Proof. By definition, the absolute value on \(\mathbb {R}\) is non-negative. Therefore \(\hspace {0.3cm} d(x,y)=|x-y|\geq 0\).

Now, since \(|x-y|=0\) if and only if \(x=y\). It follows that \(d(x,y)=0\) if and only if \(x=y\).

To verify the third axiom, we have \(\hspace {0.3cm} d(x,y)=|x-y|=|-(y-x)|=|y-x|=d(y,x)\).
To verify the triangle inequality, suppose that \(x,y,z\in \mathbb {R}\) then \begin {align*} d(x,z) &=|x-z|=|x-y+y-z|\\ &\leq |x-y|+|y-z|\\ &=d(x,y)+d(y,z) \end {align*}

Hence \((\mathbb {R},d)\) is a metric space. □

Proof.

i

\begin {align*} d(u,v) &=|(u_1+iu_2)-(v_1+iv_2)| =|(u_1-v_1)+i(u_2-v_2)|\\ &=\sqrt {(u_1-v_1)^2+(u_2-v_2)^2}\geq 0 \end {align*}

ii

\(\hspace {0.2cm} d(u,v)=|u-v|=0\iff u_1-v_1=0, u_2-v_2=0\implies u_1=v_1,u_2=v_2\implies u=v\)

iii

\begin {align*} d(u,v) &=|(u_1+iu_2)-(v_1+iv_2)| =|-((v_1+iv_2)-(u_1+iu_2))|\\ &=|(v_1+iv_2)-(u_1+iu_2)|\\ &=d(v,u) \end {align*}

iv

Let \(u,v,w\in \mathbb {C}\),then \begin {align*} d(u,w) &=|(u_1+iu_2)-(w_1+iw_2)|\\ &=|(u_1+iu_2)-(v_1+iv_2)+(v_1+iv_2)-(w_1+iw_2)|\\ &\leq |(u_1+iu_2)-(v_1+iv_2)|+|(v_1+iv_2)-(w_1+iw_2)|\\ &=d(u,v)+d(v,w) \end {align*}

Hence \((\mathbb {C},d)\) is a metric space. □

Proof. If \(x=(x_1,x_2),y=(y_1,y_2)\in \mathbb {R}^2\), we see that \(|x_1-y_1|\geq 0\) and \(|x_2-y_2|\geq 0\) so that \[d(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}\geq 0.\] Also, \(\max \{|x_1-y_1|,|x_2-y_2|\}=0\) if and only if \(|x_1-y_1|=0\) and \(|x_2-y_2|=0\), which means that \(\max \{|x_1-y_1|,|x_2-y_2|\}=0\) if and only if \(x_1=y_1\) and \(x_2=y_2\). Thus \(d(x,y)=0\) if and only if \(x=y\).

We show that \(d(x,y)=d(y,x)\). Since \(\max \{|x_1-y_1|,|x_2-y_2|\}=\max \{|y_1-x_1|,|y_2-x_2|\}\), we have that \(d(x,y)=d(y,x)\).

To show the triangle inequality holds, let \(x=(x_1,x_2), y=(y_1,y_2), z=(z_1,z_2)\in \mathbb {R}^2\). Then \begin {align*} d(x,z) &=\max \{|x_1-z_1|,|x_2-z_2|\}\\ &=\max \{|x_1-y_1+y_1-z_1|,|x_2-y_2+y_2-z_2|\}\\ &\leq \max \{|x_1-y_1|+|y_1-z_1|,|x_2-y_2|+|y_2-z_2|\}\\ &\leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|y_1-z_1|,|y_2-z_2|\}\\ &=d(x,y)+d(y,z). \end {align*}

Hence \((\mathbb {R}^2,d)\) is a metric space. □

Solution. All four axioms must be checked. Write \(z=(z_1,z_2)\) for a third point.

Non-negativity

Each term is an absolute value, so \(d(x,y)\geq 0\) always.

\(d(x,y)=0\) if and only if \(x=y\)

A sum of two non-negative numbers is zero only if both are, so \(d(x,y)=0\) forces \(|x_1-y_1|=0\) and \(|x_2-y_2|=0\), that is \(x_1=y_1\) and \(x_2=y_2\), which is \(x=y\). Conversely \(d(x,x)=0+0=0\).

Symmetry

\(|a-b|=|b-a|\) for real \(a,b\), so \[d(x,y)=|x_1-y_1|+|x_2-y_2|=|y_1-x_1|+|y_2-x_2|=d(y,x).\]

Triangle inequality

Apply the triangle inequality for real numbers to each coordinate: \[|x_i-z_i|\leq |x_i-y_i|+|y_i-z_i|,\qquad i=1,2 .\] Adding the two, \[d(x,z)=\sum _{i=1}^{2}|x_i-z_i| \leq \sum _{i=1}^{2}|x_i-y_i|+\sum _{i=1}^{2}|y_i-z_i|=d(x,y)+d(y,z).\]

All four hold, so \((\mathbb {R}^2,d)\) is a metric space. This \(d\) is the taxicab metric: it measures distance as a car would travel on a square grid, and it is genuinely different from the Euclidean metric — the points at distance \(1\) from the origin form a diamond, not a circle.

Solution. In each case the first three axioms are immediate and the triangle inequality is the one that needs an argument.

1. The Euclidean metric on \(\mathbb {R}^2\)

Non-negativity holds because a square root of a sum of squares is non-negative, and \(d(x,y)=0\) forces \((x_1-y_1)^2+(x_2-y_2)^2=0\), hence \(x_1=y_1\) and \(x_2=y_2\). Symmetry holds because \((x_i-y_i)^2=(y_i-x_i)^2\). For the triangle inequality put \(u=x-y\) and \(v=y-z\); then \(x-z=u+v\) and the claim is \(||u+v||\leq ||u||+||v||\), which follows from the Cauchy–Schwarz inequality: \[||u+v||^2=||u||^2+2u\cdot v+||v||^2\leq ||u||^2+2||u||\,||v||+||v||^2 =\big (||u||+||v||\big )^2 .\] So \((\mathbb {R}^2,d)\) is a metric space.

2. The Euclidean metric on \(\mathbb {R}^n\)

Every step above is written without reference to the number of coordinates, so the same argument applies verbatim with sums running to \(n\).

3. The space \(l_p^n\)

Here the triangle inequality \[\left [\sum _{i=1}^{n}|x_i-z_i|^p\right ]^{\frac 1p}\leq \left [\sum _{i=1}^{n}|x_i-y_i|^p\right ]^{\frac 1p}+ \left [\sum _{i=1}^{n}|y_i-z_i|^p\right ]^{\frac 1p}\] is exactly Minkowski’s inequality, proved in Section 1, applied to the \(n\)-tuples \(x-y\) and \(y-z\). The other three axioms are as before. Note that \(p\geq 1\) is needed: for \(p<1\) Minkowski’s inequality reverses, and the \(l_{1/2}\) counterexample of Section 1 shows the triangle inequality genuinely fails.

Case 2 is the special case \(p=2\) of case 3, so the three examples are one theorem seen at three levels of generality.