Give a recursive definition of
a) The set of odd positive integers.
b) The set of positive integers powers of 3.
Solution for a)
$ a^0 =1$
$ a^n = 2n+1 $
Is that right ? and how can i find b) ?
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For the first, I would use $a_0=1$, $a_{n+1}=a_n+2$. Perhaps this will suggest what to do for the second question. – André Nicolas Dec 05 '13 at 05:04
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You are welcome. At the risk of confusing you, let me say that the definition $a_n=2n+1$ is technically recursive. But I am basically sure that the answer I gave is the intended one. – André Nicolas Dec 05 '13 at 05:20
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1Yeah your answer is right. But actually when i saw the test's answers which the prof wrote i found that he wanted it to be like that "$ 3 ∈ s , if x ∈ s ;then; 3 * x ∈ s" $ – Out Of Bounds Dec 05 '13 at 05:24
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That is an alternate "set-theoretic" way of saying $a_0=1$, $a_{n+1}=3a_n$. – André Nicolas Dec 05 '13 at 05:29
2 Answers
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a) $a_0 = 1$; $a_{n+1} = a_n + 2$
This is recursive because here you have terms referring to 'previous' terms, until we reach the base case $a_0$. Your definition of $a_n = 2n+1$ is not recursive because you're not recursing to a base case; your definition is just a straightforward computation.
b) $a_0 = 1$; $a_{n+1} = a_n \cdot 3$
This gives us $1, 3, 9, \ldots$ so it works. (Noting that $3^0=1$.)
Newb
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Oh. I may have misread your question. I thought you meant the positive integers that are powers of three: $1,3,9,27,\ldots$. Did you mean the positive integers raised to the third power: $1^3,2^3,3^3,\ldots$? How you phrase your question is important! – Newb Dec 05 '13 at 05:16
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It was a question on my test and it says exactly like that " set of positive integers powers of 3 " – Out Of Bounds Dec 05 '13 at 05:17
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I thought powers of 3 means that every integer raised to the third power. – Out Of Bounds Dec 05 '13 at 05:18
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The question is unclear. Does it mean the set of positive integers that ARE powers of three or does it mean the set of positive integers, RAISED TO the power of three? Ask your teacher. – Newb Dec 05 '13 at 05:18
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@SteveODonnell $1$ is a positive integer that is a power of of 3, namely $3^0$. – Newb Dec 05 '13 at 08:00
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An issue that the accepted answer doesn't address is that the question asks for a set, not a sequence.
The proper answer is:
Base step:
$3 \in S$
Recursive step:
$x \in S \implies 3x \in S$
This reads that 3 is in set $S$. If x is in $S$, then $3x$ is also in $S$, thus recursively defining this set because it builds $\{\space3,\space9\space(3(3)),\space27\space(3(9)),\space81\space(3(27))\space...\space\}$
Henry
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