Equality of Ordered Tuples
Theorem
Let $a = \tuple {a_1, a_2, \ldots, a_n}$ and $b = \tuple {b_1, b_2, \ldots, b_n}$ be ordered tuples.
Then:
- $a = b \iff \forall i: 1 \le i \le n: a_i = b_i$
That is, for two ordered tuples to be equal, all the corresponding elements have to be equal.
Proof
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
- $\tuple {a_1, a_2, \ldots, a_n} = \tuple {b_1, b_2, \ldots, b_n} \iff \forall i: 1 \le i \le n: a_i = b_i$
$\map P 1$ is true, as this just says $\tuple {a_1} = \tuple {b_1} \iff a_1 = b_1$ which is trivial.
Basis for the Induction
$\map P 2$ is the case:
- $\tuple {a_1, a_2} = \tuple {b_1, b_2} \iff a_1 = b_1, a_2 = b_2$
which has been proved in Equality of Ordered Pairs.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\tuple {a_1, a_2, \ldots, a_k} = \tuple {b_1, b_2, \ldots, b_k} \iff \forall i: 1 \le i \le n: a_i = b_i$
Then we need to show:
- $\tuple {a_1, a_2, \ldots, a_{k + 1} } = \tuple {b_1, b_2, \ldots, b_{k + 1} } \iff \forall i: 1 \le i \le n: a_i = b_i$
Induction Step
This is our induction step:
\(\ds \tuple {a_1, a_2, \ldots, a_{k + 1} }\) | \(=\) | \(\ds \tuple {b_1, b_2, \ldots, b_{k + 1} }\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \tuple {a_1, \tuple {a_2, \ldots, a_{k + 1} } }\) | \(=\) | \(\ds \tuple {b_1, \tuple {b_2, \ldots, b_{k + 1} } }\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a_1\) | \(=\) | \(\ds b_1\) | |||||||||||
\(\ds \land \ \ \) | \(\ds \tuple {a_2, \ldots, a_{k + 1} }\) | \(=\) | \(\ds \tuple {b_2, \ldots, b_{k + 1} }\) | Basis for the Induction |
But from the induction hypothesis we have that:
- $\tuple {a_2, \ldots, a_{k + 1} } = \tuple {b_2, \ldots, b_{k + 1} } \iff \forall i: 1 \le 2 \le k + 1: a_i = b_i$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N_{>0}: \tuple {a_1, a_2, \ldots, a_n} = \tuple {b_1, b_2, \ldots, b_n} \iff \forall i: 1 \le i \le n: a_i = b_i$
$\blacksquare$
Examples
Ordered Triple
Let:
- $\tuple {a_1, a_2, a_3}$ and $\tuple {b_1, b_2, b_3}$
be ordered triples.
Then:
- $\tuple {a_1, a_2, a_3} = \tuple {b_1, b_2, b_3}$
- $\forall i \in \set {1, 2, 3}: a_i = b_i$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.15$: Sequences
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 8$: Cartesian product of sets
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): Appendix $\text{A}.10$: Finite Sequences
- 2002: Thomas Jech: Set Theory (3rd ed.) ... (previous) ... (next): Chapter $1$: Pairing