
#1
Apr1512, 08:00 PM

P: 54

1. The problem statement, all variables and given/known data
I am using Spivak's Calculus and just finished the third exercise in part 1. It was a very easy exercise but it seems that Spivak makes some assumptions. The problem is as stated: If x[itex]^{2}[/itex]=y[itex]^{2}[/itex], then either x=y or x=y. Prove it. The proof was relatively simple (by factoring out (xy)(x+y) from x[itex]^{2}[/itex] y[itex]^{2}[/itex] and showing that either (x+y)=0 or (xy)=0 or both). What I had problems with was that it was assumed that a(0)=0. So, I will now propose a proof that a(0)=0 for all real a using the properties listed in the book. All I need is a confirmation that my proof haw no flaw. Thank you. 2. Relevant equations 3. The attempt at a solution Introduce (a*0). (a*0)+(a*0)=0+(a*0). By the distributive law a(0+0)=(a*0)+0 By the existence of the additive identity a*0=(a*0)+0. Subtracting (a*0) from both sides yields (a*0)(a*0)=0 And by the existence of additive identity 0=0 Which is true. Is this valid? 



#2
Apr1512, 08:09 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898





#3
Apr1612, 06:23 AM

P: 54

Please reread the edited proof. I made an error when I first posted.




#4
Apr1612, 07:18 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898

Verification of Simple Proof
Same errors I noted before. You seem to have payed no attention to my first post.




#5
Apr1612, 07:33 AM

P: 615

The easiest way to do it would probably be just to give it some modus tollens imo
I'm not quite sure what you're trying to do by playing about with 0 here... 



#6
Apr1612, 01:45 PM

P: 54

Okay. I will redo it here and explain myself better. In Spivak's book, he lists 12 properties of the real numbers. I will use those properties to prove that for any real a
(a*0)=0. Begin with the assumption that the above statement is false. Since (a*0) is in R, then by the closure under addition property, (a*0)+(a*0) must be in R. Therefore, we can add (a*0) to both sides of the above equation and we shall have (a*0)+(a*0)=0+(a*0). (1) Now, the left side of the equation can be factored out by the distributive law to yield (a(0+0))=(a*0)+(a*0). Thus, (1) becomes: (a(0+0))=0+(a*0). (2) But, 0+0=0 and the left side becomes (a(0+0))=(a(0)). Thus, for (2), we have: (a(0))=0+(a(0)). (3) But, by property 3 in Spivak's book, for every real k, there exists it's additive inverse denoted k such that k+(k)=0. Therefore, by adding the negative of (a*0) to both sides of equation (3), we get (a(0))+((a(0)))=0+(a(0))+((a(0))). (4) Again, by Spivak's third stated property, we have: (a(0))+((a(0))=0 and equation (4) becomes 0=0+0. (5) But it is clear from Spivak's second property that 0+0=0 and equation (5) becomes 0=0. But, this is obviously true, and since valid manipulation was made to our initial equation (a*0)=0, then our initial assumption was false and, therefore, (a*0)=0 must be true. I really hope this clarifies things. 



#7
Apr1612, 03:11 PM

Mentor
P: 16,698





#8
Apr1612, 03:18 PM

P: 54

Hello micromass. I did the proof through contradiction. If we begin with the assumption that the equation is false and manipulate the equation so that everything we do is justifiable, we should end up with a result that varifies the assumption. However, what I got (after showing after steps of manipulation) is that 0=0. Now, since this is obviously true, the INITIAL assumption that the equation was false is in its own right false. Therefore, the equation must be true. I believe that I had actually done this problem in Apostol's calculus book. As far as I cam recall, this was an adequate proof.




#9
Apr1612, 03:23 PM

Mentor
P: 16,698

Yes, I get what you were trying to do.
But if you assume the equation is false, then you cannot use that in your proof. You can only use things in a proof that you know are true or that you assumed are true. 



#10
Apr1612, 03:25 PM

Mentor
P: 16,698

It's like this:
I want to prove that 2+2=5. Assume by contradiction that 2+2=5 is not true. So take 2+2=5. Multiply both sides with 0, then we get 0*(2+2)=0*5. We get that 0=0. This is true, so 2+2=5 must be true. This is essentially what you did, but it is of course not valid. 



#11
Apr1612, 03:33 PM

P: 54




Register to reply 
Related Discussions  
Simple Proof  Calculus & Beyond Homework  2  
Simple dot product [inner product] (verification)  Calculus & Beyond Homework  1  
Simple limit (verification)  Calculus & Beyond Homework  9  
attempt at proof by contradiction need verification  Precalculus Mathematics Homework  6  
Verification of simple inequality proof  Precalculus Mathematics Homework  5 