# Is Induction the Key to Proving n^2 > n+1 for n ≥ 2?

• Jef123
In summary: I have come across a problem where I am not given a condition to prove something. I am only given a statement and I am asked to prove it to be true.
Jef123
1. Prove that n2 > n+1 for all n≥2

2. First, (2)^2>2+1...4>3

Now for the induction step, (n+1)2 > 2n+2

Can I just show that n2+2n+1 > 2n+1+1 because 2n+1 = 2n+1 for both sides and n2>1 for all n>1

I'm pretty sure what I did is not acceptable fro this to be considered proved

Note that the statement you want to prove is $n^2 > n+1 \text{ for } n \ge 2$. Denote it by $A_n$.

You've shown the anchor'' step - that $A_2$ is true. Remember that you need to show whenever $A_n$ is true that means $A_{n+1}$ is true. Your $A_{n+1}$ is

$$(n+1)^2 > (n+1) + 1 \text{ or } (n+1)^2 > n + 2$$

-- you do not have $2n + 2$ on the right, so your approach isn't correct from that point.

So, n2+2n+1 > n+2

Can I show that (n+1)(n+1)>(n+1)+1 by the fact that (n+1)>1, thus (n+1)2>n+2

Try looking at $(n+1)^2 - (n+2)$. Expand and use your answer inequality.

1 person
Jef123 said:
So, n2+2n+1 > n+2

Can I show that (n+1)(n+1)>(n+1)+1 by the fact that (n+1)>1, thus (n+1)2>n+2
What property of natural numbers allows you to make such a statement? (Of course we expect that the statement is true, because we're asked to prove it.)

Basically, you need to show that n2+2n+1 > n+2. Don't assume it's true.

The way I like to state the approach to proving the inductive step is as follows.

Assume that n2 > n+1 for n = k, for some k ≥ 2 .

From that show that n2 > n+1 when n = k+1 .

In other words:
Assuming that k2 > k+1, for some k ≥ 2

Show that it follows that (k+1)2 > k+2 .​

You apparently know that (k+1)2 = k2 + 2k + 1.

Therefore, the first step you might want to take is to add 2k + 1 to both sides of the inequality,
k2 > k+1​

1 person
So by adding 2k+1 to both sides of the inequality, I get k2+2k+1>3k+2

Since I am assuming that n2>n+1, then isn't it proved that (k+1)2>3k+2?

I think I am sort of confused with the process of induction. Correct me if I am wrong, but the only way to prove this (with induction) is to assume that n is true and then prove that n+1 is the successor of n, regardless if n is actually true? Meaning, I am just trying to show some algebraic manipulation that demonstrates this fact?

Jef123 said:
So by adding 2k+1 to both sides of the inequality, I get k2+2k+1>3k+2

Since I am assuming that n2>n+1, then isn't it proved that (k+1)2>3k+2?
Correct.

However, that is not quite what you need for the inductive step. For that you need to show that (k+1)2>k+2 .

Hint: Use the fact that greater than is transitive.

So, 3k+2>k+2 because 3k+2-(k+2) = 2k>0 for all k>0.

Then the relation holds for (k+1)2>k+2?

For all k>0, yes, and you only needed to prove it for all k>2.

1 person
(n+1)^2 -n =n^2 +n+2 > n+1 > 2 for n≥2 and hence, adding n to the inequality yields (n+1)^2 > n+2 and it will follow That n^2>n+1

1 person
Jef123 said:
1. Prove that n2 > n+1 for all n≥2

2. First, (2)^2>2+1...4>3

Now for the induction step, (n+1)2 > 2n+2

Can I just show that n2+2n+1 > 2n+1+1 because 2n+1 = 2n+1 for both sides and n2>1 for all n>1

I'm pretty sure what I did is not acceptable fro this to be considered proved

If you are allowed to not use induction there is a very easy proof that ##x^2 > x+1## for all real numbers ##x \geq 2##:
$$0 < (x-1)^2 = x^2 - 2x + 1 \Longrightarrow x^2 > 2x-1 = x+1 +(x-2) \geq x+1.$$

1 person
drjohnsonn said:
(n+1)^2 -n =n^2 +n+2 > n+1 > 2 for n≥2 and hence, adding n to the inequality yields (n+1)^2 > n+2 and it will follow That n^2>n+1
There are three things wrong here:
- (n+1)^2 -n =n^2 +n+2 is incorrect.
- You stated n^2 +n+2 > n+1 > 2 for n≥2 as a fact.
- The title of the thread is "Induction proof of n^2>n+1."

