# 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: 6,189 Hi Nano-Passion! Just guessing here... Are we talking about a group homomorphism? Or about the distributive property of a ring or field? Or...
P: 5,935

## 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: 6,189 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: 6,189
 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.
$$a(0)+a(0) = a(0)$$
 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.
 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.