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Verification of Simple Proof

by kripkrip420
Tags: proof, simple, verification
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kripkrip420
#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?
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HallsofIvy
#2
Apr15-12, 08:09 PM
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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
#3
Apr16-12, 06:23 AM
P: 54
Please reread the edited proof. I made an error when I first posted.

HallsofIvy
#4
Apr16-12, 07:18 AM
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Verification of Simple Proof

Same errors I noted before. You seem to have payed no attention to my first post.
genericusrnme
#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
#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
#7
Apr16-12, 03:11 PM
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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
#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
#9
Apr16-12, 03:23 PM
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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
#10
Apr16-12, 03:25 PM
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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
#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.


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