PreCalculus Show that 2|[(n+2)^2-n^2] for all integers n

PeachBlossom

PeachBlossom

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My question is, did i do the induktion correctly? Is it right that it should be 4?

Because in my first statement 8/2=4

Thanks for help in advance! Appreciate it.

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I would like to ask, is there a simple method to remember how to do induction.

I have a huge problem with this step.

There are allot of different questions that I have on induction. And I get very lost and stuck because all the questions look very different. When I go to a new question I struggle with induction every time.

Please any tips what to do. Is there any rule for induction to solve any problem no matten how they look like.

Im struggeling allot with this subject. Only got questions from teacher on paper with barely any explenation how to do this and nothing about this in my book.

Recomment me a site or something. Anything that can improve me on this topic.

Appreciate the help in advance!

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1. State the proposed rule for a value "n."
2. Find a "base case," usually it's some small number like n = 0 or n = 1.
3. Suppose that the rule is true for some whole number n = k. Show that the rule works for the case k + 1 as well.

Usually at some point you will use the rule for n = k in step 3 to simplify the problem in showing the k + 1 case.

Otherwise there aren't any hard and fast rules. It depends a lot on the specific problem.

Armed with this why don't you try some of these problems out (one at a time!) and post your efforts.

-Dan
 
Thanks! For tips. Yeah thats what I am doing. Trying to solve one problem at a time by myself and then post it.
 
I have three comments. First comment: double check your math:

\(\displaystyle [(k+1) + 2]^2 \neq (k+1)^2 + 2^2\)

You appear to be thinking that there's a general rule:

\(\displaystyle (a + b)^2 = a^2 + b^2\)

But that's absolutely not true! Consider:

\(\displaystyle (2 + 3)^2 = 5^2 = 25 \text{ but } 2^2 + 3^2 = 4 + 9 = 13\)

Rather the real rule is:

\(\displaystyle (a + b)^2 = a^2 + 2ab + b^2\)

Which you can verify by expanding the left-hand side (FOIL perhaps?).

Second comment: Once that error's corrected and you show that the proof does work for \(n = k + 1\), you're not done. The problem wants you show it's true for all integers. You've only shown it works for the positive integers. You'd need to add another induction step to show it also holds for \(n = k - 1\).

Third comment: Does the problem actually even require you to use induction? Or can you prove it by any means you see fit? If you don't have to use induction, then just simplify the expression in terms of \(n\) to show that:

\(\displaystyle (n+2)^2-n^2 = 4(n+1)\)

And you're done.
 
As ksdhart2 said, (n+2)2 - n2 = [(n+2)+n][(n+2)-n] = (2n+2)(2) which 2 clearly divides. Induction is not needed at all.
 
Sorry if im going to sound a bit dumb. To be honest I don't understand at all this topic, realizing it now, im struggeling very bad. Have tried to search the internet and watched youtube videos, for 2 whole days, nothing is really helping. Don't know what to do for me to understand this properly. Wan't to understand this kind of math really bad. Im stuck on every question on elements on number theory. :(
 
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Sorry if im going to sound a bit dumb. To be honest I don't understand at all this topic, realizing it now, im struggeling very bad. Have tried to search the internet and watched youtube videos, for 2 whole days, nothing is really helping. Don't know what to do for me to understand this properly. Wan't to understand this kind of math really bad. Im stuck on every question on elements on number theory. :(

I dont know know when to use induction and when not to use it. Dont know how to properly approach this types of questions. Have no Idea what to do..

.




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Is someone can recommend me where to start at the beginning, so I could understand this topic. I would appreciate it.
 
Is someone can recommend me where to start at the beginning, so I could understand this topic. I would appreciate it.
 
Sure.
Proof:
Let n=4 24 = 4! > 2^4 =16. This is true!

Now assume that for n=k that k!>2^k (where k>4) [this is the easy part because you do not have to do any work]

It must be shown for n=(k+1) that (k+1)!>2^(k+1) (note that you are allowed to use the assumption above)

Now it is your turn to try to do some of the work. So try to show that (k+1)!>2^(k+1)
I will give you two hints: (k+1)! = (k+1)*k! and 2^(k+1) = 2*2^k
 
When should you use induction? That is a good question. You should think about using induction only if you are trying to prove something for the natural numbers. The theorem does not have to start at 0 or 1. It could be prove something for n>17, for example. There are some exception to what I said but don't worry about that for now. I used the word think above meaning that you might get away from doing induction and use an easier method. How can you tell? The answer is not very favorable to you but the the answer is from experience.

