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Is the following proposition true or false? Justify your conclusion with a counterexample or a proof. For each integer \(a, 3\) divides \(a^{3}+23 a\).

Short Answer

Expert verified
The proposition is true. We tested all possible cases of \(a \pmod{3}\) (0, 1, and 2) and found that the expression \(a^3 + 23a \equiv 0 \pmod{3}\) holds true in all cases. Therefore, for each integer \(a\), 3 divides \(a^3 + 23a\).

Step by step solution

01

Rewrite the expression in terms of modulo 3

By rewriting the expression in terms of modulo 3, we can make it easier to identify a pattern or potential counterexamples. Given the proposition, we try to see if \(a^3 + 23a \equiv 0 \pmod{3}\) for all integer values of a. Replacing 23 with its remainder when divided by 3, we get the simplified expression \(a^3 + 2a \equiv 0 \pmod{3}\).
02

Analyze properties of modulo 3

There are three possible values for a modulo 3, namely, 0, 1, and 2 since a can be any integer. In the next steps, we will check each possible case to see if the given expression holds true.
03

Test the case when \(a \equiv 0 \pmod{3}\)

If \(a \equiv 0 \pmod{3}\), then a is divisible by 3 and can be written as a = 3k where k is an integer. Substituting a = 3k into the simplified expression, we get \((3k)^3 + 2(3k) \equiv 0 \pmod{3}\). Since both terms on the LHS retain a factor of 3, we can conclude that the expression holds true for this case.
04

Test the case when \(a \equiv 1 \pmod{3}\)

If \(a \equiv 1 \pmod{3}\), then a can be written as a = 3k + 1 where k is an integer. Substituting a = 3k + 1 into the simplified expression, we get \((3k+1)^3 + 2(3k+1) \equiv 0 \pmod{3}\). Expanding the cube and simplifying, we get \(27k^3 + 27k^2 + 9k + 1 + 6k + 2 \equiv 0 \pmod{3}\). Dividing all terms by 3, we obtain \(9k^3 + 9k^2 + 3k + 1 \equiv 0 \pmod{3}\). Then, using the fact that any multiple of 3 is congruent to 0 modulo 3, we are left with \(1\equiv 0 \pmod{3}\), which means that the expression holds true for this case as well.
05

Test the case when \(a \equiv 2 \pmod{3}\)

If \(a \equiv 2 \pmod{3}\), then a can be written as a = 3k + 2 where k is an integer. Substituting a = 3k + 2 into the simplified expression, we get \((3k+2)^3 + 2(3k+2) \equiv 0 \pmod{3}\). Expanding the cube and simplifying, we get \(27k^3 + 54k^2 + 36k + 8 + 6k + 4 \equiv 0 \pmod{3}\). Dividing all terms by 3, we obtain \(9k^3 + 18k^2 + 12k + 4 \equiv 0 \pmod{3}\). Then, using the fact that any multiple of 3 is congruent to 0 modulo 3, we are left with \(4\equiv 0 \pmod{3}\), which means that the expression holds true for this case as well.
06

Conclusion

Since the proposition \(a^3 + 23a \equiv 0 \pmod{3}\) has been tested and found to be true for all possible cases of a modulo 3 (0, 1, and 2), we can conclude that the original proposition is true.

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Key Concepts

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

Counterexample in Mathematics
In mathematics, a counterexample is an example that disproves a proposition or theory. It is a specific case for which the proposition does not hold, thereby showing that the proposition is false. Counterexamples are particularly useful in the field because they can comprehensively refute a universal claim. For instance, if someone suggests that all birds can fly, providing an example of a flightless bird, such as an ostrich, would serve as a counterexample to that claim.

When a mathematician is presented with a statement that claims something is true for all cases, finding even one counterexample is enough to invalidate the statement. In our exercise, such a counterexample would be an integer for which the expression \(a^3 + 23a\) is not divisible by 3. The absence of a counterexample, after testing all cases with mathematical rigor, as shown in the step-by-step solution, supports the truth of the proposed statement.
Modulo Arithmetic
Modulo arithmetic, also known as clock arithmetic, involves the remainder after division of one number by another. It is denoted by the phrase 'mod' and plays a pivotal role in various branches of mathematics, including number theory, cryptography, and computer science. The concept is relatively straightforward: if you have \(a \equiv b \pmod{n}\), it means that when \(a\) is divided by \(n\), it leaves the same remainder as \(b\) does when it is divided by \(n\).

