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

by Nano-Passion
Tags: friend, proof
P: 1,306
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
 P: 252 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)".
 HW Helper P: 6,187 Hi Nano-Passion! Just guessing here... Are we talking about a group homomorphism? Or about the distributive property of a ring or field? Or...
 Sci Advisor P: 6,056 Friend's proof that n(0) = 0 My guess is simply multiplication. n(0) means nx0, where x is multiply.
P: 1,306
 Quote by 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)".
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?

 Quote by 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...
Hey!

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

 Quote by mathman My guess is simply multiplication. n(0) means nx0, where x is multiply.
Precisely!
 HW Helper P: 6,187 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.)
P: 906
 Quote by Nano-Passion 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 Rk, we get that multiplying the 0-vector by any scalar is also the 0-vector.
P: 252
 Quote by Nano-Passion 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} 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 :) )
HW Helper
P: 6,187
 Quote by DivisionByZro (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}
P: 252
 Quote by I like Serena You can use for instance \begin{array}{l}...\end{array}
Thanks! +1
P: 1,306
 Quote by 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 :) )
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.
P: 1,306
 Quote by 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.)
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.
 P: 252 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.
 P: 252 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.
P: 1,306
 Quote by 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.
No my concern was this part actually:

$$a(0)+a(0) = a(0)$$
 P: 252 "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.
P: 1,306
 Quote by 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.
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
Attached Thumbnails

 P: 1,622 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.

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