Mathematical induction help

[lc312]

Hi, i have problem with solving this issue, and the question is: "Prove that for every odd ? ∈ ℕ number
?^2 + 2? + 3 is even". I mean I did solve that but I am not shure if that is correct.

n^2+2n+3

Part 1.
n=1
1^2+2*1+3
1+2+3=6 =>3*2=6

Part 2.
n=k
k^2+2k+3=2*a, a ∈ ℕ

Part 3
n=k+1
(k+1)^2+2*(k+1)+3
k^2+2k+1+2k+2+3
(k^2+2k+3)+1+2k+2
2a+2*(k+1)
2*(a+k+1)


I dont know if this is the right solution but i would be verry grateful if you can help to solve this issue.
I dont have any ideas anymore.
Many thanks.

 

[Romsek]

If you're not instructed to use induction then for this problem I wouldn't use it.

[MATH]\text{$n$ odd implies that $n^2$ is also odd. $2n$ is always even. 3 is odd}\\ \text{Thus we have the sum of 2 odd numbers and an even number which is always even.}[/MATH]
Maybe they want you to also prove the assertion that [MATH]\text{$n$ odd $\rightarrow n^2$ odd}[/MATH]
[MATH]\text{$n$ odd $\rightarrow n = 2k+1,~k\in \mathbb{Z}$}\\ \text{$(2k+1)^2 = 4k^2 + 4k + 1 = 4(k^2+k)+1 = 2(2(k^2+k))+1$ i.e. is odd}[/MATH]

 

[Romsek]

If you really want to use induction you would do it as follows

[MATH]P_1 = 1^2 + 2(1) + 3 = 6 \text{ is even}\\ \text{6 is even so $P_1=True$}[/MATH]
[MATH]\text{Assume $P_n$ is $True$ and evaluate $P_{n+1}$}\\ (n+1)^2 + 2(n+1) + 3 = \\ n^2 + 2n + 1 + 2n + 2 + 3 = \\ (n^2 + 2n + 3) + (2n+3)\\ \text{as $P_n$ holds we have an odd number plus $2n$, an even number, plus 3, an odd number, which always sum to an even number}[/MATH]
[MATH]\text{Thus $P_{n+1} = True$ and by induction the statement holds for every $n \in \mathbb{N}$}[/MATH]

 

[lc312]

If you really want to use induction you would do it as follows

[MATH]P_1 = 1^2 + 2(1) + 3 = 6 \text{ is even}\\ \text{6 is even so $P_1=True$}[/MATH]
[MATH]\text{Assume $P_n$ is $True$ and evaluate $P_{n+1}$}\\ (n+1)^2 + 2(n+1) + 3 = \\ n^2 + 2n + 1 + 2n + 2 + 3 = \\ (n^2 + 2n + 3) + (2n+3)\\ \text{as $P_n$ holds we have an odd number plus $2n$, an even number, plus 3, an odd number, which always sum to an even number}[/MATH]
[MATH]\text{Thus $P_{n+1} = True$ and by induction the statement holds for every $n \in \mathbb{N}$}[/MATH]

Yes I had to use the inducition because for this exercise.
Thank you very much, so in the part when we assume that Pn is true, I dont have to put anything I just leave it like that or?
I want to know beacuse I want to be shure if I understood that.

 

[Steven G]

Hi, i have problem with solving this issue, and the question is: "Prove that for every odd ? ∈ ℕ number ?
^2 + 2? + 3 is even". I mean I did solve that but I am not shure if that is correct.

n^2+2n+3

Part 1.
n=1
1^2+2*1+3
1+2+3=6 =>3*2=6

Part 2.
n=k
k^2+2k+3=2*a, a ∈ ℕ

Part 3
n=k+1
(k+1)^2+2*(k+1)+3
k^2+2k+1+2k+2+3
(k^2+2k+3)+1+2k+2
2a+2*(k+1)
2*(a+k+1)


I dont know if this is the right solution but i would be verry grateful if you can help to solve this issue.
I dont have any ideas anymore.
Many thanks.

