Friend's proof that n(0) = 0

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.

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

Well you skipped a step in between

[tex]n(0)[/tex]
which is the same as
[tex]n(0+0)[/tex]
[tex]n(0)+n(0)[/tex]

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

[tex]n(1)[/tex]
[tex]n(1+0)[/tex]
[tex]n(1)+n(0)[/tex]

Wait wait.. that would imply..
[tex]n(1)=-n(0)[/tex]

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!

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

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

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:
[tex]
\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}
[/tex]
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

You can use for instance

DivisionByZro

Thanks! +1

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

[tex]a(0+0) = a(0) + a(0)[/tex]
[tex]a(0)+a(0) = a(0)[/tex]
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

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):

[tex]
\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}
[/tex]

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

No my concern was this part actually:

[tex]a(0)+a(0) = a(0)[/tex]

DivisionByZro

"No my concern was this part actually:

[tex]
a(0)+a(0) = a(0)
[/tex]

"
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:

[tex]
a\cdot (0+0) = a\cdot 0 + a\cdot 0 = a\cdot 0
[/tex]

Which, if you remove the middle part, says:

[tex]
a\cdot (0+0) = a\cdot 0
[/tex]

There is no mistake here.

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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