Simple Proof From Loomis and Sternberg's Calculus

[Bashyboy]

Hello everyone,

I am currently working through Loomis and Sternberg's Advanced Calculus, and am having difficulty with rather simple proofs. Despite this, I am NOT having much difficultly with proofs that require more sophisticated mathematics. For instance, I am trying to prove that if ab = 0, then either a = 0 or b = 0. I have tried to add zero in manifold ways, yet have not been successful in proving this claim.

I have two questions: could someone provide me with some hints; and is this an early indication of my failure as a mathematician? To me, problems of this sort are not truly going to display your ability as a mathematician, as they require merely exhausting all possibilities. Does anyone else share this sentiment?

Thank you.

 

[micromass]

is this an early indication of my failure as a mathematician?

Not at all. I have often found such "easy" proofs to be the hardest.

Anyway, what axioms can you use?

 

[pasmith]

Hello everyone,

I am currently working through Loomis and Sternberg's Advanced Calculus, and am having difficulty with rather simple proofs. Despite this, I am NOT having much difficultly with proofs that require more sophisticated proofs. For instance, I am trying to prove that if ab = 0, then either a = 0 or b = 0. I have tried to add zero in manifold ways, yet have not been successful in proving this claim.

One axiom of the real numbers is that the non-zero reals are closed under multiplication; in other words, if you multiply any two non-zero real numbers together, you get a non-zero real number.

 

[MarneMath]

Have you tried to use the contraposition? For me, this makes the proof a bit more mangable to work with. Also, don't feel too bad about 'simple' proofs being difficult. I think a lot of times these 'simple' proofs require a subtleness that is often easy to overlook.

 

[Bashyboy]

micromass, that is heartening to know. I attached a picture containing the properties that I am permitted to use.

 

[micromass]

micromass, that is heartening to know. I attached a picture containing the properties that I am permitted to use.

OK, so you are working in a vector space. And you need to prove that if $\alpha\boldsymbol{v}=\boldsymbol{0}$, then either $\alpha=0$ or $\boldsymbol{v}=\boldsymbol{0}$.

Have you tried multiplying with $\alpha^{-1}$?

 

[Bashyboy]

But do I truly know what alpha inverse is? That was my first thought, to multiply by the inverse element, but according to the axioms, I do not know such a thing as of yet.

 

[verty]

It took me a while to see it, look carefully at the domain of the scalar multiplication operation. This is rather tricky, I must admit.

 

[Bashyboy]

Is it R x V?

 

[micromass]

But do I truly know what alpha inverse is? That was my first thought, to multiply by the inverse element, but according to the axioms, I do not know such a thing as of yet.

But $\alpha$ is a real number. That structure has different axioms.

This axiom system defines addition and scalar multiplication on the set $V$ to make a vector space. But in order to have a vector space, you need another structure first. You need $\mathbb{R}$ which already has a addition and multiplication. So you can use this structure.

 

[Bashyboy]

Oh, okay. In that case, if we use the properties of R, then it is relatively simple to show that the statement ax = 0 implies a = 0, but, in comparison, it is greatly difficult to show that v = 0.

 

[micromass]

Oh, okay. In that case, if we use the properties of R, then it is relatively simple to show that the statement ax = 0 implies a = 0

But this is false.

 

[Bashyboy]

Whoops, I got the two cases mixed up.

 

[Bashyboy]

Okay, so I understand how to show that av = 0 implies that v = 0, by multiplying by a^-1; but how might I show the statement av = 0 implies a = 0?

 

[verty]

Oh, okay. In that case, if we use the properties of R, then it is relatively simple to show...

If the scalars were from $\mathbb{Z}/6\mathbb{Z}$, this wouldn't be provable. We could have 3<2,2> = <0,0> but neither part is zero.

 

[micromass]

Okay, so I understand how to show that av = 0 implies that v = 0

No, this is not what you need to show and it isn't true.

but how might I show the statement av = 0 implies a = 0?

It doesn't and this is not what you need to show.

 

[Bashyboy]

Then what am I suppose to show?

 

[BruceW]

Okay, so I understand how to show that av = 0 implies that v = 0, by multiplying by a^-1

Is this true? think more about it.

edit: maybe you are just not explaining fully what you mean. But anyway, it is important to be precise in what you say.

 

[verty]

Bashyboy, I saw in an older post what you were using Spivak before, but now you are using this book. Both of these are pretty difficult books. Why are you choosing these tough books? I mean this proof was difficult, it was almost like a review question. You may be biting off more than you can chew here, or that anyone can chew without some preparatory exposure.

 

[BruceW]

we are allowed to assume the scalars are the reals. So it is not that difficult.

edit: although, I don't know if the rest of the book will be too difficult. I have not read either of them.

 

[Bashyboy]

I am just trying to prove the claim that I gave in my first post, that if av = 0, then either a = 0 or v = 0

 

[Bashyboy]

Bashyboy, I saw in an older post what you were using Spivak before, but now you are using this book. Both of these are pretty difficult books. Why are you choosing these tough books? I mean this proof was difficult, it was almost like a review question. You may be biting off more than you can chew here, or that anyone can chew without some preparatory exposure.

