Chapter 4: Problem 47
Show that if \(a, b, k,\) and \(m\) are integers such that \(k \geq 1\) \(m \geq 2,\) and \(a \equiv b(\bmod m),\) then \(a^{k} \equiv b^{k}(\bmod m)\)
Short Answer
Expert verified
If \( a \equiv b(\bmod \ m) \), then \( a^k \equiv b^k(\bmod \ m) \) for all integers \( k \geq 1 \) by mathematical induction.
Step by step solution
01
- Understand the given congruence
Given that integer values of a and b satisfy the condition: \[ a \equiv b(\bmod \ m) \]This means that when a is divided by m, it leaves the same remainder as b when divided by m.Mathematically, this can be expressed as: \[ a - b = qm \]for some integer q.
02
- Raise both sides to the power k
Raise both sides of the congruence \( a \equiv b(\bmod \ m) \) to the power of k: \[ a^k \equiv b^k(\bmod \ m) \]To show this, we need to use induction.
03
- Use mathematical induction
Base Case: For \( k = 1 \): Since \( a \equiv b(\bmod \ m) \), it is true that \( a^1 \equiv b^1(\bmod \ m) \). The base case holds.Inductive Step: Assume for some \( k = k_0 \) that \( a^{k_0} \equiv b^{k_0}(\bmod \ m) \).We need to show that it holds for \( k = k_0 + 1 \), i.e., \( a^{k_0 + 1} \equiv b^{k_0 + 1}(\bmod \ m) \).
04
- Inductive step
Using the induction hypothesis: \[ a^{k_0} \equiv b^{k_0}(\bmod \ m) \]Multiplying both sides by a, we get: \[ a^{k_0} \cdot a \equiv b^{k_0} \cdot a(\bmod \ m) \]since \( a \equiv b(\bmod \ m) \).Replace a with b in the product of congruence: \[ a^{k_0} \cdot a \equiv b^{k_0} \cdot b(\bmod \ m) \]Thus, \[ a^{k_0 + 1} \equiv b^{k_0 + 1}(\bmod \ m) \]
05
- Conclusion
Since the base case is true and the inductive step holds, by the principle of mathematical induction, the statement is true for all integers \( k \geq 1 \).Therefore, we have shown that if \( a \equiv b(\bmod \ m) \), then \[ a^k \equiv b^k(\bmod \ m) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Congruence Relation
Congruence relation is a fundamental concept in modular arithmetic.
It describes how two numbers leave the same remainder when divided by a third number, known as the modulus.
If we say that two integers, for example, a and b, are congruent modulo m, we write this as: \(\text{a} \equiv \text{b} \ (\bmod \ \text{m}) \).
Mathematically, it means that when you subtract b from a, the result is a multiple of m, or \(\text{a} - \text{b} = \text{qm}\text{ for some integer q}\).
This relationship is useful when solving problems where numbers need to be simplified and compared under a modular system.
Applying this understanding becomes especially handy in more advanced proofs involving integer powers.
It describes how two numbers leave the same remainder when divided by a third number, known as the modulus.
If we say that two integers, for example, a and b, are congruent modulo m, we write this as: \(\text{a} \equiv \text{b} \ (\bmod \ \text{m}) \).
Mathematically, it means that when you subtract b from a, the result is a multiple of m, or \(\text{a} - \text{b} = \text{qm}\text{ for some integer q}\).
This relationship is useful when solving problems where numbers need to be simplified and compared under a modular system.
Applying this understanding becomes especially handy in more advanced proofs involving integer powers.
Using Mathematical Induction
Mathematical induction is a powerful technique used to prove statements about integers.
The process is akin to falling dominos; if you can knock down the first one, and show that each one knocks down the next, all will fall.
Here is how it works:
- **Base Case:** Verify that the statement holds for the initial value, usually k = 1.
- **Inductive Step:** Assume the statement holds for some integer k = k_0 (This is called the induction hypothesis). Then, prove it holds for k = k_0 + 1.
The idea is if the base case is true, and if the statement holding for k _0 implies it also holds for k _0 + 1, the statement is true for all integers k starting from the base case.
This principle was used to prove that \(\text{a}^{k} \equiv \text{b}^{k}\text{(mod } \text{m})\). Initial value was checked for k = 1. The hypothesis was then assumed for some arbitrary k = k_0, and shown valid for k_0 + 1.
The process is akin to falling dominos; if you can knock down the first one, and show that each one knocks down the next, all will fall.
Here is how it works:
- **Base Case:** Verify that the statement holds for the initial value, usually k = 1.
- **Inductive Step:** Assume the statement holds for some integer k = k_0 (This is called the induction hypothesis). Then, prove it holds for k = k_0 + 1.
The idea is if the base case is true, and if the statement holding for k _0 implies it also holds for k _0 + 1, the statement is true for all integers k starting from the base case.
This principle was used to prove that \(\text{a}^{k} \equiv \text{b}^{k}\text{(mod } \text{m})\). Initial value was checked for k = 1. The hypothesis was then assumed for some arbitrary k = k_0, and shown valid for k_0 + 1.
Working with Integer Powers
Integer powers extend basic arithmetic into repeated multiplication.
For example, \(\text{a}^{k}\) means multiplying a by itself k times.
In modular arithmetic, working with powers of integers can simplify complex calculations. For instance, when proving \(\text{a}^{k} \equiv \text{b}^{k}\text{(mod }\text{m})\), it involves both power operations and modular reductions.
Since we know that \( \text{a} \equiv \text{b}\text{(mod }\text{m})\), by definition \(\text{a}\text{ and } \text{b}\) leave the same remainders when divided by m. Utilizing this property, we can raise them to any integer power k and they will maintain the same congruence modulo m. This is made precise using the steps of mathematical induction.
The concept of raising both sides of a congruence to power k works because of the inherent properties of congruence relations which distribute through multiplication.
For example, \(\text{a}^{k}\) means multiplying a by itself k times.
In modular arithmetic, working with powers of integers can simplify complex calculations. For instance, when proving \(\text{a}^{k} \equiv \text{b}^{k}\text{(mod }\text{m})\), it involves both power operations and modular reductions.
Since we know that \( \text{a} \equiv \text{b}\text{(mod }\text{m})\), by definition \(\text{a}\text{ and } \text{b}\) leave the same remainders when divided by m. Utilizing this property, we can raise them to any integer power k and they will maintain the same congruence modulo m. This is made precise using the steps of mathematical induction.
The concept of raising both sides of a congruence to power k works because of the inherent properties of congruence relations which distribute through multiplication.