The solution to this question below is a bit confusing to me. If anyone would be kind enough to explain how the inductive claim works and why it is true for $3a_{n+1} + 5b_{n+1} = n+1$ This would help me a lot, thank you in advance.
For all non-negative integers $n$ such that $n ≥ 8$, there exist non-negative integers $a_n$ and $b_n$ such that $3a_n+5b_n = n$
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discrete-mathematics
induction
proof-explanation
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0Outside links are frowned upon. Please put all relevant info in the question itself, and use LaTeX formatting to improve readability. Also, your title could be more descriptive. – 2017-01-02
2 Answers
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HINT:
- $A_{8}=\{3,5\}$
- $A_{9}=\{3,3,3\}$
- $A_{10}=\{5,5\}$
- $A_{n}=A_{n-3}\cup\{3\}$
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Claim : For every natural number $n\ge 8$, there exist nonnegative integers $a$ and $b$ with $3a+5b=n$
Base case $n=8$ : Take $a=1$ and $b=1$
Suppose, we have $3a+5b=n$
For $b\ge 1$, we have $3(a+2)+5(b-1)=3a+5b+1=n+1$
For $a\ge 3$, we have $3(a-3)+5(b+2)=3a+5b+1=n+1$
So, the induction step works whenever $b\ge 1$ or $a\ge 3$. The remaining case is $a<3$ and $b=0$, but the corresponding numbers are $0,3,6$, which are smaller than $8$