Is n(0) Always Equal to 0? A Debate on Friend's Creative Proof

[Nano-Passion]

I'm sorry if this is in the wrong place..

My friend had an odd yet creative "proof" that n(0) = 0. I was arguing that it wasn't conclusive but I wanted to make sure because I'm starting to think otherwise. But then again I'm a newcomer to proofs so take my words with a grain of salt.

n(0) = n(0+0) because 0+0=0. ergo, n(0)+n(0)=0. n(0)=-n(0)
because n≠-n then n(0) must equal 0

consider n(0) = n(0-0), then n(0)-n(0) must equal 0

 

[DivisionByZro]

What exactly is "n(0)"? What type of function is this? I'm not familiar with the notation.

In any case, I can, for the moment, only speculate that you can't really split up "n(0)" into "n(0) = n(0) + n(0)".

 

[I like Serena]

Hi Nano-Passion!

Just guessing here...
Are we talking about a group homomorphism?
Or about the distributive property of a ring or field?
Or...

 

[mathman]

My guess is simply multiplication. n(0) means nx0, where x is multiply.

 

[Nano-Passion]

What exactly is "n(0)"? What type of function is this? I'm not familiar with the notation.

In any case, I can, for the moment, only speculate that you can't really split up "n(0)" into "n(0) = n(0) + n(0)".

Well you skipped a step in between

$$n(0)$$
which is the same as
$$n(0+0)$$
$$n(0)+n(0)$$

This kind of seems logical to me, kind of like saying

$$n(1)$$
$$n(1+0)$$
$$n(1)+n(0)$$

Wait wait.. that would imply..
$$n(1)=-n(0)$$

Which can't be true. So therefore his logic is inconsistent, correct?

Hi Nano-Passion!

Just guessing here...
Are we talking about a group homomorphism?
Or about the distributive property of a ring or field?
Or...

Hey!

My apology, I meant simple multiplication. Some number n times 0 --> n(0).

My guess is simply multiplication. n(0) means nx0, where x is multiply.

Precisely!

 

[I like Serena]

Ah, okay, that would be the distributive property of a field.
(Math mumbo jumbo for the same thing ;)

But isn't it already generally known that a number times zero is zero?
Why proof it?
(Unless you want to proof it for a ring or a field.)

 

[Deveno]

n(0) = n(0+0) because 0+0=0. ergo, n(0)+n(0)=0...(snip)

the part after "ergo" hasn't been proven yet. all that has been proven is:

n*0 + n*0 = n*0

however, this in fact does imply that n*0 = 0:

subtract n*0 from both sides, and we get:

n*0 = 0.

(doing this uses implicitly that addition is cancellative: that a+b = c+b implies a = c. this is always the case if every number (or whatever we are dealing with) has an additive inverse, but is also true for just the non-negative integers).

this same "proof" holds for more general things than just numbers (integers). for example, if "n" represents a real number, and "0" is a 0-vector in R^k, we get that multiplying the 0-vector by any scalar is also the 0-vector.

 

[DivisionByZro]

This kind of seems logical to me, kind of like saying

$$n(1)$$
$$n(1+0)$$
$$n(1)+n(0)$$

Wait wait.. that would imply..
$$n(1)=-n(0)$$

Which can't be true. So therefore his logic is inconsistent, correct?

But you didn't have "n(1)+n(0)" equal to anything to start with! You simply added an equal sign after; it's not really consistent.

If you did want a short demonstration that a*0 = 0 for all a, then consider this:
$$\begin{align}<br /> Let \ a, b, c \ \epsilon \ \mathbb{R},\ and\ consider \ the \ postulate: \\<br /> a\cdot (b+c) = a\cdot b + a\cdot c \\<br /> \ It \ follows \ that: \\<br /> a\cdot (0+0) = a\cdot 0 + a\cdot 0 = a\cdot 0 \\<br /> \ We \ can \ then \ subtract \ a\cdot 0 \ from \ both \ sides \ to \ obtain: \\<br /> a\cdot 0 = 0 \\<br /> \end{align}$$
As required.

Is this kind of what you wanted to see?

(I can't seem to align my TeX to the left. Help would be appreciated :) )

 

[I like Serena]

(I can't seem to align my TeX to the left. Help would be appreciated :) )

You can use for instance

\begin{array}{l}...\end{array}

 

[DivisionByZro]

You can use for instance

\begin{array}{l}...\end{array}

Thanks! +1

 

[Nano-Passion]

But you didn't have "n(1)+n(0)" equal to anything to start with! You simply added an equal sign after; it's not really consistent.

