Proving basic math principles?

Question

I've been thinking how to prove some of these basic "formulas" but most of them I don't know how: $$\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}$$ exponentials such as $$(a^b)\cdot (a^c)=a^{b+c}$$ (I know this makes sense for integers but how do you prove it for all numbers?), and how you can multiply any numbers in any order and get the same result (again this makes sense for 2 or maybe 3 numbers, but how do you know it holds true for any amount). Thanks

Answer 1

As with all "basic facts" in math, we take certain basic assumptions to be axioms. Definitions are similar to axioms in that vein, in that they are not theorems but definitions which make explicit what certain basic things mean. Without basic axioms and primitive definitions, we'd have no building blocks from which to construct theorems, or to prove more complex operations, functions, sets, structures, etc.

Multiplication of two rational numbers $x = \dfrac ab, \; y = \dfrac cd$ is defined by $$xy = \dfrac ab \cdot \dfrac cd = \dfrac {a\cdot c}{b \cdot d}.$$

Of course, we'd want to appeal to our definition of rational numbers, which requires an appeal to the definition of integers. We also need to be clear about what we mean by "multiplying two numbers"

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Similarly, we usually define, for real numbers $a, b, c$, (another definition required) $$\large \;(a^b)\cdot (a^c) = (a)^{b+c}.\;$$ This can be depicted for integers a, b, c, by:

$$\large a^b \cdot a^c \; = \;\underbrace{a \cdot a \cdot a \cdot}_{\large b \;factors} \cdots \underbrace{\cdot a \cdot a \cdot a}_{\large c\; factors}\quad = \quad \underbrace{a \cdot a\cdot \cdots \cdot a \cdot a}_{\large (b+c) \;factors} = a^{b+c}$$

And similarly, $\large\;\;(a^b)^c = a^{b\cdot c}$, which depicted with integers $a, b, c$:

$$ \large (a^b)^c = \underbrace{a^b\cdot a^b \cdot \cdots \cdot a^b \cdot a^b}_{\large c\; factors} = \underbrace{\underbrace{a \cdot a \cdots \cdot a}_{\large b\; factors}\cdot \underbrace{ a \cdot a \cdots a}_{\large b \;factors}\cdots\cdots \underbrace{ a \cdot a \cdots a}_{b \;factors}}_{\large c \;factors} $$ $$=\large\underbrace{a \cdot a\cdot a \cdots \cdots a \cdot a}_{\large b c\;factors}\quad = \quad a^{bc}$$

Answer 2

Let's just prove these statements for rational numbers, because proving them for real numbers would require a formal definition of real numbers (which can get rather technical.)

The statement $(a/b)(c/d) = ab/cd$ is usually taken as the definition of multiplication for rational numbers, so there is nothing to prove.

If you accept that the identity $a^ma^n = a^{m+n}$ is true for natural numbers, then following the usual definition of exponentiation for rational bases as $(a/b)^m = a^m/b^m$ where $a$, $b$, and $m$ are integers, and the usual definition of multiplication of rational numbers as above, you can prove the desired identity for rational bases as follows: $$(a/b)^m(a/b)^n = (a^m/b^m)(a^n/b^n) = (a^ma^n/b^mb^n) = a^{m+n}/b^{b+n} = (a/b)^{m+n}$$

Proving it for rational exponents is a bit trickier, but perhaps you get the idea. It's important to notice that we are not proving formulas for scratch, but rather from simpler formulas that we already know. At some point the process will have to stop and we will just have to accept a few simple statements as axioms.

If you accept that multiplication is associative and commutative, that is, $(xy)z = x(yz)$ and $xy = yx$, then you can prove that the product of $n$ numbers $x_1,\ldots,x_n$ does not depend on their order. For example, you can apply commutativity to switch the $x_1$ and the $x_2$, and then again to switch the $x_2$ and the $x_3$, and so on to get whatever rearrangement you want (by not writing parentheses we are implicitly using associativity.) Formally this is a proof by induction: if we can make $k$ switches without changing the product then we can also make $k+1$ switches without changing the producct, so "by induction" we can make any number of switches.