For example if you must prove that n(n+1) is even for positive integers you can do an induction proof but I would not.
Here is my proof. If n is even, then (n+1) is odd and the product is even. If n is odd, then the next integer n+1 must be even and n(n+1) is even. So the product n(n+1) is always even for n being a positive integer
 
Jomo said:
Do you see a (minor) problem with problem number 1?
Click to expand...
Actually I was mistaken. I thought that 24=4! is less than 2^4 =16. And I DID get 24 and 16. I guess I will be away for awhile sitting in the corner for 4!+2^6 minutes.
 
PeachBlossom said:
Sorry if im going to sound a bit dumb. To be honest I don't understand at all this topic, realizing it now, im struggeling very bad. Have tried to search the internet and watched youtube videos, for 2 whole days, nothing is really helping. Don't know what to do for me to understand this properly. Wan't to understand this kind of math really bad. Im stuck on every question on elements on number theory. :(
Click to expand...
You need to go through a few examples. If you think you understand how it's done try a few problems. The more the better - quantity of work becomes quality of understanding.
 
Number theory is a field where proofs by induction are very common because proofs by induction cover an infinite number of cases. For many theorems, there are alternative proofs, but that is not true for all theorems.

I gave an answer to another of your posts and gave some suggestions on how to find such proofs. Finding a proof is a different process from presenting the proof.

My first question to you is whether the whole process makes intuitive sense to you? If it doesn't, then you will have difficulty understanding the proofs or how to find them.

Second, as far as I am concerned the proof itself has two steps rather than three

Step 1: Prove the proposition is true if n = 1.

Step 2: Given that it is true for natural number k, prove that it is true for k + 1. You do not have to prove it true for k.

But knowing what two steps must be presented does not help you at all in discovering how to do step 2. Here are some things that I do.

I figure out how to re-state the terms relevant to the proposition that will involve k + 1 in terms of k. To give an example, let's use this problem. The relevant term for this problem is

[MATH]\{(k + 1) + 2\}^2 - (k +1)^2 = (k + 3)^2 - (k + 1)^2 = k^2 + 6k + 9 - (k^2 + 2k + 1) =[/MATH]
[MATH]4k + 8.[/MATH]
We are no longer talking about (k + 1) but an expression in k. We can thus use what we know about k. (We do so for this specific problem in the line marked with three asterisks.) Furthermore, it is obvious that 4k + 8 is evenly divisible by 2, but let's follow the formal proof. We start by defining what we mean by evenly divisible.

[MATH]u \ | \ \text {natural number } v \iff \exists \text { natural number } w \text { such that } u * w = v.[/MATH]
*** [MATH]4k + 8 = 2(2k +4) \text { and } 2k + 4 \in \mathbb N \text { because } k \in \mathbb N.[/MATH]
[MATH]\therefore 2 \ | \ 4k + 8 \implies 2 \ | \ \{(k + 1) + 2\}^2 - (k + 1)^2.[/MATH]
Now this approach will not always work, but it works in a large number of cases. And notice that your formal steps do not explicitly talk about re-stating things in (k + 1) in terms of expressions involving k and using what you know about k.
 
For \(\displaystyle (n+2)^2+ n^2\) divisible by 2, I would not use "induction" at all. Instead, by basic algebra, \(\displaystyle (n+ 2)^2+ n^2= n^2+ 2n+ 4+ n^2= 2n^2+ 2n+ 4= 2(n^2+ n+ 2)\).
 
PeachBlossom said:
Is someone can recommend me where to start at the beginning, so I could understand this topic. I would appreciate it.
Click to expand...
Let's look at the first problem.

According to my proposed list:
1. State the proposed rule for a value "n."

So we need to find a value of n such that [math]n! > 2^n[/math]. For n = 4 we get [math]4! > 2^4 \implies 24 > 16[/math], which is true.

2. Is someone can recommend me where to start at the beginning, so I could understand this topic. I would appreciate it.

So we hav a value of k = n such that [math]k! > 2^k[/math].

At this point I'm going to make a comment: You could have done thess steps yourself. They are baby steps and simply provide a framework for starting these. Please actually try to follow advice before you say "Is someone can recommend me where to start at the beginning, so I could understand this topic. I would appreciate it. " I already gave you the starting point and you didn't do it!

3. Suppose that the rule is true for some whole number n = k. Show that the rule works for the case k + 1 as well.

So look at [math](k + 1)! > 2^{k + 1}[/math]. Typically you need to use statement in step 2, ie. we know that there is some k such that [math]k! > 2^k[/math] for some value of k. This leads to the idea:
[math](k + 1)! = (k + 1)k![/math] and [math]2^{k + 1} = 2 \cdot 2^{k}[/math]
So if we know that [math]k! > 2^k[/math] is [math](k + 1)k! > 2 \cdot 2^k[/math]?

Please do what you can with it and let us know how it turns out.

-Dan
 
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