Is my proof by induction of even number arithmetic progression correct?
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
asked 2 days ago
NickNick
183
|
show 2 more comments
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
asked 2 days ago
NickNick
183
I think it's right.
– Michael Rozenberg
2 days ago
3
First one is wrong.
– Thomas Shelby
2 days ago
3
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.
– lulu
2 days ago
1
You can't make arguments using equation $(1) $,as you haven't proved it.
– Thomas Shelby
2 days ago
The first attempt isn't valid.
– Peter
2 days ago
|
show 2 more comments
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
asked 2 days ago
NickNick
183
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 2 days ago
NickNick
183
asked 2 days ago
NickNick
183
asked 2 days ago
NickNick
183
asked 2 days ago
NickNick
183
asked 2 days ago
NickNick
183
183
I think it's right.
– Michael Rozenberg
2 days ago
3
First one is wrong.
– Thomas Shelby
2 days ago
3
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.
– lulu
2 days ago
1
You can't make arguments using equation $(1) $,as you haven't proved it.
– Thomas Shelby
2 days ago
The first attempt isn't valid.
– Peter
2 days ago
|
show 2 more comments
I think it's right.
– Michael Rozenberg
2 days ago
3
First one is wrong.
– Thomas Shelby
2 days ago
3
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.
– lulu
2 days ago
1
You can't make arguments using equation $(1) $,as you haven't proved it.
– Thomas Shelby
2 days ago
The first attempt isn't valid.
– Peter
2 days ago
I think it's right.
– Michael Rozenberg
2 days ago
I think it's right.
– Michael Rozenberg
2 days ago
3
3
First one is wrong.
– Thomas Shelby
2 days ago
First one is wrong.
– Thomas Shelby
2 days ago
3
3
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.
– lulu
2 days ago
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.
– lulu
2 days ago
1
1
You can't make arguments using equation $(1) $,as you haven't proved it.
– Thomas Shelby
2 days ago
You can't make arguments using equation $(1) $,as you haven't proved it.
– Thomas Shelby
2 days ago
The first attempt isn't valid.
– Peter
2 days ago
The first attempt isn't valid.
– Peter
2 days ago
|
show 2 more comments
0
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I think it's right.
– Michael Rozenberg
2 days ago
3
First one is wrong.
– Thomas Shelby
2 days ago
3
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.
– lulu
2 days ago
1
You can't make arguments using equation $(1) $,as you haven't proved it.
– Thomas Shelby
2 days ago
The first attempt isn't valid.
– Peter
2 days ago