If I multiply $a$ by $0$, it means $a$ is added $0$ times, but I am not able to visualize $0$ times!
Why Multiplication of zero by any number is zero?
619 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 3 best solutions below
On
If you whant the rigorous answer here it is: consider the ring $A$, we want to prove $a \cdot 0=0$ for any $a \in A$. Since $0$ is the neuter element for the addition we have $0+0=0$, thus $a \cdot 0= a\cdot (0+0)=a \cdot 0+a \cdot 0$ by distributivity of the product, now we add $-a \cdot 0$ to both hands of the equation to get $0=a \cdot 0$. If $A$ is not commutative then we can prove $0 \cdot a$ in the exact same way.
On
Natural numbers are intimately related to finite sets. Many of the arithmetic properties of the natural numbers are reflections of simple counting involving finite sets. So, interpreting your question as one about natural numbers, let's note that give any two (finite or infinite) sets $A,B$ we can form a new set called their cartesian product: $A\times B=\{(a,b)\mid a\in A,b\in B\}$. Intuitively, this is the set consists of all possible pairs, the first from $A$, the second from $B$. Now, any finite set $A$ can be assigned a natural number $|A|$, simply the number of elements in it. It is not hard to become convinced that $|A\times B|=|A|\cdot |B|$. Then to answer the question what is $0\cdot a$ let's find sets $A$ and $B$ with $|A|=0$ and $|B|=a$. There is only one set with $|A|=0$ and that is the empty set $A=\{\}$. So, $0\cdot a=|A|\cdot |B|=|A\times B|$. But, when $A$ is empty so is the set $A\times B$, no matter what $B$ is (how can we form pairs of which the first thing must come from a set having nothing in it?). To conclude: $0\cdot a=|\{\}|=0$.
Exercise: which other rules of arithmetic can be seen to be reflections of facts about finite sets?
One argument would be to point out that, for example $$ 7\times 0 = \underbrace{0+0+0+0+0+0+0}_{7\text{ times}} = 0 $$ and since multiplication of positive numbers satisfy $a\times b=b\times a$ we want the same rule to hold for zero.
A different argument is that we want the distributive law to keep working when we introduce zero: $$ 1 \times 7 = (1+0)\times 7 = (1\times 7)+(0\times 7) $$ which is only possible if $0\times 7$ is $0$.