Proof by contradiction - any non-zero number divided by itself is 1

[musicgold]

Homework Statement

This is not a homework problem. I was explaining my kid the concept of mathematical rigour and the importance of proofs. She threw a curved ball at me - how I can prove that every number divided by itself is 1.

Relevant Equations

I tried using proof by contradiction to prove it but couldn't do it. Is this because the statement is a corollary of the Multiplative identity posulate (of the famous 13 posultates)?

Proof by contradiction (for some reason the LaTeX code is not working for me. Sorry)

Lets assume that A, B, and C are non-zero real numbers; A = B ; and C is not equal to 1.

A/ B = C
A = B x C

But BxC could be equal to B, if and only if C =1

Also, could you recommend a book where I can learn more about such basic proofs?

Thanks.

 

[RPinPA]

Yes, I'd say that the multiplicative identity can be used here.

If C = 1, the multiplicative identity, then B x C = B

Using the definition of division (y is defined to be z/x with x nonzero, if xy = z), then C = B/B.

 

[FactChecker]

In abstract algebra, the multiplicative identity is defined and denoted as '1'. Then the multiplicative inverse of x is defined as the number, b, such that x*b= multiplicative identity = 1. Then b is denoted as '1/x'. So it is more of a definition and notation than a proof.

 

[WWGD]

Following up on Factchecker, rewording : $\mathbb R -\{0\}$ is a Field, so that each number has a ( multiplicative) inverse. Call 1/x the inverse of x . It follows x(1/x)=x/x=1. More generally, if F =(S, +,*) is a field; S is the underlying set and +, * are respectfully addition and multiplication , then (S, *) is an Abelian group.

 

[Hornbein]

A/ B = C
A = B x C

That's cheating.

A/B = C
B x A/B = B x C
A x B/B = B x C
A x 1 = B x C Using your claim.
A = B x CIt's hard to do partly because I don't know what is allowed. But using your claim is definitely a no-no.

If you know that you have a group then B/B = 1 is part of the definition of a group so there isn't anything to prove. I think this is why I can't prove your claim: B x 1/B = 1 is the definition of 1/B. How does one prove a definition? You have to show that it doesn't lead to a contradiction. It's not obvious to me how to do that.

 

[FactChecker]

Suppose the multiplicative inverse of x is defined as '1/x' and is not something to prove. The existence and uniqueness of the multiplicative inverse are things that still should be proven. The proof of uniqueness is easy. The formal proof of existence might be much harder. But I don't think that those are the proofs that the OP was asking about.

 

[pbuk]

I think that field theory and the rest of abstract algebra is probably a bit too advanced for the OP (although maybe not for his daughter). I'd start with letting $\frac x x = 1 + a$ and proceed
$$ \begin{align}
\frac x x & = 1 + a \\
\frac x x x & = (1 + a) x \\
... \\
a & = 0 \\
\frac x x & = 1 + 0 \\
\frac x x & = 1
\end{align}$$
Now can you get from $\frac x x x = (1 + a) x$ to $a = 0$?

 

[PeroK]

$\frac x x$ is by definition the number $y$ such that $x = xy$.

By inspection $y = 1$ is the solution for any nonzero $x$.

More formally, $\frac 1 x \equiv x^{-1}$ and the proposition is true by definition.

 

[PAllen]

To make this something other than a definition, you would need some definition of the real numbers that does not include a statement about multiplicative identity (i.e. a nonstandard, but hopefully 'intuitive' and consistent alternate set of axioms). Then, proof of A/A=1 for nonzero A would be part of verifying the equivalence of this alternate axiom set with the standard one. I am doubtful this would be satisfying to your daughter.

 

[FactChecker]

One line of thought that is not steeped in theory is this:
1) n items divided among n people gives you one per person. So $n_{integer}/n_{integer}=1$.
2) x.yz/x.yz = xyz/xyz = xyz items divided among xyz people gives you one per person. So $x_{finiteDigits}/x_{finiteDigits} = 1$
3) $y_{infiniteDigits}/y_{infiniteDigits}$ = as close as we want to a $x_{finiteDigits}/x_{finiteDigits}$ = 1.

 

[DaveC426913]

1) n items divided among n people gives you one per person.

How do you know this to be true, though?
Circular reasoning? Begging the question? Straight up intuition?

It's the opposite of mathematical rigour. The opposite of what the OP is asking for.

 

[PeroK]

How do you know this to be true, though?
Circular reasoning? Begging the question? Straight up intuition?

It's the opposite of mathematical rigour. The opposite of what the OP is asking for.

This is getting silly! There is a bijection between any two sets of order $n$.

 

[DaveC426913]

There is a bijection between any two sets of order $n$.

OK, is bijection a valid function in a mathematical proof?

If so, that is the axiom to with the OP can prove the conjecture, and all the rest is overkill.

 

[PeroK]

OK then that is the axiom to with the OP can prove the conjecture, and all the rest is overkill.

No, because we need to consider real numbers. My post #8 covers it.

Moreover, the first part of post #8 covers the case for natural numbers. In other words: $\frac n n = 1$ because $n = n\cdot 1$. That's your first definition of what it means to divide. In general:

If $a = b \cdot c$ then $\frac a b = c$. That's the best starting point, IMO.

 

[DaveC426913]

No, because we need to consider real numbers. My post #8 covers it.

OK, I'm simply pointing out that FactChecker's logic is either
- not rigorous (relying on unproven statements),
or
- needlessly complex (because you can just use the bijection function in place of his logic).

 

[FactChecker]

How do you know this to be true, though?
Circular reasoning? Begging the question? Straight up intuition?

It's the opposite of mathematical rigour. The opposite of what the OP is asking for.

I guess that is true.
This gives an intuitive place to start. I might call it an axiom, except that it might be provable by induction from "One item divided among one person gives one item per person."
So here is a question. Can you prove this?
Given the axiom: One item divided among one person gives one item per person.
Use induction to prove: N items evenly divided evenly among N people gives one item per person.

 

[PeroK]

I guess that is true.
This gives an intuitive place to start. I might call it an axiom, except that it might be provable by induction from "One item divided among one person gives one item per person."
So here is a question. Can you prove this?
Given the axiom: One item divided among one person gives one item per person.
Use induction to prove: N items evenly divided evenly among N people gives one item per person.

In set theory, we might start with the definition that a set of order $N$ is one with a bijection to the set $I_N = \{1, 2, \dots N\}$.

Then, any two sets of order $N$ automatically have a bijection between them. That implies one and only one item from a set of $N$ items may be assigned to every person in a set of $N$ people.

 

[DaveC426913]

Given the axiom: One item divided among one person gives one item per person.

The problem is, that is little better than the OP's original conjecture:

"...every number divided by itself is 1..."

If one finds your axiom acceptable as a self-evident axiom, then might as well find the OP's conjecture to be as self-evident, i.e. without further need for a proof.

(Sorry if it seems like I'm pounding on you, I'm not. But I do think it strikes at the heart of what a proof is. And will be a very useful "anti-example" for the OP to consider.)

 

[FactChecker]

The problem is, that is little better than the OP's original conjecture:

"...every number divided by itself is 1..."

If one finds your axiom acceptable as a self-evident axiom, then might as well find the OP's conjecture to be as self-evident, i.e. without further need for a proof.

(Sorry if it seems like I'm pounding on you, I'm not. But I do think it strikes at the heart of what a proof is. And will be a very useful "anti-example" for the OP to consider.)

I disagree. I think the statement that one item given to one person leaves him with one item is significantly more obvious than the general case. I think that it is worthy of being a self-evident axiom. I can't imagine that anyone would challenge it.