Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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![### Mathematical Induction Proof
**Objective:**
Prove that \(6^n -1\) is divisible by 5 for all natural numbers \(n\).
**Step 1: Base Case**
Show true for \(n = 1\):
\[ 6^1 - 1 = 5 \]
Since 5 is clearly divisible by 5, the base case holds true.
**Step 2: Inductive Hypothesis**
Assume the statement is true for \(n = k\). That is, assume:
\[ 6^k - 1 \text{ is divisible by } 5 \]
or equivalently,
\[ 6^k - 1 = 5m \]
for some integer \(m\).
**Step 3: Inductive Step**
We need to show the statement is true for \(n = k + 1\). That is, we need to prove:
\[ 6^{k+1} - 1 \text{ is divisible by } 5 \]
Starting with the left hand side:
\[ 6^{k+1} - 1 \]
Rewrite \(6^{k+1}\) as \(6 \cdot 6^k\):
\[ 6 \cdot 6^k - 1 \]
Now, use the inductive hypothesis:
\[ 6 \cdot (5m + 1) - 1 \]
Distribute 6 in the brackets:
\[ 6 \cdot 5m + 6 \cdot 1 - 1 = 30m + 6 - 1 = 30m + 5 \]
Factor 5 out of the expression:
\[ 5(6m + 1) \]
Since the result is \(5\) multiplied by an integer \((6m + 1)\), it is divisible by \(5\).
**Step 4: Conclusion**
By the principle of mathematical induction, \(6^n - 1\) is divisible by 5 for all natural numbers \(n\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c88b4b6-4152-4350-b43c-eadd21488152%2F680f9144-a4d0-4106-9be8-d470c4d05bbb%2F8vjbsgi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Mathematical Induction Proof
**Objective:**
Prove that \(6^n -1\) is divisible by 5 for all natural numbers \(n\).
**Step 1: Base Case**
Show true for \(n = 1\):
\[ 6^1 - 1 = 5 \]
Since 5 is clearly divisible by 5, the base case holds true.
**Step 2: Inductive Hypothesis**
Assume the statement is true for \(n = k\). That is, assume:
\[ 6^k - 1 \text{ is divisible by } 5 \]
or equivalently,
\[ 6^k - 1 = 5m \]
for some integer \(m\).
**Step 3: Inductive Step**
We need to show the statement is true for \(n = k + 1\). That is, we need to prove:
\[ 6^{k+1} - 1 \text{ is divisible by } 5 \]
Starting with the left hand side:
\[ 6^{k+1} - 1 \]
Rewrite \(6^{k+1}\) as \(6 \cdot 6^k\):
\[ 6 \cdot 6^k - 1 \]
Now, use the inductive hypothesis:
\[ 6 \cdot (5m + 1) - 1 \]
Distribute 6 in the brackets:
\[ 6 \cdot 5m + 6 \cdot 1 - 1 = 30m + 6 - 1 = 30m + 5 \]
Factor 5 out of the expression:
\[ 5(6m + 1) \]
Since the result is \(5\) multiplied by an integer \((6m + 1)\), it is divisible by \(5\).
**Step 4: Conclusion**
By the principle of mathematical induction, \(6^n - 1\) is divisible by 5 for all natural numbers \(n\).
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