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A naïve definition of exponentiation is that, it is just a 'repeated multiplication'. In this sense, $b$ in $a{\wedge}b$ should be an integer. Then we introduced
${\displaystyle \exp(x):=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}=1+x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{6}}+{\frac {x^{4}}{24}}+\cdots }$ and ${\displaystyle \ln(1+x)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}x^{k}}{k}}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots }$
With the help of these, we can extend the above $b$ to real numbers as $a{\wedge}b:=exp(b{\cdot}ln(a))$

On a similar note, a naïve definition of multiplication is that, it is a 'repeated addition'. In this sense, $b$ in $a*b$ should be an integer. Now, with only the knowledge of addition & subtraction of two real numbers and multiplication of a real number with an integer, how can I arrive at multiplication of two real numbers?

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    $\begingroup$ You can't. Multiplication of reals is not repeated addition, and exponentiation of reals is not repeated multiplication. These simple notions only apply when working with simple numbers like integers. Real numbers are part of analysis, not elementary algebra, which is why things are much more complicated. $\endgroup$
    – peek-a-boo
    Mar 16, 2022 at 10:42
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    $\begingroup$ You can first define multiplication of rational numbers by $\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}$. The function $f:\mathbb R^2\to\mathbb R$ given by $f(x,y)=x\cdot y$ is the unique continuous function which agrees with the definition of multiplication of rationals. $\endgroup$
    – Joe
    Mar 16, 2022 at 11:07
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    $\begingroup$ Here is an equivalent formulation of the above definition. Let $x,y\in\mathbb R$, and let $(x_n)$ and $(y_n)$ be any two sequences of rational numbers such that $x_n\to x$ and $y_n\to y$. It can be proven that regardless of which sequences we choose, the sequence $(x_n\cdot y_n)$ has the same limit $l$. We can define $x\cdot y$ as $l$. $\endgroup$
    – Joe
    Mar 16, 2022 at 11:12
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    $\begingroup$ What is your definition of a real number? $\endgroup$
    – Somos
    Mar 16, 2022 at 11:13
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    $\begingroup$ It's quite confusing to write $a^b$ as a wedge product $a \wedge b$... $\endgroup$ Mar 16, 2022 at 11:24

1 Answer 1

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As Joe has said in the comments, $\Bbb R$ is defined to be a complete ordered field. And you can prove that any two complete ordered fields $R_1$ and $R_2$ are isomorphic: There is a one-to-one mapping between all the elements of $R_1$ and all elements of $R_2$ which preserves addition, multiplication, and the ordering. Effectively they are the same ordered field, but with different elements playing the parts. So up to isomorphism, the real numbers $\Bbb R$ are the only complete ordered field.

But neither this definition nor the isomorphism theorem prove that there actually is a complete ordered field. To show that such a field exists, there are several possible constructions of sets and operators that meet all the conditions of a complete ordered field. They generally follow variations of this outline (with the individual sets sometimes obtained from their predecessors by significantly different processes):

