algebra-precalculusinequalitynotation
1.What does $a $< $b $<c really mean?
Somebody says that it means "3 distinct numbers".
and for example they said 1$\le$a$\lt$b$\lt$c$\le$N means 3 numbers between 1 and N.
2.We want to say "3 distinct numbers";
Then Why don't we use 1$\le$a$\ne$b$\ne$c$\le$N instead of 1$\le$a$\lt$b$\lt$c$\le$N ?
3.If we want to show 3 numbers that are not necessarily distinct, It is not possible to do that except with $\le$. am I right?
Would you please explain about all of these 3 questions? Thank you.
You were probably taught that “exponentiation is repeated multiplication”:
$$b^x = \underbrace{b\times b\times b\times\cdots\times b}_{x\text{ times}}$$
From this simple definition, you can observe two properties:
For example:
We can then definite exponentation over more general sets of numbers in a way that these two properties continue to hold.
From the above rule for addition of exponents, we obtain a rule for subtraction of exponents: $b^{x - y} = {b^x \over b^y}$, because then $b^{(x - y) + y} = b^{x-y} \cdot b^y = {b^x \over b^y} \cdot b^y = b^x$ as expected. This lets us expand the domain of exponents to include zero and negative integers:
$$b^0 = b^{y-y} = {b^y \over b^y} = 1,\; b \ne 0$$ $$b^{-y} = b^{0-y} = {b^0 \over b^y} = {1 \over b^y},\; b \ne 0$$
If you assume that the multiplicate property of exponents holds for rationals, then $\left(b^{1 \over n}\right)^n = b^{{1 \over n} \cdot n} = b^1 = b$. So $b^{1 \over n}$ is a number whose $n$th power is $b$. In other words,
$$b^{1 \over n} = \sqrt[n]{b},\; b \ge 0$$
And $b^{m \over n} = \left(b^{1 \over n}\right)^m = (\sqrt[n]{b})^m$.
For example, $4096^{5/12} = \left(\sqrt[12]{4096}\right)^5 = 2^5 = 32$.
I still haven't answered your question of what $2^\pi$ means. But at this point, we can calculate $2^x$ for $x$ aribitrarily close to $\pi$.
As $x$ approaches $\pi$, $2^x$ approaches a limit, which is approximately $8.824977827076287$. For the sake of making $2^x$ continuous, we define $2^{\pi}$ to be equal to this limit.
(Note that there's nothing special about decimal fractions. I could have used the sequence $[3, {22 \over 7}, {333 \over 106}, {355 \over 113}, \ldots ]$ of best rational approximations, but that would have been less obvious.)
However, taking the trillionth root of huge powers of a number isn't very practical for calculation. A more useful method is to use logarithms.
$\log_c y$ is defined as the number $x$ such that $c^x = y$. From the two basic properties of exponentation, you can obtain the identities:
And from the latter, you get $$b^x = c^{x \cdot \log_c b}.$$ This means that if you have an exponential and logarithm function for one value of $c$, you can calculate them for any value for $b$.
Typical choice of $c$ are:
So, if you wanted to calculate $2^{\pi}$, you'd actually calculate $10^{\pi \cdot \log_{10} 2}$ or $e^{\pi \cdot \ln 2}$. And that would typically be done with the assistance of a logarithm table or a slide rule.
In Calculus, you'll learn about Taylor series, and the well-known ones for $e^x$, sine and cosine:
What happens when you plug $x = i \theta$ into the Taylor series for $e^x$?
\begin{align} e^{i \theta} & = 1 + i \theta + \frac{(i \theta)^2}{2} + \frac{(i \theta)^3}{6} + \frac{(i \theta)^4}{24} + \frac{(i \theta)^5}{120} + \frac{(i \theta)^6}{720} + \frac{(i \theta)^7}{5040} + \frac{(i \theta)^8}{40320} + \cdots \\[10pt] & = 1 + i \theta + i^2 \frac{\theta^2}{2} + i^3 \frac{\theta^3}{6} + i^4 \frac{\theta^4}{24} + i^5 \frac{\theta^5}{120} + i^6 \frac{\theta^6}{720} + i^7 \frac{\theta^7}{5040} + i^8 \frac{\theta^8}{40320} + \cdots \\[10pt] & = 1 + i \theta - \frac{\theta^2}{2} - i \frac{\theta^3}{6} + \frac{\theta^4}{24} + i \frac{\theta^5}{120} - \frac{\theta^6}{720} - i \frac{\theta^7}{5040} + \frac{\theta^8}{40320} + \cdots \\[10pt] & = \left( 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \frac{\theta^6}{720} + \frac{\theta^8}{40320} - \dots\right) + i \left(\theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \frac{\theta^7}{5040} + \cdots \right) \\[10pt] & = \cos\theta + i \sin\theta \end{align}
This is called Euler's formula, and it lets us extend exponentiation to the complex numbers:
$$e^{x+iy} = e^x \cdot e^{iy} = e^x (\cos{y} + i \sin{y})$$
So far, I've been assuming throughout that the base is a positive real number. But what if it too is complex.
Using Euler's formula above, we can take the logarithm of a (nonzero) complex number, provided that we express it in polar coordinates.
$$\log(\cos\theta + i\sin\theta) = i\theta$$ $$\log(r(\cos\theta + i\sin\theta)) = \log r + i\theta$$
And with this definition, we can define $b^x = e^{x \log b}$, just like we can with real numbers.
Except that there's one point of ambiguity: Trig functions are periodic. So it would be more accurate to say:
$$\log(r(\cos\theta + i\sin\theta)) = \log r + i(\theta + 2\pi n), n \in \mathbb{Z}$$
If you want a single “principal” value for the logarithm, then you need to constrain the angle within the interval $[0, 2\pi)$ or $[-\pi, \pi)$ or something else with a width of $2\pi$. The choice of this interval is called a “branch” of the complex logarithm.
Best Answer
A popular way to introduce numbers without any ordering is "Let $a,b,c \in (1,N)$". Or you could say "Let $a,b,c \in (1,N)$ be distinct" if you want to specify they're all different values.
The reason we can write $a = b = c$ or $a < b < c$ and it makes sense is because the binary relations $=$ and $<$ are transitive. E.g. $a < b$ and $ b < c$ implies $a < c$. The binary relation $\ne$ for example is not transitive. $1 \ne 2$ and $2 \ne 1$ does not imply that $1 \ne 1$. So statements like $a \ne b \ne c$ don't make sense.