First of all you need to write equal signs. That way your teacher can mark where your equality is not valid.

k^2+2k+1+2k+2+3
(k^2+2k+3)+1+2k+2

What happened to the 1?

You seem to be forgetting that n is odd! Now if n=k is odd, then n=k+1 is even. You know this!! I would say that it is true for n= 2k+1 and then prove it true for the next allowable n which is n=2k+3.

 

[Steven G]

If you really want to use induction you would do it as follows

[MATH]P_1 = 1^2 + 2(1) + 3 = 6 \text{ is even}\\ \text{6 is even so $P_1=True$}[/MATH]
[MATH]\text{Assume $P_n$ is $True$ and evaluate $P_{n+1}$}\\ (n+1)^2 + 2(n+1) + 3 = \\ n^2 + 2n + 1 + 2n + 2 + 3 = \\ (n^2 + 2n + 3) + (2n+3)\\ \text{as $P_n$ holds we have an odd number plus $2n$, an even number, plus 3, an odd number, which always sum to an even number}[/MATH]
[MATH]\text{Thus $P_{n+1} = True$ and by induction the statement holds for every $n \in \mathbb{N}$}[/MATH]

I think it is corner time for you. You say that (n^2 + 2n + 3) is odd but it is even and that makes the entire sum odd, not even! You fell for what I initially fell for and did not make sure that n is odd. The difference is that I did not put my initial error in writing. Make sure that you stay in the corner for an odd number of minutes.

 

[Steven G]

Since the OP was told an incorrect proof (sorry Romsek) I will post a valid one.

n=1
1^2+2*1+3
1+2+3=6 which is even.

n=2k+1: Assume (2k+1)^2 + 2(2k+1) + 3 = 4k^2 +8k + 6 is even (which of course is obvious as 4k^2 +8k + 6 = 2(2k^2 + 4k + 1))

n= 2k+3: Show that (2k+3)^2 + 2(2k+3) + 3 is even
(2k+3)^2 + 2(2k+3) + 3 = 4k^2 + 16k + 18 = 2(2k^2 + 8k + 9) which is even. Or as Romsek would say, (2k+3)^2 is odd, 2(2k+3) is even and 3 is odd making the sum = odd + even + odd = even.

Note that sometimes you do not need to use the induction hypothesis!
I would never have done this proof by induction unless I was instructed to do so.

 

[Steven G]

Actually you do not need to prove that if a is odd then a^2 is odd to prove this theorem without induction.
n^2+2n+3 = (n^2 + 2n) + 3 = n(n+2) + 3. Now if n is odd we have n+2 is odd. Now an odd times an odd is odd. Adding 3 to an odd makes it even.

 

[JeffM]

Yes I had to use the inducition because for this exercise.
Thank you very much, so in the part when we assume that Pn is true, I dont have to put anything I just leave it like that or?
I want to know beacuse I want to be shure if I understood that.

What do you mean you "don't have to put anything." Romsek put quite a bit in after that. What in the world are you asking?

 

[lc312]

Since the OP was told an incorrect proof (sorry Romsek) I will post a valid one.

n=1
1^2+2*1+3
1+2+3=6 which is even.

n=2k+1: Assume (2k+1)^2 + 2(2k+1) + 3 = 4k^2 +8k + 6 is even (which of course is obvious as 4k^2 +8k + 6 = 2(2k^2 + 4k + 1))

n= 2k+3: Show that (2k+3)^2 + 2(2k+3) + 3 is even
(2k+3)^2 + 2(2k+3) + 3 = 4k^2 + 16k + 18 = 2(2k^2 + 8k + 9) which is even. Or as Romsek would say, (2k+3)^2 is odd, 2(2k+3) is even and 3 is odd making the sum = odd + even + odd = even.

Note that sometimes you do not need to use the induction hypothesis!
I would never have done this proof by induction unless I was instructed to do so.

Thank you very much for your help.
I was instructed to do this with induction thats the reason.

 

[Steven G]

Thank you very much for your help.
I was instructed to do this with induction thats the reason.

Doing it by induction is fine but do you understand that letting n=k and n=k+1 is not correct for this proof?