Let \( n\in \mathbb{N}\) be natural number and \( x \in \mathbb{R}\) a real number.
We call \(x^n\) a number \(x\) multiplied \(n\) times by itself:
$$ x^n = \hspace{0.2em} \underbrace {x \times x \times x \times x \times x ...} _\text{ \(n\) factors } $$
In a general way, we define:
$$\forall (x, n) \in \mathcal{D}(x^n), \ f(x) = x^n $$
$$ with \enspace \mathcal{D}(x^n) = \Biggl[(x = 0) \land (n \in \mathbb{R}) \Biggr] \lor \underbrace{ \overbrace { \Biggl[ \forall x \in \hspace{0.05em} \mathbb{R}^*, \enspace \forall n \in \frac{\mathbb{Z}}{2\mathbb{N} + 1 } \Biggr] } ^\text{even and odd functions} \lor \overbrace { \Biggl[ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace \forall n \in \frac{\mathbb{Z}}{2\mathbb{N}^*} \Biggr] } ^\text{ square root } } _\text{\(n \ \in \ \mathbb{Q}\) under an irreducible form} \lor \underbrace{ \Biggl[ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace \forall n \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \mathbb{Q} \Bigr] \Biggr] } _\text{irrational numbers}$$
In all demonstrations, the case where \((x=0)\) had been set aside:
for convenience reasons;
to avoid to indeterminate form \(0^0\).
Since the natural logarithm does not accept negative values, we will use a sign generator to compensate. Therefore, we will set down:
$$ \forall x^n \in \hspace{0.05em} \mathcal{D}(x^n), $$
$$ x^n = \left( \frac{x}{|x|}\right)^{n} \ e^{ln|x^{n}|} \qquad (1) $$
$$ \forall (x^a,y^a) \in \hspace{0.05em} \mathcal{D}(x^n)^2, $$
$$ x^a y ^a = (xy)^a $$
$$ \forall (x^a,y^a) \in \hspace{0.05em} \mathcal{D}(x^n)^2, \ y \neq 0, $$
$$ \frac{x^a}{y^a} = \left( \frac{x}{y} \right)^a $$
$$ \forall (x^a, x^b) \in \mathcal{D}(x^n)^2, $$
$$ x^a x^b = x^{a+b} $$
$$ \forall (x^a, x^b) \in \mathcal{D}(x^n)^2, \ x \neq 0, $$
$$ \frac{x^a}{x^b} = x^{a-b} $$
$$ \forall x^a \in \mathcal{D}(x^n), \ \forall b \in \mathbb{R}, $$
$$ (x^a)^b = (x^b)^a = x^{ab} $$
A number raised to the power of zero
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, $$
$$ x^0 = 1 $$
$$ \forall x^a \in \mathcal{D}(x^n), $$
$$\frac{1}{x^a} = x^{-a}$$
$$ \forall x \in \hspace{0.05em} \mathbb{R}^+, $$
$$ x^{\frac{1}{2}}= \sqrt{x} $$
$$ \forall a \in \hspace{0.05em} \mathbb{Z}^*, \ \Bigl[ \forall x \in \hspace{0.05em} \mathbb{R}_+, \ n \in \{2a\} \Bigr] \lor \Bigl[ \forall x \in \hspace{0.05em} \mathbb{R}, \ n \in \{2a + 1\} \Bigr], $$
$$ x^{\frac{1}{n}}= \sqrt[n]{x} $$
Recap table of the formulas of the powers of x
Click on the title above to access the recap table.
Let \( (x, y) \in \hspace{0.05em} \bigl[ \mathbb{R}^* \bigr]^2 \) two non-zero real numbers and \( a\in \hspace{0.05em} \mathbb{N} \) a natural number.
$$ (xy)^a = \hspace{0.2em} \underbrace {xy \times xy \times xy \times xy \times xy ...} _\text{ \(a\) factors } $$
The product of two number being commutative:
$$ (xy)^a = \hspace{0.2em} \underbrace {x \times x \times x\times x \times x...} _\text{ \(a\) factors } \ \times \ \underbrace {y \times y \times y \times y \times y ...} _\text{ \(a\) factors } $$
Now, considering the fundamental definition of a power of \(x\):
$$ x^n = \hspace{0.2em} \underbrace {x \times x \times x \times x \times x ...} _\text{ \(n\) factors } $$
We then have,
$$ (xy)^a = x^a \times y^a $$
And as result,
$$ \forall (x,y) \in \hspace{0.05em} \bigl[ \mathbb{R}^* \bigr]^2, \ \forall a \in \mathbb{N}, $$
$$ x^a y ^a = (xy)^a $$
The division of a number \(a\) by a number \(b\) being just the multiplication of \(a\) by the inverse of \(b\), the previous formulas also work well for exponents in \( \mathbb{Z}\).
