Induction proof verification?
$begingroup$
Using induction, I have to prove that $n^2+n$ is divisible by 2.
Here's how I did it, and I wanted to know if this is considered valid. I started with $(k+1)^2+(k+1)$, and after simplifying and factoring, I got $(k+2)(k+1)$. From this, can't I just say that if $k$ is odd, then the term $(k+1)$ becomes an even number, and when an even number is multiplied by an odd number, you get an even number which is always divisible by $2$. Then, the same logic can be applied for when $k$ is even and added to the term $(k+2)$? $Q.E.D$...?
proof-verification induction
asked Dec 10 '18 at 1:38
Jon dueJon due
858
$endgroup$
add a comment |
$begingroup$
Using induction, I have to prove that $n^2+n$ is divisible by 2.
Here's how I did it, and I wanted to know if this is considered valid. I started with $(k+1)^2+(k+1)$, and after simplifying and factoring, I got $(k+2)(k+1)$. From this, can't I just say that if $k$ is odd, then the term $(k+1)$ becomes an even number, and when an even number is multiplied by an odd number, you get an even number which is always divisible by $2$. Then, the same logic can be applied for when $k$ is even and added to the term $(k+2)$? $Q.E.D$...?
proof-verification induction
asked Dec 10 '18 at 1:38
Jon dueJon due
858
$endgroup$
add a comment |
$begingroup$
Using induction, I have to prove that $n^2+n$ is divisible by 2.
Here's how I did it, and I wanted to know if this is considered valid. I started with $(k+1)^2+(k+1)$, and after simplifying and factoring, I got $(k+2)(k+1)$. From this, can't I just say that if $k$ is odd, then the term $(k+1)$ becomes an even number, and when an even number is multiplied by an odd number, you get an even number which is always divisible by $2$. Then, the same logic can be applied for when $k$ is even and added to the term $(k+2)$? $Q.E.D$...?
proof-verification induction
asked Dec 10 '18 at 1:38
Jon dueJon due
858
$endgroup$
Using induction, I have to prove that $n^2+n$ is divisible by 2.
Here's how I did it, and I wanted to know if this is considered valid. I started with $(k+1)^2+(k+1)$, and after simplifying and factoring, I got $(k+2)(k+1)$. From this, can't I just say that if $k$ is odd, then the term $(k+1)$ becomes an even number, and when an even number is multiplied by an odd number, you get an even number which is always divisible by $2$. Then, the same logic can be applied for when $k$ is even and added to the term $(k+2)$? $Q.E.D$...?
proof-verification induction
proof-verification induction
asked Dec 10 '18 at 1:38
Jon dueJon due
858
asked Dec 10 '18 at 1:38
Jon dueJon due
858
asked Dec 10 '18 at 1:38
Jon dueJon due
858
asked Dec 10 '18 at 1:38
Jon dueJon due
858
asked Dec 10 '18 at 1:38
Jon dueJon due
858
858
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are many ways to prove this statement, but you are instructed to do so by induction. This means you must use the inductive hypothesis, i.e. will require you "capturing" something of the form $k^2+k$.
In this case:
$$(k+1)^2+(k+1) = (k^2+k)+2(k+1)$$
answered Dec 10 '18 at 1:42
David PetersonDavid Peterson
8,87621935
$endgroup$
2
$begingroup$
+1 I wish teachers would require induction only where it's the method of choice ...
$endgroup$
– Ethan Bolker
Dec 10 '18 at 1:47
add a comment |
$begingroup$
Base Case is trivial.
We assume true for $n=k$, that is: $k^2+k=2m, min Bbb Z$
Then show for $n=k+1$:
$$(k+1)^2+k+1=k^2+2k+1+k+1=k^2+3k+2$$
$$=(k^2+k)+2k+2=2m+2k+2=2(m+k+1)$$
answered Dec 10 '18 at 1:45
Rhys HughesRhys Hughes
6,7271530
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add a comment |
$begingroup$
$begin{align}{bf Hint} f(k) &= (k+1),k\
Rightarrow f(k+1) &= (k+1)(k+2)
end{align}$
$,Rightarrow f(k)$ & $f(k+1)$ have equal parity since $,k,$ & $,k+2,$ do. $ $
Therefore $,f(k),$ even $,Rightarrow, f(k+1),$ even
answered Dec 10 '18 at 1:57
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
$begingroup$
Notice no computation is needed this way (except adding 1).
