# Friend's proof that n(0) = 0

1. Dec 10, 2011

### 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.

2. Dec 10, 2011

### 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)".

3. Dec 10, 2011

### 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...

4. Dec 10, 2011

### mathman

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

5. Dec 10, 2011

### Nano-Passion

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?

Hey!

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

Precisely!

6. Dec 10, 2011

### 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.)

7. Dec 10, 2011

### Deveno

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 Rk, we get that multiplying the 0-vector by any scalar is also the 0-vector.

8. Dec 10, 2011

### DivisionByZro

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} Let \ a, b, c \ \epsilon \ \mathbb{R},\ and\ consider \ the \ postulate: \\ a\cdot (b+c) = a\cdot b + a\cdot c \\ \ It \ follows \ that: \\ a\cdot (0+0) = a\cdot 0 + a\cdot 0 = a\cdot 0 \\ \ We \ can \ then \ subtract \ a\cdot 0 \ from \ both \ sides \ to \ obtain: \\ a\cdot 0 = 0 \\ \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 :) )

9. Dec 10, 2011

### I like Serena

You can use for instance
Code (Text):
\begin{array}{l}...\end{array}

10. Dec 10, 2011

### DivisionByZro

Thanks! +1

11. Dec 10, 2011

### Nano-Passion

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.

12. Dec 10, 2011

### Nano-Passion

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.

13. Dec 10, 2011

### 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.

14. Dec 10, 2011

### DivisionByZro

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

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

Last edited: Dec 10, 2011
15. Dec 10, 2011

### Nano-Passion

No my concern was this part actually:

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

16. Dec 10, 2011

### 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.

17. Dec 10, 2011

### Nano-Passion

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

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18. Dec 11, 2011

### 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.

19. Dec 11, 2011

### I like Serena

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.

Last edited: Dec 11, 2011
20. Dec 11, 2011

### Fredrik

Staff Emeritus
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

21. Dec 11, 2011

### Fredrik

Staff Emeritus
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.

Last edited: Dec 11, 2011
22. Dec 11, 2011

### DivisionByZro

Yes, that's what I meant. :D

23. Dec 11, 2011

### jgens

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).

24. Dec 11, 2011

### I like Serena

Assuming those existences, how are they defined then?

25. Dec 11, 2011

### jgens

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.