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Chapter 5: Problem 9

Use strong induction to prove that \(\sqrt{2}\) is irrational. [Hint: Let \(P(n)\) be the statement that \(\sqrt{2} \neq n / b\) for any positive integer \(b . ]\)

Short Answer

Expert verified
Strong induction shows \(\sqrt{2} eq \frac{n}{b}\) for any positive integer \(b\), proving \(\sqrt{2}\) is irrational.

Step by step solution

01

- Define the hypothesis

Let us define the statement that we need to prove by strong induction. Let the statement be: \(P(n)\): \(\sqrt{2} eq \frac{n}{b}\) for any positive integer \(b\).
02

- Base case

For strong induction, we need to prove the base case where \(P(1)\) is true. Consider the base case \(n = 1\): We need to show that \(\sqrt{2} eq \frac{1}{b}\) for any positive integer \(b\). Since \(\sqrt{2} = 1.414...\), it is clear that there is no positive integer \(b\) such that \(\sqrt{2} = \frac{1}{b}\). Hence, \(P(1)\) is true.
03

- Inductive step

Assume that the statement is true for all integers up to \(k\), i.e., assume that \(P(n)\) is true for all positive integers \(n\) such that \(1 \leq n \leq k\). This means \(\sqrt{2} eq \frac{n}{b}\) for any positive integers \(n\) and \(b\) within this range. Now we need to show that \(P(k+1)\) is also true: We must show that \(\sqrt{2} eq \frac{k+1}{b}\) for any positive integer \(b\).
04

- Proof by contradiction

Assume for contradiction that \(\sqrt{2} = \frac{k+1}{b}\) for some positive integer \(b\). Then \(\sqrt{2}\) can be written in the form \(\frac{a}{d}\) where \(a\) and \(d\) are positive integers with no common factors (for simplicity, take the fraction in its simplest form). Then: \[ \sqrt{2} = \frac{a}{d} \] Squaring both sides: \[ 2 = \frac{a^2}{d^2} \] Thus, \[ a^2 = 2d^2 \] This implies that \(a^2\) is even, which means \(a\) must also be even because the square of an odd number is odd.
05

- Analyze evenness

Let \(a = 2m\) where \(m\) is some integer. Substituting into the equation above: \[ (2m)^2 = 2d^2 \] \[ 4m^2 = 2d^2 \] \[ 2m^2 = d^2 \] This implies that \(d^2\) is also even, and so \(d\) must be even as well. Therefore, both \(a\) and \(d\) are even, which means they have a common factor of 2. This contradicts our assumption that \(\frac{a}{d}\) was in its simplest form.
06

- Conclusion

Since assuming that \(\sqrt{2}\) is rational leads to a contradiction, it follows that \(\sqrt{2}\) cannot be expressed as \(\frac{a}{d}\) for any positive integers \(a\) and \(d\). Hence, \(\sqrt{2}\) is irrational. By the principle of strong induction, we have shown that \(P(n)\) is true for all positive integers \(n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

irrational numbers
Irrational numbers are real numbers that cannot be expressed as a simple fraction or ratio of two integers. Unlike rational numbers, which can be written in the form \(\frac{a}{b}\), where both \(a\) and \(b\) are integers, irrational numbers have non-repeating and non-terminating decimal expansions. Examples of irrational numbers include \(\pi\), \(e\), and \(\sqrt{2}\).Let's consider \(\sqrt{2}\). If you try to write \(\sqrt{2}\) as a fraction, say \(\frac{n}{b}\), you鈥檒l quickly realize that no such pair of integers \(n\) and \(b\) exists that can perfectly represent \(\sqrt{2}\). This is what makes \(\sqrt{2}\) irrational.The proof often uses properties of squares and their evenness to show contradictions, emphasizing how \(\sqrt{2}\) does not fit the mold of rational numbers by failing the criteria outlined for these.
mathematical induction
Mathematical induction is a powerful proof technique often used to show statements are true for all natural numbers. It consists of two main steps:
  • Base Case: Validate the statement for the initial value, often \(n = 1\) or \(n = 0\).
  • Inductive Step: Assume the statement is true for some arbitrary positive integer \(k\) (the induction hypothesis), and use this assumption to prove the statement for \(k+1\).
In the context of proving that \(\sqrt{2}\) is irrational, strong induction involves assuming the statement is true for all integers up to some \(k\) and then proving it for \(k+1\). This approach allows for a clearer domino effect, ensuring that if the initial case holds, the truth propagates through all successive integers.
proof by contradiction
Proof by contradiction is a fascinating method used in mathematical proofs to establish the truth of a statement by showing that assuming the contrary leads to a contradiction. Here's how it works:
  • Assume the Opposite: Begin by assuming that the statement you want to prove is false.
  • Find a Contradiction: Using logical steps, demonstrate that this assumption leads to a contradiction or an impossibility.
In proving the irrationality of \(\sqrt{2}\), we assume that \(\sqrt{2}\) is rational, meaning it can be written as \(\frac{a}{d}\) in its simplest form with no common factors. Through algebraic manipulation, we show that both \(a\) and \(d\) must be even, contradicting our initial condition of \(\frac{a}{d}\) being in simplest form. This contradiction implies that our assumption was false, thereby proving \(\sqrt{2}\) is irrational.
integer properties
Understanding integer properties is fundamental to many proof techniques in mathematics, including the proof showing that \(\sqrt{2}\) is irrational. Some critical properties include:
  • Even and odd numbers: The evenness or oddness of an integer is vital. An even number can be written as \(2m\), and an odd number as \(2m + 1\).
  • Factorization: Any integer can be uniquely factorized (except for the order) into prime numbers, known as the Fundamental Theorem of Arithmetic.
  • Simplicity of Fractions: A fraction \(\frac{a}{b}\) is in its simplest form if \(a\) and \(b\) have no common divisors other than 1.
In the proof by contradiction to show the irrationality of \(\sqrt{2}\), we use these properties. We conclude that both \(a\) and \(d\) must be even if \(\sqrt{2} = \frac{a}{d}\), contradicting the simplest form condition and reinforcing the irrationality of \(\sqrt{2}\).

