If $(a,b)$ satisfies this relation then so does

${\displaystyle (b,a),\hspace{0.5em}(-a,b)\hspace{0.5em}(-b,a)\hspace{0.5em}(-a,-b),\hspace{0.5em}(-b,-a),\hspace{0.5em}(a,-b),\hspace{0.5em}\mathrm{and}\hspace{1em}(b,-a).}$

Hence this graph is the square joining the points (1,0), (0,1), (-1,0), (0,-1) and (1,0) in this order.

Alternatively the graph is the line segments (where $-1<x<1$ and $-1<y<1$) given by: $x+y=1$ in the first quadrant; $-x+y=1$ in the second quadrant; $-x-y=1$ in the third quadrant and $x-y=1$ in the fourth quadrant. (2) As $2^{1/n}$ must be greater than 1 we take $2^{1/n}=1+\delta$ where $\delta >0$. Expanding by the Binomial Theorem, because all the terms are positive, we have:

${\displaystyle 2=(1+\delta {)}^n>1+n\delta .}$

Hence $1>n\delta$ and so $\delta <1/n$. It follows that

${\displaystyle 1<2^{1/n}<1+\frac{1}{n}=\frac{n+1}{n}}$

and so

${\displaystyle \frac{n}{n+1}<\frac{1}{2^{1/n}}<1.}$

(3) Fixing one value of $n$ in $x^n+y^n=1$:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{y^n}& =1-x^n\\ \multicolumn{1}{c}{{\mathrm{ny}}^{n-1}\frac{\mathrm{dy}}{\mathrm{dx}}}& =-{\mathrm{nx}}^{n-1}\\ \multicolumn{1}{c}{\frac{\mathrm{dy}}{\mathrm{dx}}}& =-\frac{x^{n-1}}{y^{n-1}}<0.\end{array}}$

As $x$ and $y$ are positive in the first quadrant we see that the derivative is negative which shows that $y$ decreases as $x$ increases.

The relation is symmetric in $x$ and $y$ so the graph is symmetric about the line $y=x$ and goes through the point $(\frac{1}{2^{1/n}},\frac{1}{2^{1/n}})$. In part (2) we showed that $\frac{1}{2^{1/n}}$ is sandwiched between $\frac{n}{n+1}$ and 1. So the graph is confined between the two squares described. (4) By a symmetry argument exactly as in part (1) the graph of $|x{|}^n+|y{|}^n=1$ is symmetrical about the lines $y=x,\hspace{0.5em}y=-x,\hspace{0.5em}x=0,\hspace{1em}y=0$.

Because $\frac{n}{n+1}\to 1$ as $n\to \infty$ we see that, as n increases, the graphs in this family are confined to a narrower and narrower band rapidly becoming almost square in shape.

In the second quadrant if $x$ is large and negative then $x^n$ is large and negative and $y^n=1-x^n$ is large and positive and $y\to -x$ as $n\to \infty$.

By a similar argument, or explaining it by the symmetry of the relation, $y\to -x$ as $n\to \infty$ in the fourth quadrant.