Let us get the expression \((1)\) back:
$$ \forall x^n \in \hspace{0.05em} \mathcal{D}(x^n), $$
$$ x^n = \left( \frac{x}{|x|}\right)^{n} \ e^{ln|x^{n}|} \qquad (1) $$
Let \( (x, y) \in \hspace{0.05em} \bigl[ \mathbb{R}^* \bigr]^2 \) two non-zero real numbers and \( a\in \hspace{0.05em} \mathbb{R} \) areal number. So,
$$ \forall (x^a, y^a) \in \hspace{0.05em} \mathcal{D}(x^n)^2, $$
$$ x^a y^a = \left( \frac{x}{|x|}\right)^{a} e^{ln|x^{a}|} \ \left( \frac{y}{|y|}\right)^{a} e^{ln|y^{a}|} $$
$$ x^a y^a = \left( \frac{x}{|x|}\right)^{a} \left( \frac{y}{|y|}\right)^{a} \ e^{ln|x^{a}|} \ e^{ln|y^{a}|} $$
Thanks to the definition of the definition set \(\mathcal{D}(x^n)\), we are sure that the sign generators are well defined. Since they can take as value either \(1\), or \(-1\), let us make a table to consider all these cases, a bit like a truth table:
$$\frac{x}{|x|}$$ |
$$\frac{y}{|y|}$$ |
$$\left(\frac{x}{|x|}\right)^{a}$$ |
$$\left(\frac{y}{|y|}\right)^{a}$$ |
$$ \left(\frac{x}{|x|}\right)^{a} \left(\frac{y}{|y|}\right)^{a}$$ |
$$ \frac{x}{|x|} \frac{y}{|y|}$$ |
$$ \left(\frac{x}{|x|} \frac{y}{|y|}\right)^{a}$$ |
|---|---|---|---|---|---|---|
$$-1$$ |
$$-1$$ |
$$(-1)^a$$ |
$$(-1)^a$$ |
$$((-1)^a)^2 = 1$$ |
$$1$$ |
$$1$$ |
$$-1$$ |
$$1$$ |
$$(-1)^a$$ |
$$1$$ |
$$(-1)^a$$ |
$$-1$$ |
$$(-1)^a$$ |
$$1$$ |
$$-1$$ |
$$1$$ |
$$(-1)^a$$ |
$$(-1)^a$$ |
$$-1$$ |
$$(-1)^a$$ |
$$1$$ |
$$1$$ |
$$1$$ |
$$1$$ |
$$1$$ |
$$1$$ |
$$1$$ |
This table shows that:
$$\forall x^n \in \mathcal{D}(x^n), \ \left(\frac{x}{|x|}\right)^{a} \left(\frac{y}{|y|}\right)^{a} = \left( \frac{x}{|x|} \frac{y}{|y|} \right)^{a} $$
So,
$$ x^a y^a = \left( \frac{x}{|x|} \frac{y}{|y|} \right)^{a} \ e^{ln|x^{a}|} \ e^{ln|y^{a}|} $$
The product of absolute values being the absolute value of the product, we do have this:
$$ x^a y^a = \left( \frac{xy}{|xy|} \right)^{a} \ e^{ln|x^{a}|} \ e^{ln|y^{a}|} $$
Now, using the formulas of logarithm and exponential functions:
$$ x^a y^a = \left( \frac{xy}{|xy|} \right)^{a} \ e^{a \ ln|x| + a \ ln|y|} $$
$$ x^a y^a = \left( \frac{xy}{|xy|} \right)^{a} \ e^{a \ ln|xy| } $$
$$ x^a y^a = \left( \frac{xy}{|xy|} \right)^{a} \ \ e^{ln|xy|^a} $$
We definitely find again the same form as the expression \((1)\). So:
$$ \forall (x^a,y^a) \in \hspace{0.05em} \mathcal{D}(x^n)^2, $$
$$ x^a y^a = (xy)^a $$
We showed that the formula works in the natural numbers set \( (\mathbb{N}) \), in the integers set \( (\mathbb{Z}) \), but also in the real numbers set \( (\mathbb{R}) \).
