1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spivak Calculus Problem 1.iii

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that if x2=y2, then x=y or x=-y.

    This is from Spivak's Calculus, problem 1(iii).

    2. Relevant equations
    Distributive law. If a, b, and c are any numbers, then a(b+c)=a*b+a*c.
    Existence of additive inverse. If a is any number, then a+0=0+a.
    Existence of multiplicative inverse. For every number a≠0, there is a number a-1 such that a*a-1=a-1*a=1.

    3. The attempt at a solution
    First by the existence of additive inverse,

    x2+(-y2)=y2+(-y2)=0.

    Then in an earlier problem I proved x2-y2=(x-y)(x+y) using a triple application of the distributive law and the existence of additive inverse. So,

    x2-y2=0=(x-y)(x+y).

    Now here's where I get uncomfortable. I want to use the existence of multiplicative inverses to do this

    (x-y)-1(x-y)(x+y)=1*(x+y)=0

    But multiplicative inverse has the stipulation that (x-y)≠0. So then when I do the other part

    (x+y)-1(x-y)(x+y)=1*(x-y)=0

    I get a contradiction. I must be making a mistake since this only shows that x=-y, right? Or is it just an exclusive "or" in the problem statement or something? Earlier he says to assume inclusive ors...
     
  2. jcsd
  3. Mar 18, 2012 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There are two possibilities - x=y or x is not equal to y. If x=y you are done! That was one of the solutions you had to show is possible. If they're not equal do your division to find the other solution possibility
     
  4. Mar 18, 2012 #3

    jgens

    User Avatar
    Gold Member

    You can also just use the zero product property instead of imitating its proof. Recall that (x-y)(x+y)=0 implies (x-y)=0 or (x+y)=0. The proof is pretty simple from there.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Spivak Calculus Problem 1.iii
Loading...