Problem 8 - Book of Proof

Question

Use the method of proof by contradiction to prove the following statements. (In each case, you should also think about how a direct or contrapositive proof would work. You will find in most cases that proof by contradiction is easier.) Suppose \(a, b, c \in \mathbb{Z} .\) If \(a^{2}+b^{2}=c^{2},\) then \(a\) or \(b\) is even.

Step-by-Step Solution

Answer

Proof by contradiction successfully proves that in a Pythagorean triplet, at least one of the first two integers, 'a' or 'b', must be even.

Step 1: Assumption of the Contrary

Assume the contrary of the statement we wish to prove, meaning, let's assume that both 'a' and 'b' are odd numbers.

Step 2: Mathematical Representation of Assumption

Define the odd numbers mathematically. An odd number can be written as \(2n + 1\), where n is any integer. Hence, 'a' and 'b' can be represented as \(2m + 1\) and \(2n + 1\) respectively, where 'm' and 'n' are integers.

Step 3: Substitution into the Given Equation

Replace 'a' and 'b' in the given equation, \(a^{2}+b^{2}=c^{2}\), by the defined odd numbers, i.e., \((2m+1)^2 + (2n+1)^2 = c^2\). Simplify this to \(4m^2 + 4m + 1 + 4n^2 + 4n + 1 = c^2\), which simplifies to \(4(m^2 + m + n^2 + n) + 2 = c^2\).

Step 4: Contradiction

The equation obtained in step 3 is a contradiction as the left side of the equation is even (since it's a multiple of 2), but the right side, \(c^2\), which is supposed to be equal to the left side, cannot be even unless 'c' itself is even. But if 'c' is even, this means that either 'a' or 'b' should also be even, contradicting our initial assumption that both 'a' and 'b' are odd.

Key Concepts

Mathematical Proof

Mathematical proofs are structured arguments that demonstrates the truth of a mathematical statement. These are essential to validate any mathematical claim with certainty. Proofs can be carried out in several ways, including direct proof, proof by contradiction, and proof by contrapositive, among others.

• Direct Proof: This involves proving a statement by straightforward logical deduction from known facts or premises.
• Proof by Contrapositive: This method involves proving an equivalent statement that ensures the original statement is true.
• Proof by Contradiction: This approach assumes the negation of the statement you want to prove and shows that this assumption leads to a contradiction. This contradiction implies that the original statement must be true.
  For the exercise given, proof by contradiction is used. It assumes that both numbers 'a' and 'b' are odd and shows this leads to a fallacy when substituted into the equation. Ultimately, by revealing a contradiction, it concludes that either 'a' or 'b' must be even.

Integers

Integers are whole numbers that can be positive, negative, or zero. They are represented by the symbol \(\mathbb{Z}\). This set is infinite and includes numbers like \(-3, 0, 7,\) and \(100\). Unlike fractions or decimals, integers do not have a fractional part.
Within integers, you can perform various mathematical operations like addition, subtraction, multiplication, and division (where the result is also an integer, except where dividing by zero).
In the exercise, integers are significant as the variables \(a, b,\) and \(c\) are all integers. Understanding integers enables us to define numbers as even or odd, splitting them into two basic categories that form the backbone for the proof discussed.

Even and Odd Numbers

Within the realm of integers, numbers can either be even or odd.

• Even Numbers: These are integers that can be divided by 2 without leaving a remainder. In general, even numbers have the form \(2k\), where \(k\) is any integer.
• Odd Numbers: These numbers cannot be evenly divided by 2, giving a remainder of 1. They are expressed as \(2n + 1\), where \(n\) is any integer.

In the provided exercise, understanding even and odd numbers is crucial. The starting assumption is that both \(a\) and \(b\) are odd, expressed as \(2m + 1\) and \(2n + 1\). Substituting these into the equation \(a^2 + b^2 = c^2\) reveals a contradiction unless at least one of the numbers, \(a\) or \(b\), is even. This insight is vital to employing proof by contradiction effectively.