If you did want a short demonstration that a*0 = 0 for all a, then consider this:
$$\begin{align}<br /> Let \ a, b, c \ \epsilon \ \mathbb{R},\ and\ consider \ the \ postulate: \\<br /> a\cdot (b+c) = a\cdot b + a\cdot c \\<br /> \ It \ follows \ that: \\<br /> a\cdot (0+0) = a\cdot 0 + a\cdot 0 = a\cdot 0 \\<br /> \ We \ can \ then \ subtract \ a\cdot 0 \ from \ both \ sides \ to \ obtain: \\<br /> a\cdot 0 = 0 \\<br /> \end{align}$$
As required.

Is this kind of what you wanted to see?

(I can't seem to align my TeX to the left. Help would be appreciated :) )

I just wanted to see if there is a loop hole in his proof. Something in your proof was confusing. At one point you have

$$a(0+0) = a(0) + a(0)$$
$$a(0)+a(0) = a(0)$$
How did you conclude that?? It looks that your using the property that your trying to prove, which isn't allowed.

Also, I have a proof of my own for this which I thought was pretty cool. I'll post it up here after this is taken care of.

 

[Nano-Passion]

Ah, okay, that would be the distributive property of a field.
(Math mumbo jumbo for the same thing ;)

But isn't it already generally known that a number times zero is zero?
Why proof it?
(Unless you want to proof it for a ring or a field.)

It is generally known because you have been taught it in since you were a little kid. But in mathematics everything has to be proven, which is why its so darn self-consistent and powerful.

 

[DivisionByZro]

I had that a(b+c) = ab + ac. This is a common postulate in some books. I take it for granted most of the time. Surely that's not what's being proved here. What we're trying to prove is that:
a*0 = 0 for all a.

Using b=c=0 in my second line of manipulations, a*(0+0)= a*0 + a*0.
This is close to what your friend had, except that there were two steps missing. My approach is perfectly valid and in fact is also shown in some textbooks.

 

[DivisionByZro]

Here's another way you can look at this (From a less rigorous perspective perhaps):

$$\begin{align}<br /> n\cdot x = x\cdot (n+1)-x \\<br /> \ for \ some \ x,n \ \epsilon \ \mathbb{R} \\<br /> \ Let \ n=0, \ then: \\<br /> \ 0\cdot x = x\cdot (1)-x = 0 \\<br /> As \ required. \blacksquare \\<br /> \end{align}$$

Which is actually kind of a silly way to put it. If you expand my first line you get : nx = xn.
Of course if you let n=0, then 0*x = x*0. If I was grading an assignment I'm not sure I'd mark my above "proof" correct.

 

[Nano-Passion]

I had that a(b+c) = ab + ac. This is a common postulate in some books. I take it for granted most of the time. Surely that's not what's being proved here. What we're trying to prove is that:
a*0 = 0 for all a.

Using b=c=0 in my second line of manipulations, a*(0+0)= a*0 + a*0.
This is close to what your friend had, except that there were two steps missing. My approach is perfectly valid and in fact is also shown in some textbooks.

No my concern was this part actually:

$$a(0)+a(0) = a(0)$$

 

[DivisionByZro]

"No my concern was this part actually:

$$a(0)+a(0) = a(0)$$

"
If b=c=0, then b+c=0, as you had in your opening post. The left-hand side of the fourth line of the proof is:

$$a\cdot (0+0) = a\cdot 0 + a\cdot 0 = a\cdot 0$$

Which, if you remove the middle part, says:

$$a\cdot (0+0) = a\cdot 0$$

There is no mistake here.

 

[Nano-Passion]

"No my concern was this part actually:

$$a(0)+a(0) = a(0)$$

"
If b=c=0, then b+c=0, as you had in your opening post. The left-hand side of the fourth line of the proof is:

$$a\cdot (0+0) = a\cdot 0 + a\cdot 0 = a\cdot 0$$

Which, if you remove the middle part, says:

$$a\cdot (0+0) = a\cdot 0$$

There is no mistake here.

Whoops, I don't know how I missed that. Anyhow, I guess that justifies my friend's proof. I was just wondering if his had any loopholes or not. At any rate, here is my proof, which I thought was rather interesting -- though I am biased. =p

 

[jgens]

This is one case where it becomes extremely important to make it explicit what you are assuming in your proof. If you begin with just the axioms for the real numbers or the rational numbers (or any ring for that matter), then you don't have a proof there.

