How to show that $0 \times 2 = 0$?

Question

Here is the suggested proof: $0 \cdot 2=2 \cdot 0 =(1+1) \cdot 0 = 1\cdot 0 + 1\cdot 0 = 0 + 0$. But my question lies in this step. Here is the definition of Zero: $0+a = a$ (for any number $a$) therefore: $0+5=5$ or $0+1639=1639$, but can we say $0+0 =0$ ? I mean, that "any" includes "Zero". But we have not defined zero yet.

In other words, we are using "something" before we defined that "something". These days I'm struggling with these fundamental doubts. Especially after reading some quotes from Bertrand Russell saying:" Mathematicians don't know what they're doing nor know their work is true( or something like this...) and John von Neumann saying: "When we are doing mathematics we don't know whether our works are true or not. We just get used to them" Why Russell wrote about 300 pages to show what is "$1$"? Can anyone please introduce me a book which proves math's certainty (if it exists) ?

"Here is the definition of Zero [...] But we have not defined zero yet." Hrmm.. — anon, Jul 04 '12 at 04:54
So your question is really about the coherency of the statement $0+0=0$? — , Jul 04 '12 at 05:02
Zero is not defined as that number $z$ such that $z+a = a$, $\forall a$. The definition of $z$ is way before addition. — , Jul 04 '12 at 05:02
+1 to Tim Duff's observation. The title does not reflect the question as well as answers dealing with the point if 0+0=0. — Sniper Clown, Jul 04 '12 at 07:33

score 8 · Accepted Answer · answered Jul 04 '12 at 05:02

8

If you want to ask in detail how something is proved, you need to specify what set of axioms and what rules of inference are allowed.

However, the usual situation is you have a set of axioms. Zero is a distinguished element, that is, one that is named in the axioms. It doesn't need a definition. As you say, one axiom (in PA, for example) is $\forall a: a+0=a$ This is not a definition of $0$, it is part of a definition of $+$. You can certainly substitute $0$ for $a$ to get $0+0=0$ Then you also have an axiom that $\forall a:a \cdot 0=0$. You also need the commutative law for multiplication (a theorem of PA), the distributive law (an axiom), the definition of $2$ (usually $SS0$ where $S$ is a unary function intended as the successor), and $1+1=2$ (a theorem of PA once you define $1$ as $S0$).

The most famous attempts to prove the certainty of mathematics were Hilbert's program and Russell and Whitehead's Principia Mathematica. This is an enormous area of study.

answered Jul 04 '12 at 05:02

Ross Millikan

374,822

This is what I thought of. Actually, I saw that proof(if so) in a book about Mathematics and the writer didn't mention anything about Peano (I think deliberately for simplicity). Thanks – Zeta.Investigator Jul 04 '12 at 05:14
1

The axiom $\forall a: a+0=a$ is as much part of the definition of $0$ as it is part of the definition of '$+$', namely not at all. The (nullary, unary, binary) operations like $0$ and $+$ are part of the language of PA, they are not functions defined by the axioms. The axioms describe properties of the operations that are assumed to hold, but these properties need not determine any function uniquely. Indeed (depending on the setting) there may be more than one model of PA, and operations will correspond to different functions in each model. – Marc van Leeuwen Jul 04 '12 at 14:05
$a0 = 0$ is not typically given as an axiom. One proves it by $a0 = a(0+0) = a0 + a0$ then adding $-(a0)$ to both sides. – nullUser Jul 04 '12 at 14:21

score 5 · Answer 2 · answered Jul 04 '12 at 05:03

Chris Dugale gives the answer that I would have given, but it seems that the question is actually a different one:

How can we infer $0+0=0$?
But this is no problem: We define $0$ as an element of our ring, field, set of numbers that satisfies $a+0$ for all $a$. Usually we have an axiom saying that such an element $0$ exists. Now $0+0=0$ simply follows from the definition of $0$ together that $0$ is one of the $a$'s. I don't see any problem with this. You define $0$ in some way and then prove some things about the object that satisfies this definition. Actually, we just pick a number with certain properties and then at some point show that there is only one such number. But this comes later. For your argument it is enough to pick one number $b$ such that for all $a$ we have $a+b=a$. For simplicity we denote this $b$ by $0$.

score 3 · Answer 3 · answered Jul 04 '12 at 04:51

3

Here we go. We're going to kick it up a knotch. BAM. Let $a\in \mathbb{R}$. Then, we have that $0a=(0+0)a$. This is true since 0 is the additive idenity. But then, we have $0a=0a+0a$ by distributivity. But then we have that $0a-0a=0a$. But, by the left hand side is zero. Hence, $0=0a$. Thus, since two is a real number, we have that $0*2=0$. Note that this proves that you can not divide by zero, why ?

answered Jul 04 '12 at 04:51

RougeSegwayUser

2,871

You used the fact: 0+0 = 0 . But here lies my question. Why? 2.I have a similar ambiguity about subtraction and division. So let's put aside these two for the proof

Zeta.Investigator

Jul 04 '12 at 04:56

You shouldn't need $\mathbb R$ for this. It comes much later. – Ross Millikan Jul 04 '12 at 05:03

