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Chapter 3: Problem 4

Prove that an integer \(n\) is even if and only if its last decimal digit is even.

Short Answer

Expert verified
We proved the statement in both directions using properties of integers and the definition of even numbers. We showed that if n has an even last digit, then it is indeed even; and conversely, if n is an even integer, its last digit must also be even. This proves that an integer $n$ is even if and only if its last decimal digit is even.

Step by step solution

01

Case 1: "If" - Proving an integer with an even last digit is even

Suppose the integer n has an even last digit, say d. Then, we can express n as a sum of the even last digit multiplied by a power of ten (since we are in the decimal system) and the remaining part of the number: n = 10 * k + d, where k is an integer and d 鈭� {0, 2, 4, 6, 8} Now let's prove that n is even. Recall that a number is even if it can be expressed as 2 * z, where z is an integer. Since d is an even number, we can express it as d = 2 * m, where m is an integer. Substitute it into the equation for n: n = 10 * k + 2 * m We can factor out the 2: n = 2 * (5 * k + m) Since k and m are integers, (5 * k + m) is also an integer. Let's denote this integer by z: n = 2 * z, where z is an integer Thus, we have shown that an integer with an even last digit is even.
02

Case 2: "Only If" - Proving that if an integer is even, its last digit is even

Now let's prove the opposite direction. Suppose n is an even integer. We will show that its last digit must be even. Let's use the representation of n as the sum of its last digit and the remaining part of the number: n = 10 * k + d, where k is an integer and 0 鈮� d 鈮� 9 Since n is even, we can express it as: n = 2 * m, where m is an integer Now, let's analyze the possible values of d: 1. If d is odd (d 鈭� {1, 3, 5, 7, 9}), then 10 * k + d is odd, which contradicts the assumption that n is even. 2. If d is even (d 鈭� {0, 2, 4, 6, 8}), then 10 * k + d is even, which aligns with the assumption that n is even. Thus, when n is even, the last digit (d) must also be even, which concludes the proof. In summary, we have shown that an integer n is even if and only if its last decimal digit is even.

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Key Concepts

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

Number Theory
Number theory is a branch of mathematics that deals with the properties and relationships of numbers, often focusing on integers. One of the fundamental concepts in number theory is understanding the difference between even and odd numbers. An integer is considered even if it can be divided by 2 without leaving a remainder. Conversely, an odd number will have a remainder of 1 when divided by 2.

The exercise given is a perfect example of number theory in action. It asks us to prove the statement that an integer is even if and only if its last decimal digit is even. This type of problem requires a deep understanding of how numbers function within the decimal system and how mathematical proofs can demonstrate such properties. By examining the last digit, we recognize that it holds all the necessary information to determine a number's parity (whether it is even or odd).

Number theory often provides the foundational knowledge needed for more complex areas of mathematics, like cryptography, which relies on properties of numbers for secure communication.
Mathematical Proofs
Mathematical proofs are structured arguments used to establish the truth of mathematical statements. Proofs use logical reasoning, based on accepted axioms and previously proven propositions, to provide a step-by-step verification of statements. There are different types of proofs, such as direct proofs, proof by contradiction, and proof by induction.

In our specific problem, a direct proof was used to show that an integer is even if and only if its last digit is even. This involved two main parts:
  • "If" case: Demonstrating that if the last digit of a number is even, then the whole number must be even.
  • "Only if" case: Demonstrating that if a number is even, its last digit must be even as well.
Each part of the proof was carefully constructed, beginning with assumptions and working through logical steps to reach the conclusion. This ensures that the argument is both sound and convincing. Proofs are crucial in mathematics because they provide certainty that mathematical statements are universally valid.
Decimal System
The decimal system is a base-ten numbering system, which is the most commonly used system for representing numbers. It utilizes ten digits, from 0 to 9, and is based on powers of ten. Each position in a number has a tenfold increase in value as you move from one digit to the next, from right to left. This system is intuitive and aligns well with human counting methods.

The problem we examined makes use of the decimal system to explore properties of even numbers. More specifically, it focuses on the last digit of a number to determine its parity. Why is the last digit so important? Because in the base-ten system, the last digit's position represents the ones place, which directly influences whether the number as a whole is divisible by 2.

Understanding the decimal system is crucial for many areas of mathematics, as it is foundational for arithmetic operations, number representation, and computational algorithms. Through analyzing how numbers behave in the decimal system, students strengthen their overall mathematical reasoning skills.

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Most popular questions from this chapter

Find the greatest common divisor of 4905 and \(32445 .\)

Show that if \(a \mid b\) and \(c \mid d\), then \(a c \mid b d\).

Let \(m \geq 1\). Prove that an integer \(n\) is divisible by \(5^{m}\) if and only if the integer \(k\) consisting of its last \(m\) decimal digits is divisible by \(5^{m}\). Note that \(k=n \bmod 10^{m}\).

In this exercise, we show that one can multiply two \(k\)-bit binary numbers \(A\) and \(B\) faster than in \(\mathrm{O}\left(k^{2}\right)\) steps when \(k\) is large. Make \(k\) even by prepending a 0 bit, if necessary. We may have to remove two or three leading 0 bits from the product at the end. Write the numbers in base \(b=2^{k / 2}\) as \(A=A_{1} b+A_{0}\) and \(B=B_{1} b+B_{0}\). Prove that $$ A B=\left(b^{2}+b\right) A_{1} B_{1}+b\left(A_{1}-A_{0}\right)\left(B_{0}-B_{1}\right)+(b+1) A_{0} B_{0} . $$ This formula shows that the product \(A B\) of two \(k\)-bit numbers can be formed by multiplying the three \(k / 2\)-bit numbers \(\left(A_{1}-A_{0}\right)\left(B_{0}-B_{1}\right)\), \(A_{1} B_{1}\) and \(A_{0} B_{0}\), together with simple shifting and adding operations. Note that one can multiply a binary number by \(b\) or \(b^{2}\) by shifting the bits by \(k / 2\) or \(k\) positions. This simple trick can be used recursively. Let \(T(k)\) denote the time needed to multiply two \(k\)-bit binary numbers. The formula shows that \(T(k) \leq 3 T(k / 2)+c k\), for some constant \(c\). Show that this inequality implies that \(T\left(2^{i}\right) \leq c\left(3^{i}-2^{i}\right)\), for \(i \geq 1\). Deduce from this that \(T(k) \leq 3 c \cdot 3^{\log _{2} k}=3 c k^{\log _{2} 3}\). Since \(\log _{2} 3 \approx\) \(1.585<2\), this method, which is called Karatsuba multiplication, is faster theoretically than conventional multiplication when \(k\) exceeds a threshold. In practice, when multiplying large numbers with the same length, one uses the formula recursively down to the threshold.

Prove that an integer \(n\) is divisible by 5 if and only if its last decimal digit is divisible by 5 .
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