Does a number get destroyed when it is multiplied by zero?

Question

We consider following subtraction- $$3-1=2$$ If we subtract 1 from 3 than result is 2. in This operation 1 is not destroyed but only replaced from 3 now we consider following equation- $$1\times 0=0$$ in this multiplication, what happened with 1? where 1 is replaced? does it mean 1 is destroyed?

If we say that 1 is changed into 0 after multiplication by zero than what is the meaning of changed into zero in mathematics? Is the number destroyed or replaced anywhere?

Answer 1

Numbers aren't physical so it's almost weird to say they are "destroyed" or "replaced".

It's easier to interpret it the usual way ...

3 - 1 = One less than 3 = 3 less 1 = 3 remove 1 = 2.

-1x0 = 0 times (-1) = "None" of -1 = "None" = 0. In the case here, -1 didn't disappear, it's just that you didn't have any in the first place.

Try it with 2x0 ...

2x0 = two zeros added together = zero twos = 0.

Answer 2

This is a philosophical question. It is related to the fact that $1/0$ if undefined in mathematics. There are several approaches to define divition by zero... One of them is to define every number as set with one element the number itself. And define zero as the set of all numbers. Multiplication then defined as the set of all products $ab$ where $a$ and $b$ are element of the multiplied sets... Therefore multiplication of any number by zero which group of all numbers will be group of all number which is a zero... So is the number 1 is destroyed? We can say it is not because it is embedded inside the zero... It's always been there you just didn't know....

Answer 3

If you say $O$ is zero and $S(x)$ is a function that gives you the successor of $x$, then $S(O)$ is one, $S(S(O))$ is two, etc.

Now, if you want to compute the predecessor of $x$, say $P(x)$, then, you can do it by creating a list which consumes the $S$ in an expression like $S(S(S(....)))$, until you have $S(O)$.

For example 3-1: You start with three: $S(S(S(O)))$, and you build this list: $$\begin{pmatrix} S(S(S(O))) & S(S(O)) \\ O & S(O)\end{pmatrix}$$ You have $S(O)$ on the right lower part of the list, so you have your answer, two, in the right upper part of the list.

Now, with multiplication, the idea is that the algorithm behind $a \times b$ when $a,b$ are natural numbers: $a,b \in \mathbb N$ is the following:

$a$ and $b$ are of the form $S(S(S(...)))$. For example $a=3=S(S(S(O)))$ and $b=2=S(S(O))$. If you apply $b$ to $a$ by replacing all $S$ in $a$ by the $S$ in $B$, you get: $a \times b=S(S(S(S(S(S(O))))))=6$

Now, if $b=0=O$, you do the same thing, you replace all $S$ in $a$ by $b$ and you obtain $a \times O = O$.

Answer 4

Doing arithmetic operations does not change numbers, but just produces new numbers as a result. If you compute $1+1=2$ then you did not change $1$ into $2$, you just produced the value$~2$ by adding $1$ and $1$ together. If one did assume that one of those $1$s were changed into$~2$, all kinds of strange questions would arise. Which one was changed, or did they both change? And what would $(1+1)+1$ be; if the final $1$ has in fact been changed into $2$, then the result should be $4$ rather than $3$. But this is not how mathematics works; whenever we write a the same number we mean the same value, so numbers themselves never change, and always represent the same value.

As for multiplication, it just happens that the value of $a\times b$ is always $0$ when $a=0$, regardless of what $b$ was, just like it always is $0$ if $b=0$, regardless of what $a$ was. The result being independent of one of the values, provide one knows the other to have a specific value, does not happen with addition or subtraction. However it really is not a big deal that it does sometimes happens with multiplication. Nothing gets destroyed, just ignored in some cases.