$endgroup$
– Bill Dubuque
Dec 10 '18 at 1:59
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are many ways to prove this statement, but you are instructed to do so by induction. This means you must use the inductive hypothesis, i.e. will require you "capturing" something of the form $k^2+k$.
In this case:
$$(k+1)^2+(k+1) = (k^2+k)+2(k+1)$$
answered Dec 10 '18 at 1:42
David PetersonDavid Peterson
8,87621935
$endgroup$
2
$begingroup$
+1 I wish teachers would require induction only where it's the method of choice ...
$endgroup$
– Ethan Bolker
Dec 10 '18 at 1:47
add a comment |
$begingroup$
There are many ways to prove this statement, but you are instructed to do so by induction. This means you must use the inductive hypothesis, i.e. will require you "capturing" something of the form $k^2+k$.
In this case:
$$(k+1)^2+(k+1) = (k^2+k)+2(k+1)$$
answered Dec 10 '18 at 1:42
David PetersonDavid Peterson
8,87621935
$endgroup$
2
$begingroup$
+1 I wish teachers would require induction only where it's the method of choice ...
$endgroup$
– Ethan Bolker
Dec 10 '18 at 1:47
add a comment |
$begingroup$
There are many ways to prove this statement, but you are instructed to do so by induction. This means you must use the inductive hypothesis, i.e. will require you "capturing" something of the form $k^2+k$.
In this case:
$$(k+1)^2+(k+1) = (k^2+k)+2(k+1)$$
answered Dec 10 '18 at 1:42
David PetersonDavid Peterson
8,87621935
$endgroup$
There are many ways to prove this statement, but you are instructed to do so by induction. This means you must use the inductive hypothesis, i.e. will require you "capturing" something of the form $k^2+k$.
In this case:
$$(k+1)^2+(k+1) = (k^2+k)+2(k+1)$$
answered Dec 10 '18 at 1:42
David PetersonDavid Peterson
8,87621935
answered Dec 10 '18 at 1:42
David PetersonDavid Peterson
8,87621935
answered Dec 10 '18 at 1:42
David PetersonDavid Peterson
8,87621935
answered Dec 10 '18 at 1:42
David PetersonDavid Peterson
8,87621935
8,87621935
2
$begingroup$
+1 I wish teachers would require induction only where it's the method of choice ...
$endgroup$
– Ethan Bolker
Dec 10 '18 at 1:47
add a comment |
2
$begingroup$
+1 I wish teachers would require induction only where it's the method of choice ...
$endgroup$
– Ethan Bolker
Dec 10 '18 at 1:47
2
2
$begingroup$
+1 I wish teachers would require induction only where it's the method of choice ...
$endgroup$
– Ethan Bolker
Dec 10 '18 at 1:47
$begingroup$
+1 I wish teachers would require induction only where it's the method of choice ...
$endgroup$
– Ethan Bolker
Dec 10 '18 at 1:47
add a comment |
$begingroup$
Base Case is trivial.
We assume true for $n=k$, that is: $k^2+k=2m, min Bbb Z$
Then show for $n=k+1$:
$$(k+1)^2+k+1=k^2+2k+1+k+1=k^2+3k+2$$
$$=(k^2+k)+2k+2=2m+2k+2=2(m+k+1)$$
answered Dec 10 '18 at 1:45
Rhys HughesRhys Hughes
6,7271530
$endgroup$
add a comment |
$begingroup$
Base Case is trivial.
We assume true for $n=k$, that is: $k^2+k=2m, min Bbb Z$
Then show for $n=k+1$:
$$(k+1)^2+k+1=k^2+2k+1+k+1=k^2+3k+2$$
$$=(k^2+k)+2k+2=2m+2k+2=2(m+k+1)$$
answered Dec 10 '18 at 1:45
Rhys HughesRhys Hughes
6,7271530
$endgroup$
add a comment |
$begingroup$
Base Case is trivial.