Therefore in any case,
$$\forall (x^a,y^a) \in \hspace{0.05em} \mathcal{D}(x^n)^2, $$
$$ x^a y^a = (xy)^a $$
The division of a number \(a\) by a number \(b\) being just the multiplication of \(a\) by the inverse of \(b\), the previous formulas also work well with quotients.
$$\forall (x^a,y^a) \in \hspace{0.05em} \mathcal{D}(x^n)^2, \ y \neq 0, $$
$$ \frac{x^a}{y^a} = \left(\frac{x}{y}\right)^a $$
Let \( x\in \hspace{0.05em} \mathbb{R}^* \) a non-zero real number and \( (a, b) \in \hspace{0.05em} \mathbb{N}^2 \) two natural numbers.
Performing the product \(x^a x^b \), we do have:
$$ x^a x^b = \hspace{0.2em} \underbrace {x \times x \times x \times x \times x ...} _\text{ \(a\) factors } \ \times \ \hspace{0.2em} \underbrace {x \times x \times x \times x \times x ...} _\text{ \(b\) factors }$$
$$ x^a x^b = \hspace{0.2em} \underbrace {x \times x \times x \times x \times x ...} _\text{ \((a + b)\) factors } $$
Now, considering the fundamental definition of a power of \(x\):
$$ x^n = \hspace{0.2em} \underbrace {x \times x \times x \times x \times x ...} _\text{ \(n\) factors } $$
We finally obtain,
$$ \forall x \in\hspace{0.05em} \mathbb{R}^*, \enspace \forall ( a, b) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ x^a x^b = x^{a+b} $$
The division of a number \(a\) by a number \(b\) being just the multiplication of \(a\) by the inverse of \(b\), the previous formulas also work well for exponents in \( \mathbb{Z}\).
Retrieving our expression \((1)\) defined earlier,we do have this:
$$ \forall x^n \in \hspace{0.05em} \mathcal{D}(x^n), $$
$$ x^n = \left( \frac{x}{|x|}\right)^{n} \ e^{ln|x^{n}|} \qquad (1) $$
So,
$$ \forall (x^a, x^b) \in \hspace{0.05em} \mathcal{D}(x^n)^2, $$
$$ x^a \ x^b= \left( \frac{x}{|x|}\right)^{a} \ e^{ln|x^{a}|} \ \left( \frac{x}{|x|}\right)^{b} \ e^{ln|x^{b}|} $$
$$ x^a \ x^b= \left( \frac{x}{|x|}\right)^{a} \left( \frac{x}{|x|}\right)^{b} \ e^{ln|x^{a}|} \ \ e^{ln|x^{b}|} $$
In the same way as before, let us make a truth table to observe the behaviour of our sign generators.
$$ \frac{x}{|x|}$$ |
$$x^a$$ |
$$x^b$$ |
$$\left(\frac{x}{|x|}\right)^{a}$$ |
$$\left(\frac{x}{|x|}\right)^{b}$$ |
$$ \left(\frac{x}{|x|}\right)^{a} \left(\frac{x}{|x|}\right)^{b}$$ |
$$x^{a+b}$$ |
$$ \left(\frac{x}{|x|}\right)^{a + b}$$ |
|---|---|---|---|---|---|---|---|
$$-1$$ |
$$odd \ function$$ |
$$odd \ function$$ |
$$-1$$ |
$$-1$$ |
$$ 1$$ |
$$even \ function$$ |
$$1$$ |
$$-1$$ |
$$odd \ function$$ |
$$even \ function$$ |
$$-1$$ |
$$1$$ |
$$ -1$$ |
$$odd \ function$$ |
$$-1$$ |
$$-1$$ |
$$even \ function$$ |
$$odd \ function$$ |
$$1$$ |
$$-1$$ |
$$ -1$$ |
$$odd \ function$$ |
$$-1$$ |
$$1$$ |
$$$$ |
$$$$ |
$$1$$ |
$$1$$ |
$$1$$ |
$$$$ |
$$1$$ |
In the first three cases, we have made two cases for even and odd functions because:
$$ \forall x^n \in \mathcal{D}(x^n), \ \frac{x}{|x|} = -1 \ \Longrightarrow \ \underbrace{\biggl[ \forall x \in \hspace{0.05em} \mathbb{R}_-^* , \enspace \forall n \in \frac{\mathbb{Z}}{2\mathbb{N} + 1} \biggr] } _\text{\( x \ < \ 0 \) and (even and odd functions)} $$
$$x^a$$ |
$$x^b$$ |
$$a$$ |
$$b$$ |
