How could I prove following exponent laws for set of real, in the given order?
1) $a^m*a^n=a^{m+n}$ CaseI a^m=a.a.a...to m factors a^n=a.a.a...to n factors a^ma^n=a.a.a...to m+n factors a^ma^n=a^(m+n) I have proved it for positive integers but I don't know how to prove it for negative integers without using 5)
But by using 5) I proved it
Let n be a negative integer and m be a positive integer and m+n>0 a^(m+n)a^-n=a^(m+n-n)=a^m [CaseI] Dividing by a^-n a^(m+n)=a^m/a^-n=a^ma^n
Let n be a negative integer and m be a positive integer and m+n<0 a^-(m+n)*a^m=a^(-m-n+m)=a^-n [Case I] Dividing by a^-(m+n)a^-n a^m/a^-n=1/a^-(m+n) a^ma^n=a^(m+n)
But I still have to prove it for cases m,n being 0, rational and real I proved for 0 by using 4) m=0 and n is a positive or negative integer L.H.S=a^0*a^n=1*a^n=a^n R.H.S=a^(0+n)=a^n L.H.S=R.H.S
n=0 and m is a positive or negative integer L.H.S=a^m*a^0=a^m*1=a^m R.H.S=a^(m+0)=a^m L.H.S=R.H.S
m=0 and n=0 L.H.S=a^0*a^0=1*1=1 R.H.S=a^(0+0)=a^0=1 L.H.S=R.H.S
So I tried to prove them without 4) and 5) but I couldn.t do it and move forward for rational and real
2) $\dfrac{a^m}{a^n}=a^{m-n}$ a^m=a.a.a...to m factors a^n=a.a.a...to n factors Let m>n a^m/a^n=(a.a.a...to m factors)/(a.a.a...to n factors)=a.a.a...to (m-n) factors=a^(m-n) For m
3) $(a^m)^n=a^{m*n}$ I proved this only for positive integers (a^m)^n=a^m.a^m.a^m...to n factrors=a^(m+m+m+...to n)=a^(m*n)
4) $a^0=1$ If I proved 1) for negative integers and 5) m=-n a^-n*a^n=a^(-n+n) a^n/a^n=a^0 a^0=1
5) $a^{-n}=\dfrac{1}{a^n}$ If I proved 1) for negative integers and 4) m=-n a^-na^n=a^(-n+n) a^-na^n=a^0=1 a^-n=1/a^n
And I don't know to do the rest for rationals and reals but proved for negative and positive integers 6) $a^{p/q}=(a^p)^{1/q}$
7) $(ab)^n=a^nb^n$
8) $(a/b)^n=\dfrac{a^n}{b^n}$
Here $a,b,m,n$ are real.
Is it necessary that $a>0, b>0$ ?
Are 4) and 5) definitions used in problems dealing with indices?
Although it is necessary to prove above over the set of real, what is meant by a term like $a^{\pi}$? Does that mean $a$ is repeated $\pi$ times?
Moreover, what is meant by $a^{-n}$ where $n$ is positive integer?
The definition of an expression $a^x$ for real numbers base $a \gt 0$ and exponent $x$ must precede any proof of laws of exponents it satisfies.
The numerous parts of this problem then require a road map, to be sure that everything is addressed. Many of the individual parts have duplicates of some kind that have been previously addressed, but it may be worth setting out a road map as Community Wiki.
We begin with the properties of real arithmetic, specifically multiplication of positive real numbers gives a unique positive real product, with real multiplication being both associative and commutative.
Then, as the poster has indicated, certain laws of exponents for positive integer exponents $x=n$ can be proved by induction, given the recursive definition that $a^1 = a$ and $a^{n+1} = a\cdot a^n$.
The definition is then extended to general integer exponents $x \in \mathbb{Z}$: $a^0 = 1$ and $a^{-n} = 1/a^n$. This extended definition involves division, so note that we are relying on $a \gt 0$ for the specific condition that $a \neq 0$ to make the definition valid. The original poster should note that cases where exponent $x$ is a negative integer are now expressed as $1/a^n$ where $n=-x$ is a positive integer.
Various laws of exponents are then to be proven for general integer exponents, using the previous properties of exponents for positive integers $x=n$ established by induction.
The next stage of extending the definition is to rational numbers $x=m/n$ as exponents. Here it is very important that base $a \gt 0$ since $a^{1/n}$ will be defined as the principal $n$th root of $a$, and this definition relies on finding unique $b \gt 0$ such that $b^n = a$, again relating back to the case of exponents $n$ that are positive integers.
The final extension of definition is the passage from rational number exponents to real number exponents. This extension is a matter of taking limits, and there is a substantial proof to be made that $a^x$ is well-defined by $\lim_{k\to \infty} a^{r_k}$ for any sequence of rational number $\{r_k\}_{k=1}^\infty$ such that $x=\lim_{k\to \infty} r_k$. That is, we need to show both that the limit exists and that we get the same result for $a^x$ whatever sequence of rational numbers converging to $x$ is used.