Unresolved,why does negative exponent turns into a fraction?

Question

As we know that, according to multiplicative identity 2 to the power of 3 means, 1*2*2*2=8;but why 2 to the power -3 is equal to the 1/8,if i think it is the reverse of multiplicative identity then from where 1 is coming? i don't want only mathematical proof,but also the intuitive story behind this? Why it is turning into a fraction and also why it positive fraction? Why minus is vanished?

Answer 1

Say $x > 0$ and $a \in \mathbb R$. For me the easiest intution is this: $$x^a x^{-a} = x^{a+(-a)} = x^0 = 1$$ but also $$\frac{x^a}{x^a} =1.$$ Setting the two equal, we see $$x^{a}x^{-a} = \frac{x^a}{x^a}$$ and then we can divide by $x^a$ on both sides to see $$x^{-a} = \frac 1 {x^a}.$$ Of course, arguements like this are essentially circular. It is more a definitional convention. Each posivite real number $x$ has a corresponding positive number which, when multiplied by $x$ gives the multiplicative identity: $1$. We denote this corresponding number by $x^{-1}$ to indicate that it inverts $x$. Then when we write $x^{-a}$ for some number $a$, by definition, we mean the inverse of $x$ raised to the $a$ power. That is $x^{-a} := (x^{-1})^a$.

Answer 2

When you are starting out with arithmetic, first you have counting, then you have addition of positive numbers, and then you have repeated addition, which is multiplication. There are obvious real world ways to give meaning to these expressions in terms of counting of objects when the operations involve positive integers. If you write $n*m=m+m+\ldots +m$ where you have added $m$ to itself $n$ times, then you can extend the formula to when $m$ is not a positive integer. But in this case, what is $m*n$ when $m$ is not a positive integer? Or when neither $m$ nor $n$ are?

At this point, these expressions don't have any meaning, and you have to decide how you want to give them meaning. What mathematicians have found is that one should look for properties that hold in the familiar case, declare that any reasonable extension will still have these properties, and use these properties to define what happens in the unfamiliar case.

So with multiplication, we have that $x(y+z)=xy+xz$, $xy=yx$, $x(yz)=(xy)z$, and $1*x=x*1=x$, and the question is, can we extend multiplication to negative/rational/real/complex numbers while maintaining these properties?

Since $0+a=a$, $xy=x(0+y)=x*0+xy$, and subtracting $xy$ from both sides, we have $x*0=0$. Thus, if we are to extend multiplication to $0$, we must have that $x*0=0$. Similarly, since $x+(-x)=0$, we can multiply $0=y(x+(-x))=yx+y(-x)$, and so $y(-x)=-yx$. So if we can extend multiplication and still have the distributive property, it forces the results to be what we (now) know them to be. It is a bit of work to make sure that all the other properties still hold.

This brings us to your particular question. What should $x^{-n}$ be? If $m,n>0$, we have $x^{m+n}=x^m x^n$, and so we ask "Can we make sense of $x^0$ and $x^{-n}$ in a way where this property continues to hold?

Since $0+n=n$, we would want $x^n=x^{0+n}=x^0x^n$, and dividing both sides by $x^n$ (assuming it is nonzero, which it is when $x\neq 0$), that gives us $x^0=1$. Similarly, $1=x^0=x^{n+(-n)}=x^nx^{-n}$, and dividing by $x^n$ we have $x^{-n}=1/x^n$. Thus, assuming our property of exponentiation continues to hold, this is how we must extend exponentiation to negative exponents. Of course, there is more work to make sure this definition actually works.

So why is it defined the way it is? Because we discovered useful properties in the case where everything was a positive number, and we wanted our extension to still have these properties. Of course, this has its limits. For example, you can't use the basic properties to make sense of $(-8)^{\pi}$ in a reasonable and unambiguous way. Like I mentioned above, after we use the properties to come up with a potential definition, we need to make sure it works, and it just doesn't for some types of exponents. But where it does work, we happily use it.