Proof of basic properties

Question

Can anyone provide proof of properties such as:

$$a(b+c) = ab+ac$$

$$(a+b)^2 = a^2+2ab+b^2$$

And exponent rules:

$$a^n \cdot a^m = a^{n+m}$$

$$(a^n)^m = a^{n \cdot m}$$

For $a, b, c \in \mathbb{R}$

I'm quite sure that the proof is the same or very similar for the first two and the last two, which is why I pick so 'many' examples.

Thanks.

Answer 1

The first equation is an axiom of $\mathbb{R}$

The second equation is an application of the first one and uses another axiom of $\mathbb{R}$, the commutative property : $$(a+b)^2 = (a+b)\cdot (a+b) = a\cdot a + a\cdot b + b\cdot a + b\cdot b = a^2+2ab+b^2$$

The two next proves are made for integers. We can extand these results saying $$\underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{\lfloor m\rfloor \textrm{ times}}\cdot a^{m-\lfloor m\rfloor} = \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}$$

The third one uses the commutative and associative properties : $$\underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}\cdot \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{n \textrm{ times}}= \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m+n \textrm{ times}}$$

The last one is the same of the third one : $${\underbrace{\underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}\cdot \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}\cdot \cdot\ldots \cdot\underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}}_{n \textrm{ times}}} = \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m\cdot n \textrm{ times}}$$

Answer 2

There is no such thing as a "proof" for the first equation , it's a thing (namely the disribution law) you want any ring/ vector space /module to have, so it's valid by definition of the respective structure.
The second one isn't even true for non-commutative rings.

On the third equation: Depending on how you define the exponential map, the first equation is just its the functional equation, and therefore true by definition.

Answer 3

The second one follows from the first. Let $x = (a + b)$. Then \begin{align}(a + b)^2 &= x(a + b) \\ &= xa + xb \\ &= (a+b)a + (a+b)b \\ &= a^2 + ba + ab + b^2 \\ &= a^2 + 2ab + b^2 \end{align}

where the last step also requires multiplication to be commutative.

The last two follow easily if you define $a^b$ = $\exp(b \log a)$. This is not circular because you can define $\exp$ by power series, or a number of other sensible possibilities.

Proving the first from first principles can be difficult or not depending on your starting point. Here's one way you might progress:

• Prove that $a(b + c) = ab + ac$ in the natural numbers (by induction, perhaps, or by observing that the LHS is the cardinality of the set $A \times (B \sqcup C)$, where $\sqcup$ means "tagged union", and the RHS is the cardinality of $A \times B \sqcup A \times C$, and those two sets are isomorphic)
• Prove that you can add additive inverses to all elements while extending multiplication in such a way that $a(b + c)$ is essentially what it is in the natural numbers but with some messing about with signs thrown in. Hence the property holds for integers too.
• Prove that if you throw in multiplicative inverses for all nonzero elements, you can define a sensible multiplication compatible with the integer multiplication, and such that most of the properties carry right over. So it's true of rational numbers too.
• Prove that taking limits preserves both products and sums, so that $a(b + c)$ in the real numbers can be written as a limit of a product of a sum of rational sequences, $a_n(b_n + c_n)$, each of which individually equals $a_n b_n + a_n c_n$, so whose limit must equal $ab + ac$.

Answer 4

I'm really disappointed to see everyone here has just said that the first one is an axiom. Yes, okay, formally it can be considered to be just an axiom, but there's a reason why. $\mathbb R$ is not just a game played with syntactical rules, it's intended to describe the real world.

The proof of the distributive law on $\mathbb R$ can be built up by proving it for $\mathbb N$, then for positive rationals, then for $\mathbb Q$, and finally for $\mathbb R$.

In $\mathbb N$ it's fairly obvious. If I have $a$ apples in one hand and $b$ apples in the other, how many do I have? I have $a + b$ apples. If I have $a$ boxes of $n$ apples each, and then $b$ boxes of $n$ apples each, how many boxes of $n$ apples each do I have? $a + b$ boxes. That is, $a\cdot n+b\cdot n=(a + b)\cdot n$.

The simplest way to prove it for positive fractions is to first prove the laws $\frac a b \frac c d=\frac {ab} {cd}$ and $\frac a b + \frac c d = \frac {ad + cb} {bd}$, and then just prove it algebraically.

For negative and positive fractions, that is, $\mathbb Q$, you have to just write down the definition of multiplication and addition in $\mathbb Q$ and verify every possible case ($a(b + c)$ for every possible combinatoin of signs of $a, b$ and $c$). The reason it works is essentially that the rules for negative numbers are defined in part specifically so that distributivity continues to work.

And for $\mathbb R$, the proof depends on how you construct $\mathbb R$, but the reason is always the same: it works for $\mathbb Q$, and everything about $\mathbb R$ is just the extension by continuity of $\mathbb Q$, so it's not surprising that many laws that work in $\mathbb Q$ continue to work in $\mathbb R$.

The commutative and associative properties of $\mathbb R$ can be proven in much the same way.

Answer 5

The properties for powers come from associativity of the multiplication for natural powers. To extend to integers and keep $$ a^{m + n} = a^m \cdot a^n $$ needs $$ a^{-m} = \frac{1}{a^{-m}} $$ To get rational powers, you want $$ \left( a^{m/n} \right)^n = a^m $$ i.e., $$ \left( a^{m/n} \right)^n = \sqrt[n]{a^m} $$ To extend this to real powers you want $a^m$ be continuous on both $a$ and $m$, the only definition that works turns out to be $$ a^m = \exp( m \ln a ) $$ Initially only natural powers are defined, it turns out there are natural extensions of the operation to integers, rationals, and reals that preserve important properties, and the notation is extended.

Something similar happens with the more familiar operations of sum and product.

The above certainly is a lot of reading modern notions into history. Initially it was just assumed that e.g. $a \cdot b = b \cdot a$ for reals by considering the areas of rectangles oriented different ways, the recognition of the need to define the operations rigorously and to prove that they comply to the familiar rules (and so it makes sense to use the familiar notations) is rather recent.

I'm on my tablet, so it is hard to search for previous questions, but it is illuminating to read the recurrent threads on defining division by zero here on MSE.