$$a+b$$ |
$$x^{a+b}$$ |
|---|---|---|---|---|---|
$$odd \ function$$ |
$$odd \ function$$ |
$$a \in \frac{2\mathbb{Z} + 1}{2\mathbb{N} + 1} $$ |
$$b \in \frac{2\mathbb{Z'} + 1}{2\mathbb{N'} + 1} $$ |
$$(a+b) \in \left[ \frac{2\mathbb{Z} + 1}{2\mathbb{N} + 1} + \frac{2\mathbb{Z'} + 1}{2\mathbb{N'} + 1} \right] = \left[ \frac{(2\mathbb{Z} + 1)(2\mathbb{N'} + 1) + (2\mathbb{Z'} + 1)(2\mathbb{N} + 1)}{(2\mathbb{N} + 1)(2\mathbb{N'} + 1)} \right] = \left[ \frac{4\mathbb{Z}\mathbb{N'} + 2\mathbb{Z} + 2\mathbb{N'} + 1 + 4\mathbb{Z'}\mathbb{N} + 2\mathbb{Z'} + 2\mathbb{N} + 1 }{4\mathbb{N}\mathbb{N'} + 2\mathbb{N} + 2\mathbb{N'} + 1} \right] = \frac{2\mathbb{Z''}}{2\mathbb{N''} + 1} $$ |
$$even \ function$$ |
$$odd \ function$$ |
$$even \ function$$ |
$$a \in \frac{2\mathbb{Z} + 1}{2\mathbb{N} + 1} $$ |
$$b \in \frac{2\mathbb{Z'}}{2\mathbb{N'} + 1} $$ |
$$(a+b) \in \left[ \frac{2\mathbb{Z} + 1}{2\mathbb{N} + 1} + \frac{2\mathbb{Z'}}{2\mathbb{N'} + 1} \right] = \left[ \frac{(2\mathbb{Z} + 1)(2\mathbb{N'} + 1) + (2\mathbb{Z'})(2\mathbb{N} + 1)}{(2\mathbb{N} + 1)(2\mathbb{N'} + 1)} \right] = \left[ \frac{4\mathbb{Z}\mathbb{N'} + 2\mathbb{Z} + 2\mathbb{N'} + 1 + 4\mathbb{Z'}\mathbb{N} + 2\mathbb{Z'} }{4\mathbb{N}\mathbb{N'} + 2\mathbb{N} + 2\mathbb{N'} + 1} \right] = \frac{2\mathbb{Z''} + 1}{2\mathbb{N''} + 1} $$ |
$$odd \ function$$ |
$$even \ function$$ |
$$odd \ function$$ |
$$a \in \frac{2\mathbb{Z}}{2\mathbb{N} + 1} $$ |
$$b \in \frac{2\mathbb{Z'} + 1}{2\mathbb{N'} + 1} $$ |
$$Idem$$ |
$$odd \ function$$ |
In the fourth and last case, we only have positive functions:
$$ \forall x^n \in \mathcal{D}(x^n), \ \frac{x}{|x|} = 1 \ \Longrightarrow \ \underbrace{ \Biggl[ x \in \hspace{0.05em} \mathbb{R}^*_+, \ n \in \frac{\mathbb{Z}}{2\mathbb{N} + 1} \Biggr] } _\text{\((x \ > \ 0) \) et (even and odd functions)} \lor \ \underbrace{ \Biggl[ x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace n \in \frac{\mathbb{Z}}{2\mathbb{N}^*} \Biggr] } _\text{square roots} \ \lor \ \underbrace{ \Biggl[ x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace n \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \mathbb{Q} \Bigr] \Biggr] } _\text{real numbers} $$
Having studied all cases possible, we definitely showed that:
$$ \forall x^n \in \mathcal{D}(x^n), \ \left(\frac{x}{|x|}\right)^{a} \left(\frac{x}{|x|}\right)^{b} = \left(\frac{x}{|x|}\right)^{a+b} $$
So,
$$ x^a \ x^b= \left( \frac{x}{|x|}\right)^{a+b} \ e^{ln|x^{a}|} \ e^{ln|x^{b}|} $$
Now, using the formulas of logarithm and exponential functions:
$$ x^a \ x^b= \left( \frac{x}{|x|}\right)^{a+b} \ e^{a \ ln|x|} \ e^{b \ ln|x|} $$
$$ x^a \ x^b= \left( \frac{x}{|x|}\right)^{a+b} \ e^{(a+b) \ ln|x|} $$
$$ x^a \ x^b= \left( \frac{x}{|x|}\right)^{a+b} \ |x|^{a+b} $$
We definitely find again the same form as the expression \((1)\). So:
As a result,
$$ \forall (x^a, x^b) \in \mathcal{D}(x^n)^2, $$
$$ x^a \ x^b = x^{a+b} $$
We showed that the formula works in the natural numbers set \( (\mathbb{N}) \), in the integers set \( (\mathbb{Z}) \), but also in the real numbers set \( (\mathbb{R}) \).
