Verification of Simple Proof


by kripkrip420
Tags: proof, simple, verification
kripkrip420
kripkrip420 is offline
#1
Apr15-12, 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 (x-y)(x+y) from x[itex]^{2}[/itex]
-y[itex]^{2}[/itex] and showing that either (x+y)=0 or (x-y)=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?
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
HallsofIvy
HallsofIvy is offline
#2
Apr15-12, 08:09 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898
Quote Quote by kripkrip420 View Post
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 (x-y)(x+y) from x[itex]^{2}[/itex]
-y[itex]^{2}[/itex] and showing that either (x+y)=0 or (x-y)=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).
"Introduce (a*0)" in what??.

By the distributive law

a(0+0)=(a*0)
That's not the distributive law- that's the fact that 0+ 0= 0. And from that it follows that a(0)+ a(0)= a(0)

By the existence of the additive identity

a=a*0
That is certainly NOT true! And I don't see any point in talking about the "existence of the additive identity" when you have already been using "0" which is defined as the additive identity.

And we have proven our proposition.

Is this valid?
No, it is not. Instead, from the distributive law a(0+ 0)= a(0)+ a(0). And since 0 is additive identity, x+ 0= x for any x. In particular 0+ 0= 0 so the left side is just a(0). Once you have a(0)= a(0)+ a(0) use the fact that a(0) has an additive inverse.
kripkrip420
kripkrip420 is offline
#3
Apr16-12, 06:23 AM
P: 54
Please reread the edited proof. I made an error when I first posted.

HallsofIvy
HallsofIvy is offline
#4
Apr16-12, 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.
genericusrnme
genericusrnme is offline
#5
Apr16-12, 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...
kripkrip420
kripkrip420 is offline
#6
Apr16-12, 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.
micromass
micromass is online now
#7
Apr16-12, 03:11 PM
Mentor
micromass's Avatar
P: 16,698
Quote Quote by kripkrip420 View Post
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
Why are you allowed to use the above equation?? You just said you assumed the equation is false.
kripkrip420
kripkrip420 is offline
#8
Apr16-12, 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.
micromass
micromass is online now
#9
Apr16-12, 03:23 PM
Mentor
micromass's Avatar
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.
micromass
micromass is online now
#10
Apr16-12, 03:25 PM
Mentor
micromass's Avatar
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.
kripkrip420
kripkrip420 is offline
#11
Apr16-12, 03:33 PM
P: 54
Quote Quote by micromass View Post
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.
I understand what you mean. But I notice that it also does not change the outcome if I change the assumption from false to true. I must have made an error somewhere else because I made a similar proof a while back and had it verified by another forum. I will post another proof once I get back from grocery shopping. Thank you for your help.


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