However, if you are talking about constructing R from N (or something along these lines), where multiplication is defined in terms of repeated addition, then you can turn what you have written up into a rigorous proof using the notion of the empty sum.

I am guessing you are just assuming the field axioms for R,Q so you would actually need to utilize your friend's method for proving this result.

 

[I like Serena]

I had that a(b+c) = ab + ac. This is a common postulate in some books.

Not a common postulate, it is called an axiom.
In mathematics a number of axioms are asserted and from there everything is proven.
This is in all books.

Another axiom that is being used is a+0=0+a=a, meaning in particular that 0+0=0.

n(0)=0 is not an axiom, so that indeed needs to be proven.
DivisionByZro's first proof is correct.

 

[Fredrik]

(I can't seem to align my TeX to the left. Help would be appreciated :) )

The align environment is fine, but you need to use & symbols to indicate where the lines are to be aligned with each other. In this case, you should probably just start each line with &. You also need to use the \text command if you don't want everything you write to be interpreted as variables.

LaTeX guide

 

[Fredrik]

At any rate, here is my proof, which I thought was rather interesting -- though I am biased. =p

This is not a valid proof. You just made an observation about the numbers 1,2,3,4, and then concluded that a similar statement should hold for 0.

I would take these assumptions as the starting point:

ℝ is a set.
0 and 1 are members of ℝ.
Addition and multiplication are both functions from ℝ×ℝ into ℝ.
For each x in ℝ, there's a member of ℝ denoted by -x.
For each x in ℝ except 0, there's a member of ℝ denoted by x^-1.

For all x,y,z in ℝ,

(1) (x+y)+z=x+(y+z)
(2) x+0=0+x=0
(3) x+(-x)=(-x)+x=0
(4) x+y=y+x
(5) (xy)z=x(yz)
(6) x1=1x=x
(7) xx^-1=x^-1x=1
(8) xy=yx
(9) x(y+z)=xy+xz
(10) (x+y)z=xz+yz

Then I would state and prove the theorem like this:

Theorem: For all x in ℝ, x0=0.

Proof: Let x be an arbitrary member of ℝ. Axiom (2) implies that 0=0+0.
\begin{align}
& x0=x(0+0) &&\text{Use axiom (9).}\\
& x0=x0+x0 && \text{Add $-x0$ to both sides.}\\
& x0+(-x0)=(x0+x0)+(-x0) && \text{Use axioms (3) and (1).}\\
& 0=x0+(x0+(-x0)) &&\text{Use axiom (3).}\\
& 0=x0+0 &&\text{Use axiom (2).}\\
& 0=x0
\end{align}

Edit: Uhh, maybe we were only supposed to prove it for integers, not real numbers.

 

[DivisionByZro]

Not a common postulate, it is called an axiom.
In mathematics a number of axioms are asserted and from there everything is proven.
This is in all books.

Yes, that's what I meant. :D

 

[jgens]

Not a common postulate, it is called an axiom.
In mathematics a number of axioms are asserted and from there everything is proven.
This is in all books.

Not true. It depends entirely on what viewpoint you take. If we want to assume the existence of R, Q or Z a priori and that they satisfy the ring axioms, then yes, it is an axiom.

However, if we actually construct these objects, then we need to prove that all of the ring axioms hold in whatever structure. In particular, if we do this for Z say, then we would be proving that the distributive property holds in Z. Hence, it is theorem in this instance, not an axiom. For this reason, I like to think of the ring (resp. field) axioms as theorems in Z (resp. Q, R).

 

[I like Serena]

Not true. It depends entirely on what viewpoint you take. If we want to assume the existence of R, Q or Z a priori and that they satisfy the ring axioms, then yes, it is an axiom.

Assuming those existences, how are they defined then?

 

[jgens]

Assuming those existences, how are they defined then?

Well, there are generally two approaches to working with number systems. One approach is to just list all of the ring/field axioms our number system satisfies and assume that some structure exists that satisfies these axioms. This is the approach that is usually taken by textbooks and I think it is a pedagogically sound way to introduce readers to number systems like R.

However, formally it is a bad practice to assume that some structure exists that satisfies all of the properties we want. How do we know such a structure exists in the first place? This is where constructing the desired number systems comes in. When we do this, that the ring/field axioms hold are theorem that we prove.

If we assume the axioms of (ZFC) set theory, we can come up with a construction of N. Assuming N, we can construct Z. Assuming Z we can construct Q. And assuming Q we can construct R. This is why I like to think of the ring/field axioms as theorems in whatever structure we work within, even though for proofs like this one I use a synthetic approach just like the OP.

