Let the zero of $F$ be $0_F$. === Base Result === First we need to show that: :$\paren {m \cdot a} \times b = m \cdot \paren {a \times b}$ This will be done by induction: For all $m \in \N$, let $\map P n$ be the proposition: :$\paren {m \cdot a} \times b = m \cdot \paren {a \times b}$ First we verify $\map P 0$. When $m = 0$, we have: {{begin-eqn}} {{eqn | l = \paren {0 \cdot a} \times b       | r = 0_F \times b       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}: $\forall x \in F: 0 \cdot x = 0_F$ }} {{eqn | r = 0_F       | c = {{Defof|Field Zero}} }} {{eqn | r = 0 \cdot \paren {a \times b}       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}: $\forall x \in F: 0 \cdot x = 0_F$ }} {{end-eqn}} So $\map P 0$ holds. ==== Basis for the Induction ==== Now we verify $\map P 1$: {{begin-eqn}} {{eqn | l = \paren {1 \cdot a} \times b       | r = a \times b       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}: $\forall x \in F: 1 \cdot x = x$ }} {{eqn | r = 1 \cdot \paren {a \times b}       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}: $\forall x \in F: 1 \cdot x = x$ }} {{end-eqn}} So $\map P 1$ holds. This is our basis for the induction. ==== Induction Hypothesis ==== Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\paren {k \cdot a} \times b = k \cdot \paren {a \times b}$ Then we need to show: :$\paren {\paren {k + 1} \cdot a} \times b = \paren {k + 1} \cdot \paren {a \times b}$ ==== Induction Step ==== This is our induction step: {{begin-eqn}} {{eqn | l = \paren {\paren {k + 1} \cdot a} \times b       | r = \paren {a + \paren {k \cdot a} } \times b       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}} }} {{eqn | r = a \times b + \paren {k \cdot a} \times b       | c = {{Field-axiom|D}} }} {{eqn | r = a \times b + k \cdot \paren {a \times b}       | c = Induction Hypothesis }} {{eqn | r = \paren {k + 1} \cdot \paren {a \times b}       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m \in \N: \paren {m \cdot a} \times b = m \cdot \paren {a \times b}$ {{qed|lemma}} The result for $m < 0$ follows directly from Powers of Group Elements. {{qed}} === Full Result === Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: :$\forall m \in \Z: \paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$ First we verify $\map P 0$. When $n = 0$, we have: {{begin-eqn}} {{eqn | l = \paren {m \cdot a} \times \paren {0 \cdot b}       | r = \paren {m \cdot a} \times 0_F       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}: $\forall x \in F: 0 \cdot x = 0_F$ }} {{eqn | r = 0_F       | c = {{Defof|Field Zero}} }} {{eqn | r = 0 \cdot \paren {a \times b}       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}: $\forall x \in F: 0 \cdot x = 0_F$ }} {{eqn | r = \paren {m 0} \cdot \paren {a \times b}       | c =  }} {{end-eqn}} So $\map P 0$ holds. ==== Full Result - Basis for the Induction ==== Next we verify $\map P 1$. When $n = 1$, we have: {{begin-eqn}} {{eqn | l = \paren {m \cdot a} \times \paren {1 \cdot b}       | r = \paren {m \cdot a} \times b       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}: $\forall x \in F: 1 \cdot x = x$ }} {{eqn | r = m \cdot \paren {a \times b}       | c = Base Result }} {{eqn | r = \paren {m 1} \cdot \paren {a \times b}       | c =  }} {{end-eqn}} So $\map P 1$ holds. This is our basis for the induction. ==== Full Result - Induction Hypothesis ==== Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\paren {m \cdot a} \times \paren {k \cdot b} = \paren {m k} \cdot \paren {a \times b}$ Then we need to show: :$\paren {m \cdot a} \times \paren {\paren {k + 1} \cdot b} = \paren {m \paren {k + 1} } \cdot \paren {a \times b}$ ==== Full Result - Induction Step ==== This is our induction step: {{begin-eqn}} {{eqn | l = \paren {m \cdot a} \times \paren {\paren {k + 1} \cdot b}       | r = \paren {m \cdot a} \times \paren {k \cdot b + b}       | c = {{Defof|Integral Multiple|subdef = Rings and Fields}} }} {{eqn | r = \paren {m \cdot a} \times \paren {k \cdot b} + \paren {m \cdot a} \times b       | c = {{Field-axiom|D}} }} {{eqn | r = \paren {m k} \cdot \paren {a \times b} + m \cdot \paren {a \times b}       | c = Induction Hypothesis }} {{eqn | r = \paren {m k + k} \cdot \paren {a \times b}       | c = Integral Multiple Distributes over Ring Addition }} {{eqn | r = \paren {m \paren {k + 1} } \cdot \paren {a \times b}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m \in \Z: \forall n \in \N: \paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$ {{qed|lemma}} The result for $n < 0$ follows directly from Powers of Group Elements. {{qed}}