Results 1 to 6 of 6

Math Help - Proof by Contradiction: (n+2)^3 = n^3 + (n+1)^3 ?

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    31

    Proof by Contradiction: (n+2)^3 = n^3 + (n+1)^3 ?

    I have been working through D. L. Johnson's "Elements of Logic via Numbers and Sets" and have come across this problem, amongst others, which has me stumped.

    "Prove by contradiction that the cube of the largest of three consecutive integers cannot be equal to the sum of the cubes of the other two."

    Assume, then (for contradiction) that the above is not the case, i.e.

     (n+2)^3 = n^3 + (n+1)^3

    for some integer n . Then

     \begin{array}{rcl}<br />
n^3 + 3n^2 \cdot (2) + 3n \cdot (2)^2 + 8 & = & n^3 + n^3 + 3n^2 + 3n + 1 \\<br />
n^3 + 6n^2 + 12n + 8 & = & 2n^3 + 3n^2 + 3n + 1 \\<br />
n^3 - 3n^2 - 9n - 9 &  = & 0<br />
\end{array} .

    I suppose that we wish to show that this is not the case for any integer n, to provide the necessary contradiction. I'm not sure how though I might do this. Any hints much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    May 2007
    Posts
    34
    Quote Originally Posted by Harry1W View Post
    I suppose that we wish to show that this is not the case for any integer n, to provide the necessary contradiction. I'm not sure how though I might do this. Any hints much appreciated.
    Don't you just solve for the polynomial's roots to show that r_{1},r_{2},r_{3}\not\in\mathbb{Z}? Are you really asking us how you solve for the roots? I only know how to solve quadratics w/o a calculator.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Harry1W View Post
    I have been working through D. L. Johnson's "Elements of Logic via Numbers and Sets" and have come across this problem, amongst others, which has me stumped.

    "Prove by contradiction that the cube of the largest of three consecutive integers cannot be equal to the sum of the cubes of the other two."

    Assume, then (for contradiction) that the above is not the case, i.e.

     (n+2)^3 = n^3 + (n+1)^3

    for some integer n . Then

     \begin{array}{rcl}<br />
n^3 + 3n^2 \cdot (2) + 3n \cdot (2)^2 + 8 & = & n^3 + n^3 + 3n^2 + 3n + 1 \\<br />
n^3 + 6n^2 + 12n + 8 & = & 2n^3 + 3n^2 + 3n + 1 \\<br />
n^3 - 3n^2 - 9n - 9 & = & 0<br />
\end{array} .

    I suppose that we wish to show that this is not the case for any integer n, to provide the necessary contradiction. I'm not sure how though I might do this. Any hints much appreciated.
    Look at where the turning points of the cubic are. So the cubic has only one root. It lies between n = 3 and n = 6. Note that the cubic is negative for n = 5 and positive for n = 6. What do you conclude?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2009
    Posts
    31
    Quote Originally Posted by mr fantastic View Post
    Look at where the turning points of the cubic are. So the cubic has only one root. It lies between n = 3 and n = 6. Note that the cubic is negative for n = 5 and positive for n = 6. What do you conclude?
    Got it. Both the turning pts are below the n-axis, so there can be only one real root. How, however, did you conclude that it must lie between n = 3 and n= 6? I can understand the reasoning in the choice of n = 3, since we know it to lie beneath the n-axis, but why n = 6? Was this just a random stab in the dark? What would have happened if the root, say lay at n = 4000/3 (i.e. a much higher rational, non-integer number)? Would one use the Newton-Raphson method, or a similar iterative process until the 'window' was sufficiently small to evaluate the value of f(x) at a few values?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    807
    Thanks
    27
    Quote Originally Posted by Harry1W View Post
    n^3 - 3n^2 - 9n - 9  = 0 .

    Here's another approach which may be fruitful (however, I can't promise, I haven't actually tried it).

    I suppose that we wish to show that this is not the case for any integer n, to provide the necessary contradiction. I'm not sure how though I might do this. Any hints much appreciated.
    One approach might be to note that the product of all 3 roots of this equation has to be 9.

    So you can test by exhaustion whether n-1, n-3, n-9 are factors of this cubic. You'll find they're not. Therefore, none of 1, 3 and 9 are roots.

    Because the sum of roots is 3, and the sum of pairs of roots taken 2 at a time is 9 (both of which are integers), I think you may be able to show that there are other limitations on the roots such that all other possible integral solutions are likewise eliminated.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by Harry1W View Post
    I have been working through D. L. Johnson's "Elements of Logic via Numbers and Sets" and have come across this problem, amongst others, which has me stumped.

    "Prove by contradiction that the cube of the largest of three consecutive integers cannot be equal to the sum of the cubes of the other two."

    Assume, then (for contradiction) that the above is not the case, i.e.

     (n+2)^3 = n^3 + (n+1)^3

    for some integer n . Then

     \begin{array}{rcl}<br />
n^3 + 3n^2 \cdot (2) + 3n \cdot (2)^2 + 8 & = & n^3 + n^3 + 3n^2 + 3n + 1 \\<br />
n^3 + 6n^2 + 12n + 8 & = & 2n^3 + 3n^2 + 3n + 1 \\<br />
n^3 - 3n^2 - 9n - 9 &  = & 0<br />
\end{array} .

    I suppose that we wish to show that this is not the case for any integer n, to provide the necessary contradiction. I'm not sure how though I might do this. Any hints much appreciated.
    In
    n^3 - 3n^2 - 9n - 9 = 0,
    notice that all the terms are obviously divisible by 3 except for n^3. So n^3 must be divisible by 3, hence n must be divisible by 3.

    Let n = 3x, where x is an integer. Then
    27 x^3 - 27 x^2 - 27 x -9 =0, whence
    3(x^3 - x^2 - x) = 1, so
    3y = 1
    where  y=x^3 - x^2 - x is an integer.

    Don't think that will work out.
    Last edited by awkward; October 3rd 2009 at 06:05 AM. Reason: wording
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof by contradiction.
    Posted in the Discrete Math Forum
    Replies: 20
    Last Post: October 6th 2011, 01:11 PM
  2. Help with Proof by Contradiction
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 27th 2010, 10:33 PM
  3. Proof By Contradiction
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: May 8th 2010, 01:29 AM
  4. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  5. Proof By Contradiction
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 15th 2007, 11:23 PM

Search Tags


/mathhelpforum @mathhelpforum