 

[Fredrik]

I prefer the other approach, i.e. to think of all structures as defined by their axioms, and to think of the explicit constructions not as "definitions", but as existence proofs. I would just like to point out that this is a matter of taste, not rigor.

 

[Nano-Passion]

This is one case where it becomes extremely important to make it explicit what you are assuming in your proof. If you begin with just the axioms for the real numbers or the rational numbers (or any ring for that matter), then you don't have a proof there.

However, if you are talking about constructing R from N (or something along these lines), where multiplication is defined in terms of repeated addition, then you can turn what you have written up into a rigorous proof using the notion of the empty sum.

I am guessing you are just assuming the field axioms for R,Q so you would actually need to utilize your friend's method for proving this result.

Okay, so my proof is right but is non-rigorous and seems assumptive? I thought the multiplication in terms of repeated addition would be generally agreed on. I guess though since it is not written as one of the basic axioms then I must put more rigor into it.. hmm I how I can do that.

This is not a valid proof. You just made an observation about the numbers 1,2,3,4, and then concluded that a similar statement should hold for 0.

I thought it would be something that would easily be agreed on that you can view multiplication as a factor of addition? That is how multiplication was introduced to me as a little kid too.

By the way, I like your proof. Thanks for sharing. My friend's proof seems to be a bit shorter though, is his proof not correct or were you attempting to put more rigor into it?

 

[jgens]

I would just like to point out that this is a matter of taste, not rigor.

It depends on what you mean here. A construction is necessary to prove that such a structure actually exists, and that is a matter of rigor, not taste.

After the construction however, you can view the structure as completely determined by the axioms and many mathematicians adopt this view. In this case, how you view the structure is a matter of taste. I think this is what you mean, so I agree with you on this point.

 

[jgens]

Okay, so my proof is right but is non-rigorous and seems assumptive?

Not quite. My point was that it depends on context. The context in which case your proof could be formalized is a construction of N where you have defined multiplication in terms of repeated addition. This is most likely not the case, so in all likelihood, your proof is incorrect. It is much more likely that you are working with a synthetic approach to number systems, where you assume all of the basic properties that you want the number system to have, and then you need to prove everything from there.

I thought the multiplication in terms of repeated addition would be generally agreed on. I guess though since it is not written as one of the basic axioms then I must put more rigor into it.. hmm I how I can do that.

It really depends on the context. But making your argument more rigorous would take a lot more work than you want since you would have to work with a construction of N from ∅. Also, it is worth noting the following: Just because you point out a pattern for a few numbers, does not mean that pattern continues. For example, you noted that 2*3 = 6 and 2 + 2 + 2 = 6 and 3 + 3 = 6 and then generalized this to make the claim that n*a is n added a times or a added n times. You would have to show that this holds for all pairs of natural numbers for this to work, and doing this would have to rely on an explicit definition of multiplication.

 

[Fredrik]

Okay, so my proof is right but is non-rigorous and seems assumptive?

When I said that your proof (the picture in post #17) wasn't valid, I meant that it was completely wrong. 0 points, may God have mercy on your soul and all that. I just didn't want to be rude. (Sorry, I couldn't resist the Billy Madison reference. It was definitely not meant as an insult, just a little joke).

To a mathematician, your proof looks like "since the next few months are in the year 2012, it must be 2012 right now". The things you said about the numbers 1,2,3,4 don't imply anything useful about the number 0.

I thought the multiplication in terms of repeated addition would be generally agreed on.

It is, but what exactly it means depends on what axioms you take as the starting point. It's especially important to be careful with this when you're talking about zero terms. What does it mean to add zero copies of x together? It's reasonable to define it by saying that the notation 0x denotes 0. But then there's nothing to prove.

My friend's proof seems to be a bit shorter though, is his proof not correct or were you attempting to put more rigor into it?

His proof is not correct. I stopped reading it at "0+0=0. ergo, n(0)+n(0)=0". It's not clear to me what he's thinking here, and in my opinion, that's enough to make a proof invalid. But it looks like he's doing something seriously wrong. For example, if the idea is to multiply the first equality by n, then the result is n0+n0=n0, not n0+n0=0. So it looks like he's using what he's trying to prove.

My proof was a more rigorous (or at least more explicit) version of DivisionByZro's proof. I just listed the axioms and explained which one I was using at each step. He didn't explain why you can cancel a term from each side, e.g. that x+z=y+z implies x=y. That can be proved as a separate theorem, but you can also just add -z and use the axioms.