Therefore in any case,
$$\forall (x^a,y^a) \in \hspace{0.05em} \mathcal{D}(x^n)^2, $$
$$ x^a \ x^b = x^{a+b} $$
The division of a number \(a\) by a number \(b\) being just the multiplication of \(a\) by the inverse of \(b\), the previous formulas also work well with quotients.
$$ \forall (x^a, x^b) \in \mathcal{D}(x^n)^2, \ x \neq 0, $$
$$ \frac{x^a}{x^b} = x^{a-b} $$
Let \( x\in \hspace{0.05em} \mathbb{R}^* \) a non-zero real number and \( (a, b) \in \hspace{0.05em} \mathbb{N}^2 \) two natural numbers.
Performing the calculation \( (x^a)^b \), we do obtain:
$$ (x^a)^b = \hspace{0.2em} \underbrace { x^a \times x^a \times x^a } _\text{ \(b\) factors } $$
$$ (x^a)^b = \hspace{0.2em} \underbrace { \hspace{0.2em} \underbrace {x \times x \times x \times x \times x ...} _\text{ \(a\) factors } \ \times \ \underbrace {x \times x \times x \times x \times x ...} _\text{ \(a\) factors } \ \times \ \underbrace {x \times x \times x \times x \times x ...} _\text{ \(a\) factors } } _\text{ \(b \times a \) factors } $$
$$ (x^a)^b = \hspace{0.2em} \underbrace { x \times x \times x \times x \times x \times x \times x \times x \times x \times x \times x \times x \times x \times x... } _\text{ \(b \times a \) factors } $$
Now, considering the fundamental definition of a power of \(x\):
$$ x^n = \hspace{0.2em} \underbrace {x \times x \times x \times x \times x ...} _\text{ \(n\) factors } $$
Then, we obtain as a result that,
$$ (x^a)^b = x^{ab} $$
By applying the same reasoning by switching \(a\) and \(b\), we do have the flowwing result:
$$ (x^b)^a = x^{ab} $$
So,
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, \ \forall (a,b) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ (x^a)^b = (x^b)^a = x^{ab} $$
The division of a number \(a\) by a number \(b\) being just the multiplication of \(a\) by the inverse of \(b\), the previous formulas also work well for exponents in \( \mathbb{Z}\).
$$ \forall x \in \hspace{0.05em} \mathbb{R}, \ \forall (a,b) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ (x^{-a})^b = \left( \frac{1}{x^a} \right)^b $$
$$ (x^{-a})^b = \left( \left(\frac{1}{x}\right)^a \right)^b $$
$$ (x^{-a})^b = \left(\frac{1}{x}\right)^{ab} $$
$$ (x^{-a})^b = \frac{1}{x^{ab}}$$
$$ (x^{-a})^b = x^{-ab} $$
So,
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, \ \forall (a,b) \in \hspace{0.05em} \mathbb{Z}^2, $$
$$ (x^a)^b = (x^b)^a = x^{ab} $$
Let us retrieve our previously defined expression \((1)\):
$$ \forall x^n \in \hspace{0.05em} \mathcal{D}(x^n), $$
$$ x^n = \left( \frac{x}{|x|}\right)^{n} \ e^{ln|x^{n}|} \qquad (1) $$
So,
$$ \forall x^a \in \hspace{0.05em}\mathcal{D}(x^n), \ \forall b \in \hspace{0.05em} \mathbb{R}, $$
$$ (x^a)^b = \left( \left( \frac{x}{|x|}\right)^{a} \ e^{ln|x^{a}|} \right)^b $$
Using the formula of the power of a product, we do obtain:
$$ (x^a)^b = \left( \left( \frac{x}{|x|}\right)^{a} \right)^b \ \left(e^{ln|x^{a}|}\right)^b $$
As we did before, let us make a truth table to observe the behaviour of our sign generators.