  1. For any set $n$, define $s(n) := n \cup \{ n \}$.
  2. There is a unique set $\Bbb N$ with the properties
    • $\emptyset = \{\} \in \Bbb N$
    • for all $n \in \Bbb N, s(n) \in \Bbb N$
    • if $A \subseteq \Bbb N$, $\emptyset \in A$ and for all $n \in A, s(n) \in A$, then $A = \Bbb N$. (The principle of Induction).
  3. Denote $0 := \emptyset, 1 := s(0), 2 := s(1)$, etc.
  4. Define addition inductively by $n + 0 = n, n + s(m) = s(n + m)$.
  5. Define multiplication inductively by $0 \times n = 0, s(m) \times n = (m \times n) + n$.
  6. Show that the operations are cummutative, associative, obey the distributive law, and have $0$ and $1$ as indentities, respectively.
  7. In $\Bbb N^2$, define $(a,b) \sim (c,d)$ if $a + c = b + d$. This is an equivalence relation. Define the integers $\Bbb Z$ as $\Bbb N^2/\sim$, the set of equivalence classes of $\sim$.
  8. Show that addition and multiplication on $\Bbb N$ preserve equivalence under $\sim$. Use this to define them on $\Bbb Z$. Show they have the same properties as on $\Bbb N$, but also $\Bbb Z$ has additive inverses.
  9. Identify each $n \in \Bbb N$ with the equivalence class $[(n,0)] \in \Bbb Z$. Use this to treat $\Bbb N \subset \Bbb Z$.
  10. On $\Bbb Z\times \Bbb Z\setminus \{0\}$, define $(a,b) \simeq (c,d)$ if $ad = bc$. This is also an equivalence relation. Define $\Bbb Q = (\Bbb Z\times \Bbb Z\setminus \{0\})/\simeq$.
  11. Show that addition and multiplication on $\Bbb Z$ preserve equivalence under $\simeq$. Use this to define them on $\Bbb Q$. Show they have all the same properties, but also $\Bbb Q$ has multiplicative inverses for all elements but $0$.
  12. Identify each $z \in \Bbb Z$ with the equivalence class $[(z,1)] \in \Bbb Q$, thus considering $\Bbb N \subset \Bbb Z \subset \Bbb Q$.
  13. Define $\Bbb Q_{\ge 0} = \{[(n,m)] \in \Bbb Q\mid n,m \in \Bbb N, m \ne 0\}$. For all $r,s \in \Bbb Q$, define $r \le s$ if $s + (-r) \in \Bbb Q_{\ge 0}$. Show that $\le$ is a total ordering of $\Bbb Q$ preserved by addition, and by multiplication by members of $\Bbb Q_{\ge 0}$, and reversed by the taking of additive or multiplicative inverses.
  14. Define a set $C \subset \Bbb Q$ to be a "cut" if $C \ne \emptyset; C \ne \Bbb Q$; for all $r,s \in \Bbb Q, r \le s$ and $s \in C \implies r \in C$; and for all $r \in C$, there is some $s \in C$ with $r < s$ (i.e., $C$ has no maximum element). For any $D\subset \Bbb Q$, define $\overline D = \{ r \in \Bbb Q \mid \exists s \in D, r < s\}$ (the strict inequality means that if $D$ has a maximum element, it will not be in $\overline D$). If $D$ is non-empty and bounded above, then $\overline D$ is a cut.
  15. Define $\Bbb R$ to the collection of all cuts of $\Bbb Q$. Identify each $r \in \Bbb Q$ with $\overline{\{r\}} \in \Bbb R$, so we can consider $\Bbb Q \subset \Bbb R$.
  16. For $x,y \in \Bbb R$, define $x + y = \{r + s\mid r \in x, s \in y\}, -x = \overline{\{-r\mid r \in \Bbb Q\text{ and } r \notin x\}}$
  17. Define $x \le y$ if $x \subset y$.
  18. if $x > 0$, define $xy = \overline{\{rs \mid r \in x, s \in y \text{ and } r\ge 0\}}$. If $x \le 0$, define $xy = -((-x)y)$.
  19. Show that $\Bbb R$ has all the properties of $\Bbb Q$ (those of an ordered field), plus the supremum property: If $A$ is a non-empty set that is bounded above, then $A$ has a least upper bound or "supremum" $y$. That is, for all $a \in A, a \le y$, and for all $x$ such that for all $a \in A, a \le x$, whe have $y \le x$. This is what it means for $\Bbb R$ to be a complete ordered field.

This proves that a complete ordered field exists. But once you've built all of that, you throw it away and instead define $\Bbb R$ to be some arbitrary complete ordered field, and all the others as the appropriate subsets of $\Bbb R$. Why? Because we do not want all the stuff that comes with this construction. You have multiple interpretations of $\Bbb N, \Bbb Z, \Bbb Q$. You have other variations of how to built these sets. You have numbers, which we want to think of as atomic objects, instead having some complicated internal structure that doesn't add anything useful to the mix. We don't need all of that. So we just consider $\Bbb R$ to be a thing in and of itself, without any of the baggage.


