Is my proof by induction of even number arithmetic progression correct?
$begingroup$
Prove that for all integers $ngeq1$, $P(n) = 2+4+6+...+2n=n^2+n$
For $P(1)$, we have that: $2cdot1=1^2+1$
Let $k$ be an integer with $kgeq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$
I must show that $P(k+1)$ is also true:
$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$
If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.
$Q.E.D$
A small change at the inductive step:
I must show that $P(k+1)$ is also true:
$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$
$Q.E.D$
I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.
discrete-mathematics induction
$endgroup$
|
show 2 more comments
$begingroup$
Prove that for all integers $ngeq1$, $P(n) = 2+4+6+...+2n=n^2+n$
For $P(1)$, we have that: $2cdot1=1^2+1$
Let $k$ be an integer with $kgeq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$
I must show that $P(k+1)$ is also true:
$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$
If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.
$Q.E.D$
A small change at the inductive step:
I must show that $P(k+1)$ is also true:
$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$
$Q.E.D$
I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.
discrete-mathematics induction
$endgroup$
$begingroup$
I think it's right.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:07
3
$begingroup$
First one is wrong.
$endgroup$
– Thomas Shelby
Jan 5 at 13:07
3
$begingroup$
I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
$endgroup$
– lulu
Jan 5 at 13:09
1
$begingroup$
You can't make arguments using equation $(1) $,as you haven't proved it.
$endgroup$
– Thomas Shelby
Jan 5 at 13:10
$begingroup$
The first attempt isn't valid.
$endgroup$
– Peter
Jan 5 at 13:14
|
show 2 more comments
$begingroup$
Prove that for all integers $ngeq1$, $P(n) = 2+4+6+...+2n=n^2+n$
For $P(1)$, we have that: $2cdot1=1^2+1$
Let $k$ be an integer with $kgeq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$
I must show that $P(k+1)$ is also true:
$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$
If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.
$Q.E.D$
A small change at the inductive step:
I must show that $P(k+1)$ is also true:
$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$
$Q.E.D$
I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.
discrete-mathematics induction
$endgroup$
Prove that for all integers $ngeq1$, $P(n) = 2+4+6+...+2n=n^2+n$
For $P(1)$, we have that: $2cdot1=1^2+1$
Let $k$ be an integer with $kgeq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$
I must show that $P(k+1)$ is also true:
$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$
If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.
$Q.E.D$
A small change at the inductive step:
I must show that $P(k+1)$ is also true:
$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$
$Q.E.D$
I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.
discrete-mathematics induction
discrete-mathematics induction
asked Jan 5 at 13:04
NickNick
183
183
$begingroup$
I think it's right.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:07
3
$begingroup$
First one is wrong.
$endgroup$
– Thomas Shelby
Jan 5 at 13:07
3
$begingroup$
I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
$endgroup$
– lulu
Jan 5 at 13:09
1
$begingroup$
You can't make arguments using equation $(1) $,as you haven't proved it.
$endgroup$
– Thomas Shelby
Jan 5 at 13:10
$begingroup$
The first attempt isn't valid.
$endgroup$
– Peter
Jan 5 at 13:14
|
show 2 more comments
$begingroup$
I think it's right.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:07
3
$begingroup$
First one is wrong.
$endgroup$
– Thomas Shelby
Jan 5 at 13:07
3
$begingroup$
I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
$endgroup$
– lulu
Jan 5 at 13:09
1
$begingroup$
You can't make arguments using equation $(1) $,as you haven't proved it.
$endgroup$
– Thomas Shelby
Jan 5 at 13:10
$begingroup$
The first attempt isn't valid.
$endgroup$
– Peter
Jan 5 at 13:14
$begingroup$
I think it's right.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:07
$begingroup$
I think it's right.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:07
3
3
$begingroup$
First one is wrong.
$endgroup$
– Thomas Shelby
Jan 5 at 13:07
$begingroup$
First one is wrong.
$endgroup$
– Thomas Shelby
Jan 5 at 13:07
3
3
$begingroup$
I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
$endgroup$
– lulu
Jan 5 at 13:09
$begingroup$
I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
$endgroup$
– lulu
Jan 5 at 13:09
1
1
$begingroup$
You can't make arguments using equation $(1) $,as you haven't proved it.
$endgroup$
– Thomas Shelby
Jan 5 at 13:10
$begingroup$
You can't make arguments using equation $(1) $,as you haven't proved it.
$endgroup$
– Thomas Shelby
Jan 5 at 13:10
$begingroup$
The first attempt isn't valid.
$endgroup$
– Peter
Jan 5 at 13:14
$begingroup$
The first attempt isn't valid.
$endgroup$
– Peter
Jan 5 at 13:14
|
show 2 more comments
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$begingroup$
I think it's right.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:07
3
$begingroup$
First one is wrong.
$endgroup$
– Thomas Shelby
Jan 5 at 13:07
3
$begingroup$
I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
$endgroup$
– lulu
Jan 5 at 13:09
1
$begingroup$
You can't make arguments using equation $(1) $,as you haven't proved it.
$endgroup$
– Thomas Shelby
Jan 5 at 13:10
$begingroup$
The first attempt isn't valid.
$endgroup$
– Peter
Jan 5 at 13:14