Mathematical Inductions
Sequences of Mathematical Statements
Sequences of statements are logical, ordered groups of statements that are important for mathematical induction.Learning Objectives
Discuss what is meant by a sequence of mathematical statementsKey Takeaways
Key Points
- A sequence is an ordered list of objects or events. Like a set, it contains members, but unlike a set, the order of the members matters.
- A sequence of statements refers to the progression of logical implications of one statement.
- Sequences of statements are important for mathematical inductions, which rely on infinite sequences of statements.
Key Terms
- natural numbers: A set of numbers sometimes described as all non-negative integers [latex](0, 1, 2,...)[/latex] and sometimes described as all positive integers [latex](1, 2, 3,...)[/latex].
- set: A collection of zero or more objects, possibly infinite in size, and disregarding any order or repetition of the objects that may be contained within it.
Proof by Mathematical Induction
Proving an infinite sequence of statements is necessary for proof by induction, a rigorous form of deductive reasoning.Learning Objectives
Use mathematical induction to prove an infinite sequence of statementsKey Takeaways
Key Points
- Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true for all natural numbers.
- It is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one.
- Proving an infinite sequence of statements can be understood in the context of the domino effect, which by nature mediates a sequential and predictable order of events.
Key Terms
- inductive reasoning: The process of making inferences based upon observed patterns, or simple repetition. Often used in reference to predictions about what will happen or does happen, based upon what has happened.
- The basis ( base case): showing that the statement holds when [latex]n[/latex] is equal to the lowest value that [latex]n[/latex] is given in the question. Usually, [latex]n=0[/latex] or [latex]n=1[/latex].
- The inductive step: showing that if the statement holds for some [latex]n[/latex], then the statement also holds when [latex]n+1[/latex] is substituted for [latex]n[/latex].
Visualization
It may be helpful to think of the domino effect. If one is presented with a long row of dominoes standing on end, one can be sure that:- The first domino will fall
- Whenever a domino falls, its next neighbor will also fall
Dominoes: Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes.
Prove the following statement: For each positive integer [latex]n[/latex],
[latex-display]\displaystyle{0 + 1 + 2 + \cdots + n = \frac{n(n+1)}{2}}[/latex-display]
Base Case: We first have to check that the statement holds for [latex]n=0[/latex]. On the left-hand side of the equation, we have only [latex]0[/latex]. On the right hand side of the equation we substitute [latex]n=0[/latex]. Thus we have [latex]\displaystyle{0 = \frac{0(0+1)}{2}}[/latex], which can be simplified to [latex]0= 0[/latex]. We have proven that the statement holds true for the base case of [latex]n=0[/latex].
Inductive Step: Assume that the statement holds for [latex]n=k[/latex], and check if it holds for [latex]n=k+1[/latex] as well. In other words, we want to show that the statement holds true when we substitute [latex]k+1[/latex] for [latex]n[/latex]:
[latex-display]\displaystyle{0 + 1 + 2 + \cdots + k + (k+1) = \frac{\left(k+1\right)\left[\left(k+1\right)+1\right]}{2}}[/latex-display]
Note that we can use the induction hypothesis that the statement holds for [latex]n=k[/latex] to rewrite the left-hand side of the equation:
[latex-display]\displaystyle{0 + 1 + 2 + \cdots + k + (k+1) = \frac{k(k+1)}{2}} + (k+1)[/latex-display]
We can now rewrite this statement and show that it equals the right-hand side of the previous statement. In other words, if we can show that the following holds true, we will show that [latex]\displaystyle{0 + 1 + 2 + \cdots + n = \frac{n(n+1)}{2}}[/latex] holds true for [latex]k+1[/latex]:
[latex-display]\displaystyle{\frac{k(k+1)}{2} + (k+1) = \frac{\left(k+1\right)\left[\left(k+1\right)+1\right]}{2} }[/latex-display]
We can rewrite [latex]\displaystyle{\frac{k(k+1)}{2} + (k+1)}[/latex] algebraically as follows:
[latex]\displaystyle{
\begin{align}
\frac{k(k+1)}{2} + (k+1) &= \frac{k(k+1) + 2(k+1)}{2} \\
&=\frac{(k+1)(k+2)}{2} \\
&=\frac{\left(k+1\right)\left[\left(k+1\right)+1\right]}{2}
\end{align}
}[/latex]
This is exactly we wanted to prove. We have shown that [latex]\displaystyle{0 + 1 + 2 + \cdots + n = \frac{n(n+1)}{2}}[/latex] holds true for any [latex]n = k + 1[/latex] if it holds true for [latex]n = k[/latex]. This completes the induction step. We conclude that the statement holds for every non-negative integer [latex]n[/latex].Licenses & Attributions
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