induction question understanding
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0
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Im wondering if this template for induction would be valid
show true for n=1
assume true for n=k
Attempt to show true for n=k+1
but at this point just replace n with k+1 and dont use the assumption that n=k.
show that the expression with n replaced by k+1 is equivalent to the one assumed for n=k.
Basically i have been asked to figure out if this is ok. My thoughts are
They assumed it for n=k ok
Then they showed it for n=k+1. Here i think what they have done is shown if true for k+1 then true for k.
But i guess it works the other way so i think its ok.
Any help would be appreciated, thanks
proof-writing
asked Nov 15 at 2:36
hitherematey
577
add a comment |
up vote
0
down vote
favorite
Im wondering if this template for induction would be valid
show true for n=1
assume true for n=k
Attempt to show true for n=k+1
but at this point just replace n with k+1 and dont use the assumption that n=k.
show that the expression with n replaced by k+1 is equivalent to the one assumed for n=k.
Basically i have been asked to figure out if this is ok. My thoughts are
They assumed it for n=k ok
Then they showed it for n=k+1. Here i think what they have done is shown if true for k+1 then true for k.
But i guess it works the other way so i think its ok.
Any help would be appreciated, thanks
proof-writing
asked Nov 15 at 2:36
hitherematey
577
It does not work the other way around. Suppose $S(n)$ is "$ n=1$". Then $S(1)$ is true. And for all $nin Bbb N$ we have $S(n+1)implies S(n).$ But it is obviously not true that $forall nin Bbb N,(S(n)).$... Forget about $k$. There is only one variable ($n$) to consider.
– DanielWainfleet
Nov 17 at 8:01
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Im wondering if this template for induction would be valid
show true for n=1
assume true for n=k
Attempt to show true for n=k+1
but at this point just replace n with k+1 and dont use the assumption that n=k.
show that the expression with n replaced by k+1 is equivalent to the one assumed for n=k.
Basically i have been asked to figure out if this is ok. My thoughts are
They assumed it for n=k ok
Then they showed it for n=k+1. Here i think what they have done is shown if true for k+1 then true for k.
But i guess it works the other way so i think its ok.
Any help would be appreciated, thanks
proof-writing
asked Nov 15 at 2:36
hitherematey
577
Im wondering if this template for induction would be valid
show true for n=1
assume true for n=k
Attempt to show true for n=k+1
but at this point just replace n with k+1 and dont use the assumption that n=k.
show that the expression with n replaced by k+1 is equivalent to the one assumed for n=k.
Basically i have been asked to figure out if this is ok. My thoughts are
They assumed it for n=k ok
Then they showed it for n=k+1. Here i think what they have done is shown if true for k+1 then true for k.
But i guess it works the other way so i think its ok.
Any help would be appreciated, thanks
proof-writing
proof-writing
asked Nov 15 at 2:36
hitherematey
577
asked Nov 15 at 2:36
hitherematey
577
asked Nov 15 at 2:36
hitherematey
577
asked Nov 15 at 2:36
hitherematey
577
asked Nov 15 at 2:36
hitherematey
577
577
It does not work the other way around. Suppose $S(n)$ is "$ n=1$". Then $S(1)$ is true. And for all $nin Bbb N$ we have $S(n+1)implies S(n).$ But it is obviously not true that $forall nin Bbb N,(S(n)).$... Forget about $k$. There is only one variable ($n$) to consider.
– DanielWainfleet
Nov 17 at 8:01
add a comment |
It does not work the other way around. Suppose $S(n)$ is "$ n=1$". Then $S(1)$ is true. And for all $nin Bbb N$ we have $S(n+1)implies S(n).$ But it is obviously not true that $forall nin Bbb N,(S(n)).$... Forget about $k$. There is only one variable ($n$) to consider.