$$ \frac{x}{|x|}$$ |
$$x^a$$ |
$$X^b$$ |
$$\left(\frac{x}{|x|}\right)^{a}$$ |
$$ \left( \left( \frac{x}{|x|}\right)^{a} \right)^b$$ |
$$x^{ab}$$ |
$$ \left(\frac{x}{|x|}\right)^{ab}$$ |
|---|---|---|---|---|---|---|
$$-1$$ |
$$odd \ function$$ |
$$odd \ function$$ |
$$-1$$ |
$$ -1$$ |
$$odd \ function$$ |
$$ -1 $$ |
$$-1$$ |
$$odd \ function$$ |
$$even \ function$$ |
$$-1$$ |
$$ 1$$ |
$$even \ function$$ |
$$ 1 $$ |
$$-1$$ |
$$even \ function$$ |
$$odd \ function$$ |
$$1$$ |
$$ 1$$ |
$$even \ function$$ |
$$ 1 $$ |
$$1$$ |
$$$$ |
$$$$ |
$$1$$ |
$$1$$ |
$$$$ |
$$1$$ |
In the three first cases, we set aside even and odd functions:
$$ \forall x^n \in \mathcal{D}(x^n), \ \frac{x}{|x|} = -1 \ \Longrightarrow \ \underbrace{\biggl[ \forall x \in \hspace{0.05em} \mathbb{R}_-^* , \enspace \forall n \in \frac{\mathbb{Z}}{2\mathbb{N} + 1} \biggr] } _\text{\( (x \ < \ 0) \) and (even and odd functions)} $$
$$x^a$$ |
$$x^b$$ |
$$a$$ |
$$b$$ |
$$a \times b$$ |
$$x^{ab}$$ |
|---|---|---|---|---|---|
$$odd \ function$$ |
$$odd \ function$$ |
$$a \in \frac{2\mathbb{Z} + 1}{2\mathbb{N} + 1} $$ |
$$b \in \frac{2\mathbb{Z'} + 1}{2\mathbb{N'} + 1} $$ |
$$(a \times b) \in \left[ \frac{2\mathbb{Z} + 1}{2\mathbb{N} + 1} \times \frac{2\mathbb{Z'} + 1}{2\mathbb{N'} + 1} \right] = \left[ \frac{(2\mathbb{Z} + 1)(2\mathbb{Z'} + 1)}{(2\mathbb{N} + 1)(2\mathbb{N'} + 1)} \right] = \left[ \frac{4\mathbb{Z}\mathbb{Z'} + 2\mathbb{Z} + 2\mathbb{Z'} + 1}{4\mathbb{N}\mathbb{N'} + 2\mathbb{N} + 2\mathbb{N'} + 1} \right] = \frac{2\mathbb{Z''} + 1}{2\mathbb{N''} + 1} $$ |
$$odd \ function$$ |
$$odd \ function$$ |
$$even \ function$$ |
$$a \in \frac{2\mathbb{Z} + 1}{2\mathbb{N} + 1} $$ |
$$b \in \frac{2\mathbb{Z'}}{2\mathbb{N'} + 1} $$ |
$$(a+b) \in \left[ \frac{2\mathbb{Z} + 1}{2\mathbb{N} + 1} \times \frac{2\mathbb{Z'}}{2\mathbb{N'} + 1} \right] = \left[ \frac{(2\mathbb{Z} + 1)(2\mathbb{Z'})}{(2\mathbb{N} + 1)(2\mathbb{N'} + 1)} \right] = \left[ \frac{4\mathbb{Z}\mathbb{Z'} + 2\mathbb{Z}}{4\mathbb{N}\mathbb{N'} + 2\mathbb{N} + 2\mathbb{N'} + 1} \right] = \frac{2\mathbb{Z''}}{2\mathbb{N''} + 1} $$ |
$$even \ function$$ |
$$even \ function$$ |
$$odd \ function$$ |
$$a \in \frac{2\mathbb{Z}}{2\mathbb{N} + 1} $$ |
$$b \in \frac{2\mathbb{Z'} + 1}{2\mathbb{N'} + 1} $$ |
$$Idem$$ |
$$even \ function$$ |
In the fourth and last case, we only have positive functions:
$$ \forall x^n \in \mathcal{D}(x^n), \ \frac{x}{|x|} = 1 \ \Longrightarrow \ \underbrace{ \Biggl[ x \in \hspace{0.05em} \mathbb{R}^*_+, \ n \in \frac{\mathbb{Z}}{2\mathbb{N} + 1} \Biggr] } _\text{\((x \ > \ 0) \) et (even and odd functions)} \lor \ \underbrace{ \Biggl[ x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace n \in \frac{\mathbb{Z}}{2\mathbb{N}^*} \Biggr] } _\text{square roots} \ \lor \ \underbrace{ \Biggl[ x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace n \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \mathbb{Q} \Bigr] \Biggr] } _\text{real numbers} $$
