Either or statement from Abstract Algebra Book

[cbarker1]

Dear Everyone,

What are the strategies from proving a either-or statements? Is there a way for me to write an either-or statement into a standard if-then statements? For example, this exercise is from Dummit and Foote Abstract Algebra 2nd, "Let $x$ be a nilpotent element of the commutative ring $R$. Prove that $x$ is either a zero or an zero divisor."

Any advice will be helpful.

Thanks
Cbarker1

 

[Olinguito]

Hi Cbarker1.

Statements of the form “$P$ or $Q$” are generally proved as follows: either assume that $P$ is false and prove that $Q$ must be true, or assume that $Q$ is false and prove that $P$ must be true.

Thus, to prove that $x$ is either zero or a zero divisor, the best thing to do is that assume that $x\ne0$ and show that $x$ is a zero divisor. Alternatively, you can assume that $x$ is not a zero divisor and show that $x$ must be $0$ – but this might be trickier. Choose whichever is the easier.

 

[cbarker1]

Hi Cbarker1.

Statements of the form “$P$ or $Q$” are generally proved as follows: either assume that $P$ is false and prove that $Q$ must be true, or assume that $Q$ is false and prove that $P$ must be true.

Thus, to prove that $x$ is either zero or a zero divisor, the best thing to do is that assume that $x\ne0$ and show that $x$ is a zero divisor. Alternatively, you can assume that $x$ is not a zero divisor and show that $x$ must be $0$ – but this might be trickier. Choose whichever is the easier.

So a proof would look this: Let $x\ne 0$. Let $x^{m-1}$ for $m\ge 1$. Then $x\cdot x^{m-1}=x^{m}=0$. Thus $x$ is a zero divisor, QED?

 

[Olinguito]

So a proof would look this: Let $x\ne 0$. Let $x^{m-1}$ for $m\ge 1$. Then $x\cdot x^{m-1}=x^{m}=0$. Thus $x$ is a zero divisor, QED?

Yes, that’s correct.

 

[cbarker1]

Yes, that’s correct.

Do I have to prove the alternative statement?

 

[Olinguito]

No.

 

[I like Serena]

Erm... formally speaking, yes, we also need to prove the other statement.

That is, 'either P or Q' is equivalent to '(if NOT P then Q) AND (if P then NOT Q)'.

You have proven that if $x\ne 0$ that then $x$ is a zero-divisor.
So formally we also need that if $x=0$ that then $x$ is NOT a zero-divisor.

Unfortunately $x=0$ is a zero-divisor.
It means that your problem statement is incorrect.
It should just have said to prove that '$x$ is a zero-divisor'.
Or equivalently to prove that '$x=0$ or $x$ is a zero-divisor'. That is, without the 'either'.

To be honest, it's not unusual that people are sloppy with the use of either-or. That does not make it correct though.

 

[cbarker1]

Erm... formally speaking, yes, we also need to prove the other statement.

That is, 'either P or Q' is equivalent to '(if NOT P then Q) AND (if P then NOT Q)'.

You have proven that if $x\ne 0$ that then $x$ is a zero-divisor.
So formally we also need that if $x=0$ that then $x$ is NOT a zero-divisor.

Unfortunately $x=0$ is a zero-divisor.
It means that your problem statement is incorrect.
It should just have said to prove that '$x$ is a zero-divisor'.
Or equivalently to prove that '$x=0$ or $x$ is a zero-divisor'. That is, without the 'either'.

To be honest, it's not unusual that people are sloppy with the use of either-or. That does not make it correct though.

How to begin the second proof? I know I have to write down "Suppose x=0..."

 

[Olinguito]

Erm... formally speaking, yes, we also need to prove the other statement.

That is, 'either P or Q' is equivalent to '(if NOT P then Q) AND (if P then NOT Q)'.

That is NOT true. The statement “either P or Q” is just equivalent to “if not P then Q”. (It is also equivalent to “if not Q then P”.) It DOES NOT imply “if P then not Q”.

That is because in logic or is taken to be inclusive, i.e. “P or Q” means “P or Q or both”.

How to begin the second proof? I know I have to write down "Suppose x=0..."

You don’t have to!

