Why does multiplying by 0 equal 0?

[LightbulbSun]

I know why it would equal to 0 if it was (0*0). But what about an actual number? Why does (100*0) equal to 0? You're not multiplying anything, but shouldn't it still equal to 100? If I have 100 cookies on the table, and I don't multiply it by anything, why do I suddenly have zero cookies on the table?

I'm just trying to gain a conceptual understanding behind the zero-factor algebraic property.

 

[Tedjn]

One possible conceptual way to think about it is to say, multiplying 2 * 100 is like having 2 groups of 100 cookies on the table. Multiplying 1 * 100 is like having 1 group of 100 cookies on the table. Multiplying 0 * 100 is like having no groups of 100 cookies on the table, hence no cookies at all.

However, I agree that this is may not be very intuitive. When dealing with 0, it is sometimes more difficult to match mathematical situations to real life situations. It is probably better to understand 0 * any number = 0 just as a consequence of several properties of numbers that we take for granted.

 

[statdad]

Remember that 0 + a = a, 0 + 0 = 0, and a - a = 0, no matter what number you use for a. Now the rules of arithmetic give this

<br /> \begin{align*}<br /> 0 \cdot a &amp; = (0 + 0)a\\<br /> &amp; = 0 \cdot a + 0 \cdot a\\<br /> \left(0 \cdot a - 0 \cdot a\right) &amp; = 0 \cdot a\\<br /> 0 &amp; = 0 \cdot a<br /> \end{align*}<br />

 

[Vid]

You actually prove this at the beginning of an undergrad analysis/theoretical calculus class.

 

[LightbulbSun]

One possible conceptual way to think about it is to say, multiplying 2 * 100 is like having 2 groups of 100 cookies on the table. Multiplying 1 * 100 is like having 1 group of 100 cookies on the table. Multiplying 0 * 100 is like having no groups of 100 cookies on the table, hence no cookies at all.

However, I agree that this is may not be very intuitive. When dealing with 0, it is sometimes more difficult to match mathematical situations to real life situations. It is probably better to understand 0 * any number = 0 just as a consequence of several properties of numbers that we take for granted.

That's actually a very good way of thinking about it.

Remember that 0 + a = a, 0 + 0 = 0, and a - a = 0, no matter what number you use for a. Now the rules of arithmetic give this

<br /> \begin{align*}<br /> 0 \cdot a &amp; = (0 + 0)a\\<br /> &amp; = 0 \cdot a + 0 \cdot a\\<br /> \left(0 \cdot a - 0 \cdot a\right) &amp; = 0 \cdot a\\<br /> 0 &amp; = 0 \cdot a<br /> \end{align*}<br />

Ah, so it's sort of a cancelation?

 

[tiny-tim]

If I have 100 cookies on the table, and I don't multiply it by anything, why do I suddenly have zero cookies on the table?

Hi LightbulbSun!

No, if you don't multiply it by anything, you still have 100 cookies on the table.

This is a language thing …

"not multiplying by anything" is not the same as "multiplying by nothing" …

"not multiplying by anything" means leaving it the same.

EDIT: I think the French don't have this problem …
they distinguish between (pardon my French! ) …
"multiplier par rien" and "ne multiplier par rien"

 

[symbolipoint]

This is far too natural to be confusing. Remember that multiplication is repeated addition. If you add 100, ZERO times, you have ZERO. If you want 100 as result, then you must add 100 ONE time.

 

[LightbulbSun]

This is far too natural to be confusing. Remember that multiplication is repeated addition. If you add 100, ZERO times, you have ZERO. If you want 100 as result, then you must add 100 ONE time.

Now it makes more sense to me. Thanks for the explanation.

 

[Hurkyl]

I know why it would equal to 0 if it was (0*0). But what about an actual number? Why does (100*0) equal to 0? You're not multiplying anything, but shouldn't it still equal to 100? If I have 100 cookies on the table, and I don't multiply it by anything, why do I suddenly have zero cookies on the table?

I'm just trying to gain a conceptual understanding behind the zero-factor algebraic property.

The problem is that natural language sort of pigeon-holes us into thinking "none" and "at least one" are conceptually different. As soon as you break through this barrier and become comfortable working with degenerate cases, stuff like this becomes easy.

