Results 1 to 5 of 5

Math Help - Prove equality (real numbers)

  1. #1
    Junior Member
    Joined
    Oct 2011
    Posts
    58

    Prove equality (real numbers)

    a is a real number.

    Demonstrate the following equality.

    \surd(a^2) = |a|

    I tried in this way but i think it is wrong.

    \surd(a^2) = a~~~(a\geq0)

    a\geq0~~so~~a = |a|

    Is it correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,513
    Thanks
    769

    Re: Prove equality (real numbers)

    You considered only the case a\ge 0. Also, since the problem is so basic, the proof has to be really detailed, and I am not sure you sufficiently explained why \sqrt{a^2}=a when a\ge 0. So, what happen when a < 0? Recall that by definition \sqrt{x} is a nonnegative number y such that y^2=x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2011
    Posts
    58

    Re: Prove equality (real numbers)

    Quote Originally Posted by emakarov View Post
    You considered only the case a\ge 0. Also, since the problem is so basic, the proof has to be really detailed, and I am not sure you sufficiently explained why \sqrt{a^2}=a when a\ge 0. So, what happen when a < 0? Recall that by definition \sqrt{x} is a nonnegative number y such that y^2=x.

    Thats the point the exercise is so basic. But hard to do.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: Prove equality (real numbers)

    Quote Originally Posted by Fabio010 View Post
    Thats the point the exercise is so basic. But hard to do.
    Any proof of this depends on what definitions you have and what other properties of real numbers you know.
    If could be this simple:
    1) if a\ge 0 then \sqrt{a^2}=a=|a|
    2) if a< 0 then \sqrt{a^2}=-a=|a|.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2011
    Posts
    58

    Re: Prove equality (real numbers)

    Quote Originally Posted by Plato View Post
    Any proof of this depends on what definitions you have and what other properties of real numbers you know.
    If could be this simple:
    1) if a\ge 0 then \sqrt{a^2}=a=|a|
    2) if a< 0 then \sqrt{a^2}=-a=|a|.
    I found some properties in internet , and i tried:


    \sqrt{a^2}=b~~b\geq0~~so~~b^2=a^2~so~(b-a)(b+a)=0

    b = a,~in~this~case~a\geq0
    b = -a, in~this~case~a\leq0

    so \sqrt{a^2}=b=|a|
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove positive real numbers a,b,c.
    Posted in the Algebra Forum
    Replies: 31
    Last Post: August 7th 2011, 03:04 PM
  2. Replies: 10
    Last Post: December 9th 2010, 04:42 AM
  3. Replies: 4
    Last Post: September 19th 2010, 03:02 PM
  4. Replies: 2
    Last Post: October 1st 2009, 11:58 AM
  5. prove that the set of real numbers is not a countable set
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 31st 2009, 04:01 PM

Search Tags


/mathhelpforum @mathhelpforum