Added:

After all that, I lost sight of the goal: exponentiation. It is also definable inductively: For any $x \in \Bbb R$:

  • Define $x^0 = 1$. (Some people insist you make an exception to this for $x = 0$, but their reasoning amounts to "if it can't be continuous, it shouldn't be defined at all", which I find less than compelling. $0^0 = 1$ is a far more useful definition choice than any other, including no definition.)
  • For $n \in \Bbb Z, n > 0$, define inductively $x^n = x(x^{n-1})$.
  • For $n \in \Bbb Z, n < 0$, define $x^n = \frac 1{x^{-n}}$. Note that this satisfies $x^n = \frac 1x(x^{n+1})$.
  • If $y > x \ge 0, n > 0$, then $y^n > x^n$. From this it can be shown that if $a > 0$, the set $\{x \ge 0 \mid x^n < a\}$ has a supremum $y$. A little estimation shows that it must be that $y^n = a$. And because $x^n$ is increasing, $y$ is the unique solution to this equation. Define $\sqrt[n]a = y$.
  • For all rational numbers $r$, let $r = \frac mn$ with $n > 0$. For $x > 0$, define $x^r = (\sqrt[n]x)^m$. This turns out to be independent of the exact representation $\frac mn$ chosen for $r$, so $x^r$ is well defined.
  • If $r > s$ and $x > 1$, then $x^r > x^s$. For any real number $a$, the set $$\{y > 0 \mid \text{there exists }r \in \Bbb Q\text{ with }r < a\text{ and }y = x^r\}$$ has a supremum $z$. Define $x^a = z$.
  • Define $1^a = 1$ for all real $a$, and if $0 < x < 1$ then $x^a = \dfrac 1{(1/x)^a}$. Finally if $a > 0$, define $0^a = 0$.

This defines $x^y$ on the set $(x,y) \in \Bbb R^2\mid x > 0 \text{ or }(x = 0\text{ and } y\ge 0)\}$ and it is continuous everywhere except at $(0,0)$. This definition is done without reference to power series. It follows naturally from the concept of exponentiation as repeated multiplication. And the only part of it that could be considered as limits is the taking of supremums, which is just a fundamental property in the definition of real numbers.

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    $\begingroup$ You lost sight of the goal twice: the OP asks how to define multiplication, not exponentiation. $\endgroup$
    – TonyK
    Mar 18, 2022 at 12:01
  • $\begingroup$ @TonyK - that was already covered. And I had intended from the start to show how both multiplication and exponentiation can be developed rigorously in a more natural fashion, arising directly from their base concepts of repeated addition and multiplication. $\endgroup$ Mar 18, 2022 at 12:14
  • $\begingroup$ Perhaps you should read the question again. $\endgroup$
    – TonyK
    Mar 18, 2022 at 13:08
  • $\begingroup$ @TonyK - perhaps you should read my answer again, which not only describes why real multiplication is "defined" by axiom, but also exactly how it can be constructed from the concept of repeated addition when building a model for the real numbers. Because I wanted to also describe how exponentiation can similarly be defined from the concept of repeated multiplication instead by Taylor series, thereby addressing what appears to be a misconception of the poster, does not somehow make my discussion of multiplication itself disappear. $\endgroup$ Mar 18, 2022 at 13:12
  • $\begingroup$ This is a two-part answer. The key statement in the first part is, "This proves that a complete ordered field exists." Perhaps it's worth reminding the reader at that point about some of the things that this entails, in particular, that one of the things we had to do in this proof was to define multiplication on the field. The definition is there, but you might miss it if you skim over too many of the details of the field construction. $\endgroup$
    – David K
    Mar 18, 2022 at 13:18

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