For example, we say \(17 \equiv 5 \pmod{3}\) because when both 17 and 5 are divided by 3, the remainder is 2. Modulo arithmetic is essential in our exercise because it simplifies the problem and allows us to categorize all integers into sets based on their remainders when divided by 3, making it easier to prove the given proposition by examining just a few cases.
Mathematical Reasoning
Mathematical reasoning is the critical and logical thought process that enables one to solve mathematical problems. It includes the ability to explore, conjecture, and reason logically, as well as the ability to use a variety of mathematical methods effectively to solve non-routine problems. This kind of reasoning is what allows mathematicians to make conjectures, devise tests or proofs, and generalize patterns.

In our context, mathematical reasoning helps us to understand why certain steps are taken in the proof process. For instance, when we determine that the proposition must be tested for all integers, mathematical reasoning guides us to the realization that we only need to test for integers modulo 3 because those three cases will cover all possible integers. This process includes abstraction (focusing on the relevant details), generalization (understanding how a result applies in broad terms), and critical thinking (evaluating the validity of each step in a proof).
Proof by Cases
Proof by cases is a mathematical method used to show that a certain proposition is true for all possible scenarios. It involves dividing the problem into a number of separate cases, with each case covering different possibilities that could occur within the problem. The individual then proves that the proposition holds true for each case, and as a result, the proof for the entire problem is established.

In our exercise, the proposition is proven by cases when we consider all possible values of \(a \pmod{3}\)—which are 0, 1, and 2. This technique acknowledges that if the expression \(a^3 + 23a\) is divisible by 3 for these three specific cases, it will be divisible by 3 for all integers, since every integer falls into one of these three modulo categories. By successfully validating each case, we can confidently assert the truth of the original proposition without needing to test every single integer individually.

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Most popular questions from this chapter

Is the following proposition true or false? For all integers \(a\) and \(b,\) if \(a b\) is even, then \(a\) is even or \(b\) is even. Justify your conclusion by writing a proof if the proposition is true or by providing a counterexample if it is false.

(a) Let \(n \in \mathbb{N}\) and let \(a \in \mathbb{Z}\). Explain why \(n\) divides \(a\) if and only if \(a \equiv 0(\bmod n)\) (b) Let \(a \in \mathbb{Z}\). Explain why if \(a \neq 0(\bmod 3),\) then \(a \equiv 1(\bmod 3)\) or \(a \equiv 2(\bmod 3)\) (c) Is the following proposition true or false? Justify your conclusion. For each \(a \in \mathbb{Z},\) if \(a \neq 0(\bmod 3),\) then \(a^{2} \equiv 1(\bmod 3)\).

Prove each of the following: (a) For each nonzero real number \(x,\left|x^{-1}\right|=\frac{1}{|x|}\). (b) For all real numbers \(x\) and \(y,|x-y| \geq|x|-|y|\) Hint: An idea that is often used by mathematicians is to add 0 to an expression "intelligently". In this case, we know that \((-y)+y=0\). Start by adding this "version" of 0 inside the absolute value sign of \(|x|\). (c) For all real numbers \(x\) and \(y, \| x|-| y|| \leq|x-y|\).

For a right triangle, suppose that the hypotenuse has length \(c\) feet and the lengths of the sides are \(a\) feet and \(b\) feet. (a) What is a formula for the area of this right triangle? What is an isosceles triangle? (b) State the Pythagorean Theorem for right triangles. \(\star\) (c) Prove that the right triangle described above is an isosceles triangle if and only if the area of the right triangle is \(\frac{1}{4} c^{2}\).

Are the following propositions true or false? Justify all your conclusions. If a biconditional statement is found to be false, you should clearly determine if one of the conditional statements within it is true. In that case, you should state an appropriate theorem for this conditional statement and prove it. (a) For all integers \(m\) and \(n, m\) and \(n\) are consecutive integers if and only if 4 divides \(\left(m^{2}+n^{2}-1\right)\) (b) For all integers \(m\) and \(n, 4\) divides \(\left(m^{2}-n^{2}\right)\) if and only if \(m\) and \(n\) are both even or \(m\) and \(n\) are both odd.

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