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Induction proof of n^2>n+1

  1. Jun 9, 2014 #1
    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
     
  2. jcsd
  3. Jun 9, 2014 #2

    statdad

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

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

    [tex]
    (n+1)^2 > (n+1) + 1 \text{ or } (n+1)^2 > n + 2
    [/tex]

    -- you do not have [itex] 2n + 2 [/itex] on the right, so your approach isn't correct from that point.
     
  4. Jun 9, 2014 #3
    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
     
  5. Jun 9, 2014 #4

    statdad

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    Try looking at [itex] (n+1)^2 - (n+2) [/itex]. Expand and use your answer inequality.
     
  6. Jun 9, 2014 #5

    SammyS

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    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​
     
  7. Jun 10, 2014 #6
    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?
     
  8. Jun 10, 2014 #7

    D H

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    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.
     
  9. Jun 10, 2014 #8
    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?
     
  10. Jun 10, 2014 #9

    D H

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    For all k>0, yes, and you only needed to prove it for all k>2.
     
  11. Jun 13, 2014 #10
    (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
     
  12. Jun 13, 2014 #11

    Ray Vickson

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    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##:
    [tex] 0 < (x-1)^2 = x^2 - 2x + 1 \Longrightarrow x^2 > 2x-1 = x+1 +(x-2) \geq x+1.[/tex]
     
  13. Jun 13, 2014 #12

    D H

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    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.
     
  14. Jun 13, 2014 #13
    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?
     
  15. Jun 13, 2014 #14

    ehild

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    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: Jun 13, 2014
  16. Jun 13, 2014 #15
    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
     
  17. Jun 13, 2014 #16

    SammyS

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    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".
     
  18. Jun 14, 2014 #17

    ehild

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    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
     
  19. Jun 14, 2014 #18

    SammyS

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    OP did pretty much do that in the OP .

    But you overall summary of how induction proves this is excellent (as usual) !
     
  20. Jun 14, 2014 #19

    ehild

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

    ehild
     
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