You're right, you don't need it. But, that's the context that I'm assuming he is working in... – RougeSegwayUser Jul 04 '12 at 05:10

0+0=0 by definition. – RougeSegwayUser Jul 04 '12 at 05:10

You could just assume the multiplicative identity is s.t. 0a=0 for any finite a. But the goal is to have as few axioms as possible and then prove 0a=0 by other more "fundamental" axioms. In this case, perhaps the most fundamental of all axioms is 0+0=0 and next we require associativity for multiplication (in fact we don't always have to, but your question obviously pre-supposes this) – Squirtle Jul 04 '12 at 05:58

@dustanalysis: No, a "multiplicative identity" is the $1$ such that $1\cdot a=a$ for any $a$. – hmakholm left over Monica Jul 04 '12 at 11:49

@ChrisDugale "Here we go. We're going to kick it up a knotch. BAM." Please save this for the stage... in other words, it's unnecessary here. – Doug Spoonwood Jul 05 '12 at 13:33

score 3 · Answer 4 · answered Jul 04 '12 at 05:01

3

A natural (no pun intended) starting point for the theory of natural numbers is the axiom system of Peano. These axioms are sufficient to guarentee the existence and uniqueness of addition as defined by recursion. Since $0$ has already been defined in the axiomatic framework, it is intelligible in the definition of $+$ to require that $0+0=0.$

answered Jul 04 '12 at 05:01

5

My friend, you are lying when you say that the pun was not intended. – RougeSegwayUser Jul 04 '12 at 06:03

Bill Michell · Answer 5 · 2012-07-04T15:18:38.723

The wording of the extended question makes this feel like a philosophy of mathematics question, rather than the more obvious one about Peano arithmetic.

The whole point of mathematics is this:

Starting from the (stated) assumption that certain things are so, can we convince ourselves beyond any reasonable doubt that other things must also necessarily be so?

Pure mathematics makes no statement about whether the initial assumptions (axioms) actually hold in the real world. Mathematics just makes statements about statements, in an attempt to convince you of the truth of other statements.

If you are not convinced by the proof, then there is more work for the proof to do.

If enough other people are convinced by the proof, then possibly the gap is within your own understanding of the proof – or perhaps you have identified a flaw in the original proof, which you may be able to persuade others to accept.

In order to help you with the step you are currently struggling with, $0 + 0 = 0$, it would help us to know where you started your journey, and what it is safe for us to assume.

Personally, I'd consider the existence of $0$ to be axiomatic (i.e. it exists because we all agree that we have to assume something in order to get started with anything†), and the statement $0 + 0 = 0$ to be true, because it is part of the definition of the + operator.

Does that mean that $0$ exists in the real world? Mathematicians don't really care about that. Physicists, engineers and accountants might, and indeed do, because it permits them to make a living building bridges that don't fall down, but it isn't a question that Pure Mathematicians can really answer.

† Maybe there is room for some mathematics even more fundamental than this, where even simpler assumptions enable us to deduce the existence of $0$, but I don't know what it is...

When modelling the natural numbers from sets, you define $0$ to be the empty set (which you've already assumed to exist when you accepted set theory). Therefore in a certain sense the existence of $0$ is deduced. — celtschk, Jul 04 '12 at 13:54

score 1 · Answer 6 · edited Jul 04 '12 at 14:15

1

I'd try to be ambiguous here, but here is a simple explanation.

The translation of $ab$ is $a + a + a\cdots (\text{b times)}$

$0 \times 2$ is 0 added to itself two times.

$ 0 + 0$

No apples added to no apples are just no apples.

$0 + 0 = 0$

Another explanation is that the number $0$ is considered as the additive identity, i.e $x + 0 = x$. Therefore, if we assume $x = 0$, then $0 + 0 = 0$.

edited Jul 04 '12 at 14:15

Cameron Buie

102,994

answered Jul 04 '12 at 13:23

P.K.

7,742

1

Of course, the "translation" breaks down when $b$ is irrational (or even a negative integer). – Cameron Buie Jul 04 '12 at 14:18

score 1 · Answer 7 · answered Jul 04 '12 at 13:48

You didn't say which axioms and already proven theorems you are using. For example, if one of your axioms is or contains $0+0=0$, it is trivially true.

Here's a proof that $0+0=0$ given that

A1 There exists at least one $a\neq 0$

A2 $0+a=a$ for all $a\neq 0$

A3 $+$ is associative, i.e. $(a+b)+c = a+(b+c)$, unconditionally

A4 From $a+c=b+c$ it follows that $a=b$

Be $a\neq 0$ (such an $a$ exists due to A1). Since for that $a$ we have $0+a=a$ (by A2), it follows that $(0+a)\neq 0$. Therefore we have $0+a =_\text{A2} 0+(0+a) =_\text{A3} (0+0)+a$. Using A4, we can therefore conclude $0+0=0$.

BTW, I now notice that A1 isn't really needed, because if there doesn't exist an $a$ with $a\neq 0$, then especially it is not possible that $0+0\neq 0$, and thus for that case $0+0=0$. If there exists an $a\neq0$, continue as above.

How to show that $0 \times 2 = 0$?

7 Answers7