Yes, I was studying Spivak for some time, but stopped because of some time constraints. I just started Loomis And Sternberg's because my professor recommended I study it.

 

[Bashyboy]

I still do not see why what I am saying is false. I am trying to show that, if av = 0, then either a = 0 or v = 0.

 

[HallsofIvy]

You want to prove that "of ab= 0 the either a= 0 or b=0.

A little earlier you said "if av= 0 then a= 0" which is NOT true.

Do this as two cases.

Case 1: a= 0. Well, we are done!

Case 2: $a\ne 0$. Since a is not 0 you can divide both sides by it. What do you get when you divide both sides by a?

 

[BruceW]

yep. What Halls said is right. You need to think of two cases. But I think Bashy has said that he has done case 2 already. So he still needs to do case 1 (using just the properties that were given in the picture he showed).

edit: in other words, he still needs to show that $0\vec{v} = \vec{0}$ (I've just used the vector sign so that the vector zero can be distinguished from the scalar zero).

 

[Bashyboy]

Case I is true because of what I proved in problem 1.4, which was to show that the product of the scalar 0 and the vector $\mathbf{a}$ is zero, $0 \mathbf{a} = 0$.

Here is my work from problem 1.4:

$0 \mathbf{a} = (0+0) \mathbf{a} \iff$ (The real number 0, when adding to some number, results in that number)

$0 \mathbf{a} = 0 \mathbf{a} + 0 \mathbf{a} \iff$ (Property S2)

Adding the element $-0 \mathbf{a}$ to each side, which is specified axiom A4:

$0 \mathbf{a} -0 \mathbf{a} = 0 \mathbf{a} + 0 \mathbf{a}-0 \mathbf{a} \iff$

$\mathbf{0} = 0 \mathbf{a}$ (Axiom A4) [Hopefully it is clear that the last zero is actually the zero vector]

Thus it has been proven, that the product of the real number 0 with some arbitrary vector $\mathbf{a}$ results in the zero vector.

Hopefully it is clear that the last zero is actually the zero vector

 

[BruceW]

ah very nice! So, I think you have done all that the proof has asked for, since you've also shown case 2. I guess there also needs to be a sentence or two to explain how the proof of the two separate cases will prove the thing that we set out to prove. But yeah, nice work. You seem better at this than you implied in your first post :)

 

[BruceW]

although, I haven't seen your proof of case 2 yet. haha.

 

[Bashyboy]

To prove case two, wouldn't I do precisely what HallsofIvy mentioned? Every non-zero real number $a$ has an associated element denoted by $a^{-1}$ such that the product of this number and $a$ results in 1, $a \cdot a^{-1} = 1$.

Hence, if I take the equation $a \mathbf{v} = \vec{0}$, and multiply both sides by $a^{-1}$:

$a^{-1} \cdot a \mathbf{v} = a^{-1} \vec{0} \iff$

$\mathbf{v} = \vec{0}$

So, it has been shown that the equation $a \mathbf{v} = \vec{0}$ leads to either a = 0 or $\mathbf{v} = 0$; and this is in the inclusive or, meaning that a = 0 and $\mathbf{v} = \vec{0}$

 

[BruceW]

To prove case two, wouldn't I do precisely what HallsofIvy mentioned? Every non-zero real number $a$ has an associated element denoted by $a^{-1}$ such that the product of this number and $a$ results in 1, $a \cdot a^{-1} = 1$.

Hence, if I take the equation $a \mathbf{v} = \vec{0}$, and multiply both sides by $a^{-1}$:

$a^{-1} \cdot a \mathbf{v} = a^{-1} \vec{0} \iff$

$\mathbf{v} = \vec{0}$

So, it has been shown that the equation $a \mathbf{v} = \vec{0}$ leads to either a = 0 or $\mathbf{v} = 0$; and this is in the inclusive or, meaning that a = 0 and $\mathbf{v} = \vec{0}$

I agree with you up to $\mathbf{v} = a^{-1} \vec{0}$ But from here you still need to show that $a^{-1} \vec{0} = \vec{0}$

 

[Bashyboy]

Claim: $x \vec{0} = \vec{0}$

Let $x$ be some arbitrary scalar in $\mathbb{R}$.

$x \vec{0} = x ( \vec{0} + \vec{0}) \iff$

$x \vec{0} = x \vec{0} + x \vec{0} \iff$

$x \vec{0} - x \vec{0} = x \vec{0} + x \vec{0} - x \vec{0} \iff$

$\vec{0} = x \vec{0}$

Because we had supposed x was arbitrary, meaning that we have not assumed anything about x beyond the fact that it is merely in $\mathbb{R}$, then this statement must be true of all real numbers.

 

[BruceW]

yeah! very nice. ok, I am now fully convinced that $a\mathbf{v}=\mathbf{0}$ implies either $a=0$ or $\mathbf{v}=\mathbf{0}$ :) It takes quite a lot of work to prove things that seem fairly intuitive. But it's kind of satisfying too, right? I'm more physics than maths, so I don't have much experience doing proofs. But I would like to learn a bit more 'proper mathematics'. In fact, I was reading about groups this week.