# Mathematical induction inequality

• Julia Maria
In summary: No, you are still trying to prove what you want to prove. You should not have "n^2+ n> 1" at all. That is not true in general.Look at what you have: "n^2+ n+ 1> n+ 2". Since you know that n^2> n+ 1, you can replace that n+ 2 by n^2 and what you have is "n^2+ n+ 1> n^2". Just subtract n^2 from both sides.

## Homework Statement

Prove; n^2 > n+1 for n = 2,3,4 by Induction

## The Attempt at a Solution

p(n)= P(2) 2^2> 2+1 --> 4>3

Induction step:
P(n+1): (n+1)^2 > (n+1) +1
(n+1)^2> n+2
n^2 + 2n + 1 > n+2 | -n
n^2 +n + 1> 2 | -1
n^2 +n > 1

Is this correct, and how do I go from here?

Hint: for what n is the inequality suppose to be valid?

For n>1. But I Still struggle to get further..

You're now supposed to apply your assumption that $n^2>n+1$ is true. If we know that $n^2>n+1$ then is $n^2+n>1$ ?

Julia Maria said:

## Homework Statement

Prove; n^2 > n+1 for n = 2,3,4 by Induction

## The Attempt at a Solution

p(n)= P(2) 2^2> 2+1 --> 4>3
Better said "for n= 2, n^2= 2^2= 4> 3= 2+ 1"

Induction step:
You have to first say "assume that for some n, n^2> n+1".
(Personally I prefer to use another letter, k, say, so as not to confuse it with the general n.

P(n+1): (n+1)^2 > (n+1) +1
No, You are asserting what you want to prove.
Instead look at just the left side: (n+1)^2= n^2+ 2n+ 1.
By the "induction hypothesis", n^2> n+ 1 so (n+1)^2> (n+1)+ 2n+ 1= 3n+ 2> n+ 2= (n+1)+1

(n+1)^2> n+2
n^2 + 2n + 1 > n+2 | -n
n^2 +n + 1> 2 | -1
n^2 +n > 1

Is this correct, and how do I go from here?

## What is mathematical induction inequality?

Mathematical induction inequality is a method used in mathematics to prove that a statement or formula is true for all natural numbers. It involves proving that the statement is true for the first natural number, and then using the assumption that it is true for a particular natural number to prove that it is also true for the next natural number. This process is repeated until the statement is proven to be true for all natural numbers.

## How is mathematical induction inequality used?

Mathematical induction inequality is typically used to prove mathematical statements or theorems that involve natural numbers. It is also commonly used in discrete mathematics and computer science to prove the correctness of algorithms and programs.

## What is the difference between strong and weak induction?

Strong induction is a variation of mathematical induction that allows the use of multiple assumptions to prove a statement for the next natural number. In contrast, weak induction only allows the use of one assumption. Strong induction is typically used when the statement being proved depends on more than just the previous natural number.

## What are the common mistakes made when using mathematical induction inequality?

One common mistake is assuming that the statement is true for all natural numbers without actually proving it for the first natural number. Another mistake is using the wrong assumption, as this can lead to an incorrect proof. It is also important to be careful when dealing with infinite sets, as mathematical induction is only applicable to finite sets.

## Are there any real-life applications of mathematical induction inequality?

Yes, mathematical induction inequality has numerous real-life applications in fields such as computer science, economics, and physics. In computer science, it is used to prove the correctness of algorithms and programs. In economics, it can be used to prove the convergence of a series of values. In physics, it is used to prove mathematical relationships and formulas.