For example, how many pennies do you have, if you have zero rows of N pennies each? (Or, as one would generally say in natural language, if you don't have any rows of N pennies each)

 

[poutsos.A]

proof ox=o

0x = 0x + 0 = 0x + [ x + (-x)] = (0x + x) + (-x) = x( 0 +1) + (-x) = x +(-x) = 0

or

0x = 0 <===> 0x + x = 0+x <=====> x( 0 + 1) = 0 + x <===> x = 0 + x <===> x=x correct so 0x=0

 

[symbolipoint]

proof ox=o

0x = 0x + 0 = 0x + [ x + (-x)] = (0x + x) + (-x) = x( 0 +1) + (-x) = x +(-x) = 0

or

0x = 0 <===> 0x + x = 0+x <=====> x( 0 + 1) = 0 + x <===> x = 0 + x <===> x=x correct so 0x=0

How does a person learn to work with such simple, low-level ideas like that to prove what would otherwise seem so natural? Fantastic!

 

[Tac-Tics]

I know why it would equal to 0 if it was (0*0). But what about an actual number? Why does (100*0) equal to 0?

Because it turns out simple and useful.

Multiplication, and any operation, is something defined by the mathematician. We could imagine a world where 0 * n = n. But it would break a lot of useful theorems. For instance, 0 * 1 + 1= 1 + 1 = 2. However, 0 * 1 + 1 = (0 + 1)*1 (by distribution), so 0 * 1 + 1 = 1, and so 1 = 2. We must conclude that addition no longer distributes over multiplication (disastrous!).

In many definitions, when you get to the lowest possible value, the definition loses its literal intuitive meaning. One example is the factorial function, where 0! = 1. Factorial is often defined as the product: 1 * 2 * ... * n, but when n = 0, this definition doesn't make sense.

(Though there are other definitions that do, this is just one example).

 

[MathematicalPhysicist]

There isn't really any reason why, that's the way it's defined.

 

[Crazy Tosser]

There isn't really any reason why, that's the way it's defined.

We have the winner.

Mathematics is a human convention. That's how zero is defined in it.

 

[Gear300]

This is far too natural to be confusing. Remember that multiplication is repeated addition. If you add 100, ZERO times, you have ZERO. If you want 100 as result, then you must add 100 ONE time.

add it to what...anything?

 

[DaveC426913]

add it to what...anything?

Yes, anything.

126,
12,378,726,387,
0,
or simply n.



i.e.:

126 + 100*0 = 126
126 + 100*1 = 126 + 100

12,378,726,387 + 100*0 = 12,378,726,387
12,378,726,387 + 100*1 = 12,378,726,387 + 100

0 + 100*0 = 0
0 + 100*1 = 0 + 100

or simply

n + 100*0 = n
n + 100*1 = n + 100

 

[Gear300]

Yes, anything.

126,
12,378,726,387,
0,
or simply n.



i.e.:

126 + 100*0 = 126
126 + 100*1 = 126 + 100

12,378,726,387 + 100*0 = 12,378,726,387
12,378,726,387 + 100*1 = 12,378,726,387 + 100

0 + 100*0 = 0
0 + 100*1 = 0 + 100

or simply

n + 100*0 = n
n + 100*1 = n + 100

So...that would mean such is true for all systems?

 

[DaveC426913]

So...that would mean such is true for all systems?

I'm not prepared to say that. I am simply clearing up the possible confusion of the question "if you 'add' 100 times 0, what are you adding it to?".

 

[slider142]

So...that would mean such is true for all systems?

If by 0, you mean the element of the system such that a + 0 = a for all elements a in the system, then yes, as the property follows directly from this statement and the distributive property of multiplication over addition, and multiplicative commutativity. Hint: Consider the equation a*a= (a + 0)(a + 0).

 

[arildno]

If by 0, you mean the element of the system such that a + 0 = a for all elements a in the system, then yes, as the property follows directly from this statement and the distributive property of multiplication over addition, and multiplicative commutativity. Hint: Consider the equation a*a= (a + 0)(a + 0).

Change that to an identity..

 

[slider142]

One thing I should have added is instead of "the element", 0 should be the unique element with the aforementioned property. Then from distribution, one gets a*(a + 0) + 0*(a + 0) = a*a + a*0 + 0*a + 0*0. Since a*a = a*a + k only for the element k = 0 by uniqueness, we have a*0 + 0*a + 0*0 = 0. Note that a + 0 = 0 implies that 0*a + 0*0 = 0*a which implies 0*0 = 0 by uniqueness. The previous equation then becomes a*0 + 0*a + 0 = 0 which implies a*0 + 0*a = 0. By commutativity, a*0 = 0*a, so a*0 = 0.