Ray Vickson said:
If you are allowed to not use induction there is a very easy proof ...
This happens a lot with homework questions about induction. Usually the students are using induction to prove X because they were told to do "prove X using induction".

However, there are some cases where the student perceived that induction was the right tool because the problem to be solved is with regard to the integers. In mathematics, there's almost always more than one way to do it. Some students have an uncanny ability of picking the hardest approach to solving some problem.

I don't think this is such a bad problem to use induction on. Sure there are better methods perhaps, but an induction can be hammered out pretty quickly.

OP, assume that k² > k + 1 {induction hypothesis}

Now, let us look at (k+1)². We can expand this, and we have:

k² + 2k + 1

Now, can you see how we can say something about this expression, using our induction hypothesis? If k² is greater than k + 1, what is k² + 2k + 1 greater than?

Prove that n2>n+1 if n≥2. It is very simple even without induction.
You can write n=2+k, k≥0.
n2-n-1=k2+4k+4-k-2-1=k2+3k+1: as k≥0, the expression is positive, ...

ehild

Last edited:
D H said:
There are three things wrong here:
- (n+1)^2 -n =n^2 +n+2 is incorrect.
- You stated n^2 +n+2 > n+1 > 2 for n≥2 as a fact.
- The title of the thread is "Induction proof of n^2>n+1."

This happens a lot with homework questions about induction. Usually the students are using induction to prove X because they were told to do "prove X using induction".

However, there are some cases where the student perceived that induction was the right tool because the problem to be solved is with regard to the integers. In mathematics, there's almost always more than one way to do it. Some students have an uncanny ability of picking the hardest approach to solving some problem.

Induction is the only method to go about proving something that I have learned, other than proof by contradiction, but this is the first chapter, so I would assume that I will learn other ways along the way. The text asked to prove by induction, but I am open to other forms of proofs, since I am not doing this for school

Jef123 said:
Induction is the only method to go about proving something that I have learned, other than proof by contradiction, but this is the first chapter, so I would assume that I will learn other ways along the way. The text asked to prove by induction, but I am open to other forms of proofs, since I am not doing this for school
Yes, there may may be (and are in this case) simpler ways to solve this problem, but in my opinion, it is a reasonable exercise for learning the steps and logic involved in doing "proof by induction".

Jef123 said:
So, 3k+2>k+2 because 3k+2-(k+2) = 2k>0 for all k>0.

Then the relation holds for (k+1)2>k+2?

The relation (k+1)2>k+2 is also true if k2>k+1.

Yes, you have proved that the relation is true for n=k+1 if it is true for n=k. You missed the first step of induction: Prove that it is true for n=2. Just plug in 2 and see.

If it is true for n=2, than it is true for n=2+1=3. If it holds for n=3, holds also for n=4 and so on, for all integers equal of greater than 2.

ehild

ehild said:
... You missed the first step of induction: Prove that it is true for n=2. Just plug in 2 and see.
...

ehild

OP did pretty much do that in the OP .

Jef123 said:
1. Prove that n2 > n+1 for all n≥2

2. First, (2)^2>2+1...4>3

Now for the induction step, (n+1)2 > 2n+2
...

But you overall summary of how induction proves this is excellent (as usual) !

SammyS said:
OP did pretty much do that in the OP .

Oh, yes, I did not notice... :shy:

ehild

## 1. How does induction proof work?

Induction proof is a mathematical technique used to prove that a statement or formula is true for all natural numbers (1, 2, 3, ...). It involves showing that the statement is true for the first natural number (the base case) and then assuming it is true for an arbitrary natural number (the inductive hypothesis) and using this assumption to prove that it is also true for the next natural number (the inductive step). This process is repeated until it is shown to be true for all natural numbers.

## 2. What is the statement being proved in "Induction proof of n^2>n+1"?

The statement being proved is that for any natural number n, n squared (n^2) will always be greater than n+1.

## 3. What is the base case for this induction proof?

The base case for this induction proof is n=1, where 1^2 is equal to 1+1 (1^2=2).

## 4. How is the inductive hypothesis used in this proof?

The inductive hypothesis assumes that the statement is true for an arbitrary natural number k. Using this assumption, we can show that the statement is also true for the next natural number, k+1. This completes the inductive step and proves that the statement is true for all natural numbers.

## 5. Is induction proof the only way to prove this statement?

No, there are other mathematical proof techniques that can be used to prove this statement. However, induction proof is a commonly used and effective method for proving statements that involve natural numbers.

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