We assume true for $n=k$, that is: $k^2+k=2m, min Bbb Z$
Then show for $n=k+1$:
$$(k+1)^2+k+1=k^2+2k+1+k+1=k^2+3k+2$$
$$=(k^2+k)+2k+2=2m+2k+2=2(m+k+1)$$
answered Dec 10 '18 at 1:45
Rhys HughesRhys Hughes
6,7271530
$endgroup$
Base Case is trivial.
We assume true for $n=k$, that is: $k^2+k=2m, min Bbb Z$
Then show for $n=k+1$:
$$(k+1)^2+k+1=k^2+2k+1+k+1=k^2+3k+2$$
$$=(k^2+k)+2k+2=2m+2k+2=2(m+k+1)$$
answered Dec 10 '18 at 1:45
Rhys HughesRhys Hughes
6,7271530
answered Dec 10 '18 at 1:45
Rhys HughesRhys Hughes
6,7271530
answered Dec 10 '18 at 1:45
Rhys HughesRhys Hughes
6,7271530
answered Dec 10 '18 at 1:45
Rhys HughesRhys Hughes
6,7271530
6,7271530
add a comment |
add a comment |
$begingroup$
$begin{align}{bf Hint} f(k) &= (k+1),k\
Rightarrow f(k+1) &= (k+1)(k+2)
end{align}$
$,Rightarrow f(k)$ & $f(k+1)$ have equal parity since $,k,$ & $,k+2,$ do. $ $
Therefore $,f(k),$ even $,Rightarrow, f(k+1),$ even
answered Dec 10 '18 at 1:57
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
$begingroup$
Notice no computation is needed this way (except adding 1).
$endgroup$
– Bill Dubuque
Dec 10 '18 at 1:59
add a comment |
$begingroup$
$begin{align}{bf Hint} f(k) &= (k+1),k\
Rightarrow f(k+1) &= (k+1)(k+2)
end{align}$
$,Rightarrow f(k)$ & $f(k+1)$ have equal parity since $,k,$ & $,k+2,$ do. $ $
Therefore $,f(k),$ even $,Rightarrow, f(k+1),$ even
answered Dec 10 '18 at 1:57
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
$begingroup$
Notice no computation is needed this way (except adding 1).
$endgroup$
– Bill Dubuque
Dec 10 '18 at 1:59
add a comment |
$begingroup$
$begin{align}{bf Hint} f(k) &= (k+1),k\
Rightarrow f(k+1) &= (k+1)(k+2)
end{align}$
$,Rightarrow f(k)$ & $f(k+1)$ have equal parity since $,k,$ & $,k+2,$ do. $ $
Therefore $,f(k),$ even $,Rightarrow, f(k+1),$ even
answered Dec 10 '18 at 1:57
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
$begin{align}{bf Hint} f(k) &= (k+1),k\
Rightarrow f(k+1) &= (k+1)(k+2)
end{align}$
$,Rightarrow f(k)$ & $f(k+1)$ have equal parity since $,k,$ & $,k+2,$ do. $ $
Therefore $,f(k),$ even $,Rightarrow, f(k+1),$ even
answered Dec 10 '18 at 1:57
Bill DubuqueBill Dubuque
210k29192645
answered Dec 10 '18 at 1:57
Bill DubuqueBill Dubuque
210k29192645
answered Dec 10 '18 at 1:57
Bill DubuqueBill Dubuque
210k29192645
answered Dec 10 '18 at 1:57
Bill DubuqueBill Dubuque
210k29192645
210k29192645
$begingroup$
Notice no computation is needed this way (except adding 1).
$endgroup$
– Bill Dubuque
Dec 10 '18 at 1:59
add a comment |
$begingroup$
Notice no computation is needed this way (except adding 1).
$endgroup$
– Bill Dubuque
Dec 10 '18 at 1:59
$begingroup$
Notice no computation is needed this way (except adding 1).
$endgroup$
– Bill Dubuque
Dec 10 '18 at 1:59
$begingroup$
Notice no computation is needed this way (except adding 1).
$endgroup$
– Bill Dubuque
Dec 10 '18 at 1:59
add a comment |
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