– DanielWainfleet
Nov 17 at 8:01
It does not work the other way around. Suppose $S(n)$ is "$ n=1$". Then $S(1)$ is true. And for all $nin Bbb N$ we have $S(n+1)implies S(n).$ But it is obviously not true that $forall nin Bbb N,(S(n)).$... Forget about $k$. There is only one variable ($n$) to consider.
– DanielWainfleet
Nov 17 at 8:01
It does not work the other way around. Suppose $S(n)$ is "$ n=1$". Then $S(1)$ is true. And for all $nin Bbb N$ we have $S(n+1)implies S(n).$ But it is obviously not true that $forall nin Bbb N,(S(n)).$... Forget about $k$. There is only one variable ($n$) to consider.
– DanielWainfleet
Nov 17 at 8:01
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Assume true for n=k.
Attempt to show true for n=k+1.
Never mind "attempt." You must do it. Show true for n=k+1 using your assumption for n=k, and you will be done.
But at this point just replace n with k+1 and don't use the assumption that n=k.
This would be a waste of time. You would simply be making another assumption and proving nothing.
Tip: If you are stuck after proving true for 1, prove it true 2, 3 and 4 as well. This may help you to better understand the problem.
answered Nov 17 at 5:17
Dan Christensen
8,49021833
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assume true for n=k.
Attempt to show true for n=k+1.
Never mind "attempt." You must do it. Show true for n=k+1 using your assumption for n=k, and you will be done.
But at this point just replace n with k+1 and don't use the assumption that n=k.
This would be a waste of time. You would simply be making another assumption and proving nothing.
Tip: If you are stuck after proving true for 1, prove it true 2, 3 and 4 as well. This may help you to better understand the problem.
answered Nov 17 at 5:17
Dan Christensen
8,49021833
add a comment |
up vote
0
down vote
Assume true for n=k.
Attempt to show true for n=k+1.
Never mind "attempt." You must do it. Show true for n=k+1 using your assumption for n=k, and you will be done.
But at this point just replace n with k+1 and don't use the assumption that n=k.
This would be a waste of time. You would simply be making another assumption and proving nothing.
Tip: If you are stuck after proving true for 1, prove it true 2, 3 and 4 as well. This may help you to better understand the problem.
answered Nov 17 at 5:17
Dan Christensen
8,49021833
add a comment |
up vote
0
down vote
up vote
0
down vote
Assume true for n=k.
Attempt to show true for n=k+1.
Never mind "attempt." You must do it. Show true for n=k+1 using your assumption for n=k, and you will be done.
But at this point just replace n with k+1 and don't use the assumption that n=k.
This would be a waste of time. You would simply be making another assumption and proving nothing.
Tip: If you are stuck after proving true for 1, prove it true 2, 3 and 4 as well. This may help you to better understand the problem.
answered Nov 17 at 5:17
Dan Christensen
8,49021833
Assume true for n=k.
Attempt to show true for n=k+1.
Never mind "attempt." You must do it. Show true for n=k+1 using your assumption for n=k, and you will be done.
But at this point just replace n with k+1 and don't use the assumption that n=k.
This would be a waste of time. You would simply be making another assumption and proving nothing.
Tip: If you are stuck after proving true for 1, prove it true 2, 3 and 4 as well. This may help you to better understand the problem.
answered Nov 17 at 5:17
Dan Christensen
8,49021833
edited Nov 17 at 5:40
answered Nov 17 at 5:17
Dan Christensen
8,49021833
answered Nov 17 at 5:17
Dan Christensen
8,49021833
answered Nov 17 at 5:17
Dan Christensen
8,49021833
8,49021833
add a comment |
add a comment |
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It does not work the other way around. Suppose $S(n)$ is "$ n=1$". Then $S(1)$ is true. And for all $nin Bbb N$ we have $S(n+1)implies S(n).$ But it is obviously not true that $forall nin Bbb N,(S(n)).$... Forget about $k$. There is only one variable ($n$) to consider.
– DanielWainfleet
Nov 17 at 8:01