Mixing
How to prove by method of induction a statement with factorials
0
$begingroup$
The question:
Prove that for all positive integers $n geq 4$, $2^{n} < n!$
The base case is fine, but how do you proceed with inductive step?
$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$
elementary-number-theory proof-verification proof-explanation
asked Jan 31 at 0:24
ForextraderForextrader
988
$endgroup$
add a comment |
0
$begingroup$
The question:
Prove that for all positive integers $n geq 4$, $2^{n} < n!$
The base case is fine, but how do you proceed with inductive step?
$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$
elementary-number-theory proof-verification proof-explanation
asked Jan 31 at 0:24
ForextraderForextrader
988
$endgroup$
$begingroup$
For a good look at induction proofs, see this great answer
$endgroup$
– Ross Millikan
Jan 31 at 0:30
1
$begingroup$
You've written the inductive step oddly. To keep the logic straight in your head, it might help to write correctly: $$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$$ or, in words, $$2^{k} < k! ,,text{implies},, 2^{k+1} < (k+1)!$$
$endgroup$
– Lee Mosher
Jan 31 at 0:32
$begingroup$
Yes you're right. Sloppy on my end. Will change.
$endgroup$
– Forextrader
Jan 31 at 0:37
add a comment |
0
0
0
$begingroup$
The question:
Prove that for all positive integers $n geq 4$, $2^{n} < n!$
The base case is fine, but how do you proceed with inductive step?
$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$
elementary-number-theory proof-verification proof-explanation
asked Jan 31 at 0:24
ForextraderForextrader
988
$endgroup$
The question:
Prove that for all positive integers $n geq 4$, $2^{n} < n!$
The base case is fine, but how do you proceed with inductive step?
$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$
elementary-number-theory proof-verification proof-explanation
elementary-number-theory proof-verification proof-explanation
asked Jan 31 at 0:24
ForextraderForextrader
988
asked Jan 31 at 0:24
ForextraderForextrader
988
edited Jan 31 at 0:37
Forextrader
asked Jan 31 at 0:24
ForextraderForextrader
988
asked Jan 31 at 0:24
ForextraderForextrader
988
asked Jan 31 at 0:24
ForextraderForextrader
988
988
$begingroup$
For a good look at induction proofs, see this great answer
$endgroup$
– Ross Millikan
Jan 31 at 0:30
1
$begingroup$
You've written the inductive step oddly. To keep the logic straight in your head, it might help to write correctly: $$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$$ or, in words, $$2^{k} < k! ,,text{implies},, 2^{k+1} < (k+1)!$$
$endgroup$
– Lee Mosher
Jan 31 at 0:32
$begingroup$
Yes you're right. Sloppy on my end. Will change.
$endgroup$
– Forextrader
Jan 31 at 0:37
add a comment |
$begingroup$
For a good look at induction proofs, see this great answer
$endgroup$
– Ross Millikan
Jan 31 at 0:30
1
$begingroup$
You've written the inductive step oddly. To keep the logic straight in your head, it might help to write correctly: $$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$$ or, in words, $$2^{k} < k! ,,text{implies},, 2^{k+1} < (k+1)!$$
$endgroup$
– Lee Mosher
Jan 31 at 0:32
$begingroup$
Yes you're right. Sloppy on my end. Will change.
$endgroup$
– Forextrader
Jan 31 at 0:37
$begingroup$
For a good look at induction proofs, see this great answer
$endgroup$
– Ross Millikan
Jan 31 at 0:30
$begingroup$
For a good look at induction proofs, see this great answer
$endgroup$
– Ross Millikan
Jan 31 at 0:30
1
1
$begingroup$
You've written the inductive step oddly. To keep the logic straight in your head, it might help to write correctly: $$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$$ or, in words, $$2^{k} < k! ,,text{implies},, 2^{k+1} < (k+1)!$$
$endgroup$
– Lee Mosher
Jan 31 at 0:32
$begingroup$
You've written the inductive step oddly. To keep the logic straight in your head, it might help to write correctly: $$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$$ or, in words, $$2^{k} < k! ,,text{implies},, 2^{k+1} < (k+1)!$$
$endgroup$
– Lee Mosher
Jan 31 at 0:32
$begingroup$
Yes you're right. Sloppy on my end. Will change.
$endgroup$
– Forextrader
Jan 31 at 0:37
$begingroup$
Yes you're right. Sloppy on my end. Will change.
$endgroup$
– Forextrader
Jan 31 at 0:37
add a comment |
1 Answer
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0
$begingroup$
Intuitively, to go from $n!$ to $(n+1)!$ you multiply by $n+1$, which is greater than $2$.
answered Jan 31 at 0:29
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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1 Answer
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0
$begingroup$
Intuitively, to go from $n!$ to $(n+1)!$ you multiply by $n+1$, which is greater than $2$.
answered Jan 31 at 0:29
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
0
$begingroup$
Intuitively, to go from $n!$ to $(n+1)!$ you multiply by $n+1$, which is greater than $2$.
answered Jan 31 at 0:29
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
0
0
0
$begingroup$
Intuitively, to go from $n!$ to $(n+1)!$ you multiply by $n+1$, which is greater than $2$.
answered Jan 31 at 0:29
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Intuitively, to go from $n!$ to $(n+1)!$ you multiply by $n+1$, which is greater than $2$.
answered Jan 31 at 0:29
Ross MillikanRoss Millikan
301k24200375
answered Jan 31 at 0:29
Ross MillikanRoss Millikan
301k24200375
answered Jan 31 at 0:29
Ross MillikanRoss Millikan
301k24200375
answered Jan 31 at 0:29
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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$begingroup$
For a good look at induction proofs, see this great answer
$endgroup$
– Ross Millikan
Jan 31 at 0:30
1
$begingroup$
You've written the inductive step oddly. To keep the logic straight in your head, it might help to write correctly: $$(2^{k} < k!) implies ( 2^{k+1} < (k+1)!)$$ or, in words, $$2^{k} < k! ,,text{implies},, 2^{k+1} < (k+1)!$$
$endgroup$
– Lee Mosher
Jan 31 at 0:32
$begingroup$
Yes you're right. Sloppy on my end. Will change.
$endgroup$
– Forextrader
Jan 31 at 0:37