrightarrow \qquad (p-1)(q-1)=1 .$$ But intuitively, I would never have guessed this relationship. Does anyone have an intuitive argument that makes this obvious?

Added later: There are two very nice answers that reduce it to $$ x+y = 1 \qquad \Leftrightarrow \qquad xy = (1-x)(1-y) .$$ They only actually show $\Rightarrow$.

One of the answers does this by reflecting around the line $x+y=1$. The other answer uses probability, but essentially boils down to saying that if $x+y=1$, then the unordered pair $\{x,y\}$ is the same as the unordered pair $\{1-x,1-y\}$, that is, a reflection which sends $(x,y)$ to $(1-y,1-x)$. So in some sense the two answers are identical, but expressed differently.

Once you know $\Rightarrow$, you can get $\Leftarrow$ as follows. Suppose $(1-x)(1-y) = xy$. Let $z = 1-y$. From $\Rightarrow$ we get $(1-z)(1-y) = zy$. Hence $$ \frac{1-x}x = \frac y{1-y} = \frac{1-z}z $$ Hence $x=z$ and so $x=1-y$.

I would like to accept both answers, but I cannot. So following principles of distribution of wealth (points), I am accepting the one that comes from the person with less points.

Answer 1

I f we have a line $x+y=1$, ....(1), then from symmetry the area of the rectangle made by $(0,0);(x,0);(0,y);(x,y)$ should be equal to the area of the rectangle made by $(1,1); (x,y) ;(x,1) ;(1,y)$.

Because it's just the transformation of the previous rectangle by taking a reflection of the previous image about $x+y=1$ line.

So, $(1-x)(1-y)=xy$.....(2)

Now, taking $\frac{1}{p}=x, \frac{1}{q}=y$ we get

$(p-1)(q-1)=pq$ ......(1)'

As, we started with $x+y=1$ that means $\frac{1}{p}+\frac{1}{q}=1$....(2)'

As (1) and (2) are equivalent so is (1)' and (2)'.

This way we can establish an intuitive relation between $\frac{1}{p}+\frac{1}{q}=1$ and $(p-1)(q-1)=pq$.

Answer 2

For the first equation to hold, if $1/p$ is close to 1, then $1/q$ is really tiny. Likewise, in the second equation, if $(p-1)$ is close to zero, then $(q-1)$ is really large.

Answer 3

I think you can get some pictorial intuition if you rewrite $\frac{1}{p}+\frac{1}{q}=1$ as $q+p=pq$. Pretending $p$ and $q$ are positive numbers greater than $1$ for now, you can draw a rectangle with side lengths $q-1$ and $p-1$ with area $A$, whatever it is. Then extend both legs by $1$ "unit", and make a bigger rectangle. You end up getting two "thinner" rectangles of area $p-1$ and $q-1$ on the sides of your original rectangle, plus a little unit square in the corner. I drew a pretty low quality picture below.

Then geometrically, if $(q-1)(p-1)=1$, i.e., $A=1$, adding the area of four pieces shows the big outer rectangle has area $q+p$, so $pq=q+p$.

Conversely, if $pq=q+p$, so that the big outer rectangle has area $q+p$, the thin upper rectangle and unit square give you $p$ units of area, the side thin rectangle gives you $q-1$ units of area, so $A$ must equal $1$ to give you the last unit of area. That is $(p-1)(q-1)=1$.

Answer 4

It's intuitive if you think of $1/p$ and $1/q$, which we'll denote by $P$ and $Q$.

1. The first equation tells you that $P+Q=1$, i.e. you can think of them as probabilities of complementary events, say $P=\operatorname{Prob}(A)$ and $Q=\operatorname{Prob}(\text{not }A)$.
2. The second equation, in terms of $P$ and $Q$ is $$\frac{1-P}{P}\cdot\frac{1-Q}{Q}= \frac{\operatorname{Prob}(\text{not }A)}{\operatorname{Prob}(A)} \cdot \frac{\operatorname{Prob}(A)}{\operatorname{Prob}(\text{not }A)}=1$$

Answer 5

Construct three curves as follows:

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

2. $y = -\frac{1}{x}$ and then slide it along the $x$-axis so that at the point $x=p$, the vertical distance from this curve to curve (1) is 1.

3. $x = a$ where $a$ is the vertical asymptote of chart (2).

By construction, at $x=p$ the distance between curve (1) and curve (2) is $\tfrac{1}{p} + \tfrac{1}{q} = 1$. And this occurs if and only if the rectangle with corners $(a,0)$ and $(p-1,p-1)$ has area $(p-1)(q-1) = 1$.

Here is the construction for $p=4$.

$\tfrac{1}{p} + \tfrac{1}{q} = 1$ if and only if $(p-1)(q-1) = 1$ for $p = 4$">

Remark ("Funky step") Turn your head 90 degrees and look at curve (2). You'll see a chart for the function $y = \frac{1}{x}$ in the 'local' coordinates where $(a,0)$ is the origin. To help things, curve (3) is provided as the $x$-axis of this 'local' coordinate system.

Why the shaded rectangle has area $(p-1)(q-1) = 1$

Any rectangle that has one corner at $(a,0)$ and its opposite corner on curve (2) will have area 1. We construct two such rectangles:

• Rectangle $R_p$ has top-left corner at $(a,0)$ and bottom-right corner at $(p-1,p-1)$. This is the shaded rectangle

• Rectangle $R_q$ has top-left corner at $(a,0)$ and bottom-right corner at the point where the line $x=p$ intersects curve (2). This is the other rectangle drawn with solid lines in the picture

Then:

1. $R_p$ has height $p-1$. Should be visually evident

2. $R_p$ has width $q$. Why: $R_q$ has height $\frac{1}{q}$ from the original construction of curve (2). It follows then that $R_q$ has width $q$.

and therefore

3. $R_q$ has width $q-1$. Why: Let $S$ be the rectangle with bottom-left corner at the intersection of $R_p$ and $R_q$, and top-left corner at $(p,\tfrac{1}{p})$. Then $S$ is a square of side length 1.

Therefore $R_p$ is a rectangle of height $p-1$, width $q-1$, and area 1.

Answer 6

You can use trigonometry.

Denote: $$\frac1p =\sin^2 x\iff p-1=\cot^2x\\ \frac1q=\cos^2x \iff q-1=\tan^2x$$