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Most popular questions from this chapter

Show that each of these proposed recursive definitions of a function on the set of positive integers does not produce a well-defined function. a) \(F(n)=1+F(\lfloor(n+1) / 2\rfloor) \quad\) for \(\quad n \geq 1 \quad\) and \(F(1)=1 .\) b) \(F(n)=1+F(n-2)\) for \(n \geq 2\) and \(F(1)=0\) c) \(F(n)=1+F(n / 3)\) for \(n \geq 3, F(1)=1, F(2)=2,\) and \(F(3)=3\) d) \(F(n)=1+F(n / 2)\) if \(n\) is even and \(n \geq 2, F(n)=1+\) \(F(n-2)\) if \(n\) is odd, and \(F(1)=1 .\) e) \(F(n)=1+F(F(n-1))\) if \(n \geq 2\) and \(F(1)=2\)

Use structural induction to show that \(n(T) \geq 2 h(T)+1\) where \(T\) is a full binary tree, \(n(T)\) equals the number of vertices of \(T,\) and \(h(T)\) is the height of \(T\).

Sometimes we cannot use mathematical induction to prove a result we believe to be true, but we can use mathematical induction to prove a stronger result. Because the inductive hypothesis of the stronger result provides more to work with, this process is called inductive loading. We use inductive loading in Exercise \(74-76\) . Suppose that we want to prove that $$ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n}} $$ for all positive integers \(n .\) a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails. b) Show that mathematical induction can be used to prove the stronger inequality $$ \frac{1}{2} \cdot \frac{3}{4} \dots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n+1}} $$ for all integers \(n\) greater than \(1,\) which, together with a verification for the case where \(n=1\) , establishes the weaker inequality we originally tried to prove using mathematical induction.

Give a recursive definition of each of these sets of ordered pairs of positive integers. Use structural induction to prove that the recursive definition you found is correct. [Hint: To find a recursive definition, plot the points in the set in the plane and look for patterns.] a) \(S=\left\\{(a, b) | a \in \mathbf{Z}^{+}, b \in \mathbf{Z}^{+}, \text { and } a+b \text { is even }\right\\}\) b) \(S=\left\\{(a, b) | a \in \mathbf{Z}^{+}, b \in \mathbf{Z}^{+}, \text { and } a \text { or } b \text { is odd }\right\\}\) c) \(S=\left\\{(a, b) | a \in \mathbf{Z}^{+}, b \in \mathbf{Z}^{+}, a+b \text { is odd, and } 3 | b\right\\}\)

Let \(S\) be the subset of the set of ordered pairs of integers defined recursively by Basis step: \((0,0) \in S .\) Recursive step: If \((a, b) \in S,\) then \((a+2, b+3) \in S\) and \((a+3, b+2) \in S\) a) List the elements of \(S\) produced by the first five appli- cations of the recursive definition. b) Use strong induction on the number of applications of the recursive step of the definition to show that \(5 | a+b\) when \((a, b) \in S .\) c) Use structural induction to show that \(5 | a+b\) when \((a, b) \in S .\)
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