 

[Overlord]

proof ox=o

0x = 0x + 0 = 0x + [ x + (-x)] = (0x + x) + (-x) = x( 0 +1) + (-x) = x +(-x) = 0

or

0x = 0 <===> 0x + x = 0+x <=====> x( 0 + 1) = 0 + x <===> x = 0 + x <===> x=x correct so 0x=0

Don't you have to already assume 0*x=0 to perform the distributive step (bold)?

 

[Hurkyl]

Don't you have to already assume 0*x=0 to perform the distributive step (bold)?

No, just that 0 + 1 = 1, and 1x = x.

 

[poutsos.A]

Don't you have to already assume 0*x=0 to perform the distributive step (bold)?

In the 1st proof No, but in the 2nd proof ,yes i do assume 0x =0 and by double implications i end with the result x = x which is true . Now we can start with x=x which is a fact and following the way backwards as shown by the double implications end with 0x =0.

Actually the double implications trick show the way to a proof

 

[tickle_monste]

If you had a bag of apples, and you put 0 apples on the table, 100 times, how many apples are on the table? 0. If you put 1 apple on the table, 100 times, there will be 100 apples on the table. Taking 1, 100 TIMES. 1 TIMES 100 = 1*100=100. Taking 0, 100 times, is equivalent to 0 times 100 = 0*100 = 0. How many apples are on the table if I put groups of 0 apples on the table 40 bazillion times?

 

[HallsofIvy]

But that is a statement about apples, not mathematics. it shows that that particular interpretation of "0 times 100 apples" is "0 apples". It does NOT show "0 * 0= 0".

 

[moon1944]

Hope it's not bad form to post to a thread that is over a year old. I also am interested in a proof of why multiplying a number by 0 gives 0. Both statdad and poutsos.A gave nice proofs in this thread that I understand clearly. However I am using the book "Introduction To Analysis" by Maxwell Rosenlicht and he uses a slightly different proof (in chap 2, 'The Real Number System'). I can't follow his final reasoning and hope someone here can help to explain it. I will outline his proof here: (in the following, the astirix (*) is the multiplication symbol)

Proof of a*0 = 0

Start by assuming 5 field axioms of real numbers as follows:
1. commutativity
2. associativity
3. distributivity
4. existence of neutral elements (definition of 1 and 0)
5. existence of additive and multiplicative inverses (definition of -a and 1/a)

He then proves several theorems, one of which I will state here because he uses it in the proof of a*0 = 0.

Theorem F3. For any a,b belonging to R the equation x+a = b has one and only one solution, namely x = b+(-a). He uses the uniqueness of this solution in his proof of a*0 = 0.

Now here is his proof of a*0 = 0 :

a*0 + a*0 = a * (0+0) = a*0 = a*0 + 0, so that a*0 and 0 are both solutions of the equation x + a*0 = a*0, hence equal, by F3.

That's it. I can follow all of his equalities but he loses me when he states that a*0 and 0 are both solutions of the equation. If they are solutions I can understand that they would be equal by F3.

Can anyone fill in the blanks between his equalities and the statement that a*0 and 0 are both solutions of the equation x + a*0 = a*0 ?

Sincere Thanks,
Bill

 

[Landau]

a*0 + a*0 = a * (0+0) = a*0 = a*0 + 0, so that a*0 and 0 are both solutions of the equation x + a*0 = a*0, hence equal, by F3.

(...) he loses me when he states that a*0 and 0 are both solutions of the equation. If they are solutions I can understand that they would be equal by F3.

The author has shown that a*0 + a*0 = a * (0+0) = a*0 = a*0 + 0. In particular:
(1) a*0 + a*0 = a*0
(2) a*0 + 0 = a*0

Consider the equation a*0 + x = a*0. By (1) we see a*0 is a solution. By (2) we see 0 is a solution.

 

[moon1944]

Beautiful...

Thank you very much, Landau.

Bill

 

[Max™]

It helps to remind yourself that zero is an additive identity, while one is a multiplicative identity.

Let's avoid the whole "but what is nothing" issue, call the additive a, and the multiplicative m.

n+a=n
n*m=n

Now, n*a can't equal n, it can't equal m either, or there is no use assigning identities in the first place.

If adding a to n, a+n, doesn't change n, then multiplying a by n, a*n, won't change a.

If multiplying n by m, n*m, doesn't change n, and adding m to n, m+n, can't produce the same result as a+n, then n+m=r, which we can call the successor to n.

m*m=m
a*m=a

m+a=m
a+a=a

When you see a zero, think "doesn't change when adding", and one is "doesn't change when multiplying".

Assigning a value is less important than the axioms you use in most cases.