Problem 9
Question
True or false: Whenever an integer \(n\) is a divisor of the square of an integer, \(m^{2},\) it follows that \(n\) is a divisor of \(m\) as well. (In symbols, \(\left.\forall n \in \mathbb{Z}, \forall m \in \mathbb{Z}, n\left|m^{2} \Longrightarrow n\right| m .\right)\) Prove your answer.
Step-by-Step Solution
Verified Answer
False. A counterexample (e.g., \(m = 6\) and \(n = 4\)) shows that the statement is not true in general.
1Step 1: Understand the problem
We need to determine whether the statement 'if an integer \(n\) is a divisor of the square of an integer, \(m^2\), then \(n\) is also a divisor of \(m\)' is always true or false.
2Step 2: Definitions
Recall the definition of divisibility. An integer \(a\) is a divisor of another integer \(b\) if there exists an integer \(k\) such that \(b = a \times k\). For instance, \(3\) is a divisor of \(9\) because \(9 = 3 \times 3\).
3Step 3: Analyze with an example
Consider \(m = 6\) and \(n = 4\). We have \(m^2 = 6^2 = 36\). Notice that \(4\) is a divisor of \(36\) because \(36 = 4 \times 9\). However, \(4\) is not a divisor of \(6\) because there is no integer \(k\) such that \(6 = 4 \times k\). This provides a counterexample.
4Step 4: Conclusion from the counterexample
Since there exists at least one example where \(n\) divides \(m^2\) but does not divide \(m\), the statement 'if \(n\) is a divisor of \(m^2\), then \(n\) is a divisor of \(m\)' cannot be true in general.
5Step 5: Determine the truth value of the statement
Given the counterexample, the provided statement is false.
Key Concepts
Integer DivisorsProof by CounterexampleDivisibility Rules
Integer Divisors
In mathematics, one concept that often appears is the idea of integer divisors. An integer divisor is a number that can divide another integer without leaving a remainder. For instance, if we say that an integer, say 3, is a divisor of another integer, 9, it means that when 9 is divided by 3, the result is a whole number, specifically 3. This can be written in the form of an equation: 9 = 3 \times 3. In general, if an integer \(a\) divides another integer \(b\), it means there exists an integer \(k\) such that: \(b = a \times k\). This property of being divisible by another number is foundational in many areas of number theory and it often helps in proving or disproving mathematical statements.
Proof by Counterexample
There are different methods used to prove mathematical statements. One such method is proof by counterexample. Proof by counterexample involves showing that a general statement is false by finding one instance where the statement does not hold. For the given exercise, we are examining the statement that 'if an integer \(n\) is a divisor of the square of an integer, \(m^2\), then \(n\) is also a divisor of \(m\).' To disprove this statement, we find a counterexample. Consider \(m = 6\) and \(n = 4\). When we square \(m\), we get: \(m^2 = 6^2 = 36\). Here, 4 is a divisor of 36 since: 36 = 4 \times 9. However, 4 is not a divisor of 6 because there is no integer \(k\) such that: 6 = 4 \times k. This single counterexample is sufficient to show that the statement is not always true, thereby proving the statement false.
Divisibility Rules
Divisibility rules are shortcuts or guidelines that help determine whether one number is a divisor of another. They simplify the process of checking divisibility without performing actual division. For example:
- A number is divisible by 2 if it ends in 0, 2, 4, 6, or 8.
- A number is divisible by 3 if the sum of its digits is divisible by 3.
- A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
- A number is divisible by 5 if it ends in 0 or 5.
- 36 is divisible by 4 because the number formed by its last two digits (36) is divisible by 4.
- 36 ends in 6, so it is not divisible by 5.
Other exercises in this chapter
Problem 8
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The trichotomy property of the real numbers simply states that every real number is either positive or negative or zero. Trichotomy can be used to prove many st
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Prove that \(\forall x \in \mathbb{R}, x \notin \mathbb{Z} \Longrightarrow\lfloor x\rfloor+\lfloor-x\rfloor=-1\).
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In an exercise in Section 3.2 we proved that the quadratic equation \(a x^{2}+b x+c=0\) has two solutions if \(a c
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