 

[I like Serena]

That is NOT true. The statement “either P or Q” is just equivalent to “if not P then Q”. (It is also equivalent to “if not Q then P”.) It DOES NOT imply “if P then not Q”.

That is because in logic or is taken to be inclusive, i.e. “P or Q” means “P or Q or both”.

Cambridge Dictionary

either-or
adjective [ before noun ]
uk ​ /ˌaɪ.ðərˈɔːr/ /ˌiː.ðərˈɔːr/ us ​ /ˌiː.ðɚˈɔːr/ /ˌaɪ.ðɚˈɔːr/
​
Used to refer to a situation in which there is a choice between two different plans of action, but both together are not possible:
It's an either-or situation - we can buy a new car this year or we can go on holiday, but we can't do both.

How to begin the second proof? I know I have to write down "Suppose x=0..."

Let's get the definition clear first.
To be honest, with the definition as I know it and apply it, your problem statement is broken.

 

[Olinguito]

Sorry, you are using a dictionary that defines the word in everyday English use, not as it is generally used in mathematics. In mathematics, this is how the word or is generally used.

In logic, or by itself means the inclusive or, distinguished from an exclusive or, which is false when both of its arguments are true, while an "or" is true in that case.

To be honest, with the definition as I know it and apply it, your problem statement is broken.

So fix your understanding of the definition!

 

[Evgeny.Makarov]

That is because in logic or is taken to be inclusive, i.e. “P or Q” means “P or Q or both”.

Simple "or" is indeed inclusive OR in mathematics, but "either... or... ", in my experience, is used in math to mean exclusive OR. And the book says "either a zero or a zero divisor". So I agree with Klaas, even though I also agree that natural language definitions are not always applicable to math. For example, Russian legal documents keep using phrases like "either theft or injury" when they mean that both things are possible at the same time.

It turns out that the book defines a zero divisor as a nonzero element $a$ of a ring $R$ if there is a nonzero element $b$ in $R$ such that either $ab = 0$ or $ba = 0$. So exclusive OR is correct in the original statement.

 

[Olinguito]

Simple "or" is indeed inclusive OR in mathematics, but "either... or... " is, in my experience, is used in math to mean exclusive OR. And the book says "either a zero or a zero divisor". So I agree with Klaas, even though I also agree that natural language definitions are not always applicable to math. For example, Russian legal documents keep using phrases like "either theft or injury" when they mean that both things are possible at the same time.

It turns out that the book defines a zero divisor as a nonzero element $a$ of a ring $R$ if there is a nonzero element $b$ in $R$ such that either $ab = 0$ or $ba = 0$. So exclusive OR is correct in the original statement.

Thanks for clearing it up.

 

[I like Serena]

Assuming the either-or is indeed intended as an exclusive OR, then the argument is as follows.

Assume $x=0$, then $x$ is nilpotent and therefore a solution to the problem.
Furthermore $x$ is not a zero-divisor since the book's definition explicitly excludes $0$.
QED

 

[I like Serena]

It turns out that the book defines a zero divisor as a nonzero element $a$ of a ring $R$ if there is a nonzero element $b$ in $R$ such that either $ab = 0$ or $ba = 0$

Urghh... so the book itself uses either-or to mean an inclusive OR, doesn't it?

Next time I'll ask for the definitions of the author of the book. And how he uses either-or assuming he is at least consistent.

 

[Olinguito]

Next time I'll ask for the definitions of the author of the book. And how he uses either-or assuming he is at least consistent.

I suggest that next time you keep a more open mind and try not to impose your own views before analysing a situation.

 

[I like Serena]

I suggest that next time you keep a more open mind and try not to impose your own views before analysing a situation.

Right back at you!

In this case I posted what I was sure was right based on my background in logic. I also checked the wiki definition of a zero-divisor.
You challenged that.
Before posting again, I searched a bit and gave a reference. While doing so I noticed that the wiki article that you referenced later, did not actually mention the either-or construction.

You did none of that. Instead you kept second guessing me, and challenge me to the point of provocation. And you tried to impose your view without proper reference.
I had actually already abandoned this thread as a bad one with no room to get to a correct mathematical resolution.
That is when Evgeny posted, and when you actually thanked him for it. So I gave it another shot. And yet here we are.