Results 1 to 6 of 6

Math Help - Not defined problem

  1. #1
    Senior Member
    Joined
    Sep 2009
    Posts
    300

    Not defined problem

    My last maths question today is this....

    "Explain what is wrong with this argument"

    x(x+1)=0 is equivalent to x(x+1)/(x+1)(divided both sides by (x+1)

    So the solution to x(x+1)=0 is x=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    Quote Originally Posted by wolfhound View Post
    My last maths question today is this....

    "Explain what is wrong with this argument"

    x(x+1)=0 is equivalent to x(x+1)/(x+1)(divided both sides by (x+1)

    So the solution to x(x+1)=0 is x=0
    if x(x+1) = 0 , then x = 0 or (x+1) = 0, making x = -1 another solution.

    division by 0 in the form of (x+1) ist verboten!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    Ok thanks , so for both solutions the equation is undefined because if I use the value of x=-1 I will be dividing by zero which is forbidden?(same with x=0 forbidden)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774
    Ok thanks , so for both solutions the equation is undefined because if I use the value of x=-1 I will be dividing by zero which is forbidden?(same with x=0 forbidden)
    Which equation? There is nothing undefined about x(x + 1) = 0. On the other hand, dividing both sides by x + 1 is in fact a statement that says,

    For all x, x(x + 1) = 0 iff x(x + 1)/(x + 1) = 0

    This statement is false for x = -1.

    That is why, when simplifying an equation, it is important to write not just a sequence of equations E1 = E'1, E2 = E'2, ..., but also the logical relationship between the them, e.g., E1 = E'1 <=> E2 = E'2 => E3 = E'3, ... In the example above,

    x(x + 1) = 0 <= x(x + 1)/(x + 1) = 0

    is OK, but

    x(x + 1) = 0 <=> x(x + 1)/(x + 1) = 0

    is not.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    Ok thanks,
    but what does this arrow mean? x(x + 1) = 0 <= x(x + 1)/(x + 1) = 0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774
    It's just my notation for "is implied by", or "if", or "follows from" whereas <=> means "if and only if".
    Last edited by emakarov; October 2nd 2010 at 01:00 PM. Reason: Added <=>
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 17th 2011, 01:46 AM
  2. Well defined
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: June 26th 2010, 02:39 PM
  3. well defined
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 4th 2010, 01:07 AM
  4. defined by
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 26th 2009, 12:55 PM
  5. Replies: 2
    Last Post: August 5th 2009, 10:20 AM

Search Tags


/mathhelpforum @mathhelpforum