This table show well that:
$$\forall x^n \in \mathcal{D}(x^n), \ \left( \left( \frac{x}{|x|}\right)^{a} \right)^b = \left(\frac{x}{|x|}\right)^{ab} $$
So,
$$ (x^a)^b = \left(\frac{x}{|x|}\right)^{ab} \ \left(e^{ln|x^{a}|}\right)^b $$
As we did before, using the formulas of logarithm and exponential functions:
$$ (x^a)^b = \left(\frac{x}{|x|}\right)^{ab} \ e^{b \ ln|x^{a}|} $$
$$ (x^a)^b = \left(\frac{x}{|x|}\right)^{ab} \ e^{ab \ ln|x|} $$
$$ (x^a)^b = \left(\frac{x}{|x|}\right)^{ab} \ e^{ ln|x^{ab}|} \Longleftrightarrow (x^a)^b = x^{ab} $$
We do find again the form of the expression \((1)\).
Also, we can find the second formula by manipulating again the initial expression:
$$ (x^a)^b = \left( \left(\frac{x}{|x|}\right)^{b} \right)^{a} \ \left(e^{ln|x^{b}|}\right)^a \Longleftrightarrow (x^a)^b = (x^b)^a $$
As a result,
$$ \forall (x^a, x^b) \in \hspace{0.05em} \mathcal{D}(x^n)^2, $$
$$ (x^a)^b = (x^b)^a = x^{ab} $$
We showed that the formula works in the natural numbers set \( (\mathbb{N}) \), in the integers set \( (\mathbb{Z}) \), but also in the real numbers set \( (\mathbb{R}) \).
So in any case,
$$ \forall (x^a, x^b) \in \hspace{0.05em} \mathcal{D}(x^n)^2, $$
$$ (x^a)^b = (x^b)^a = x^{ab} $$
Let \( x \in \hspace{0.05em} \mathbb{R}^* \) be a non-zero real number and \( a \in \hspace{0.05em} \mathbb{R} \)a real number.
With the formula of the multiplication of powers, we do have:
$$ x^a x^0 = x^{a+0} $$
$$ x^a x^0 = x^{a} $$
At this stage it seems obvious that,
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, $$
$$ x^0 = 1 $$
Let \( x\in \hspace{0.05em} \mathbb{R^*} \) be a non-zero real number and \(a \in \hspace{0.05em} \mathbb{R}\) a rela number.
Let us find to what on a inverse of a power looks like. We already know that :
$$x^a \times \frac{1}{x^a} = 1 $$
But, \(1 = x^0\), so:
$$x^a \times \frac{1}{x^a} = x^0 $$
With the formula of the product of a power, we then have:
$$x^a \times x^{-a} = x^0 $$
So, by indentification,
$$ \forall x^a \in \mathcal{D}(x^n), $$
$$ \frac{1}{x^a} = x^{-a}$$
Let us find what is the power \(p\) of any square root. We already know that:
$$ \forall x \in \hspace{0.05em} \mathbb{R}^+, \ \sqrt{x} \ \sqrt{x}= x $$
Avec the formula of the product of a power:
$$ \forall x^p \in \hspace{0.05em} \mathcal{D}(x^n)^2, \ x^p \ x^p = x^{2p} $$
Now combining these two formulas:
$$ x^1 = x^{2p} \Longrightarrow 2p = 1 \Longrightarrow p = \frac{1}{2} $$
So,
$$ \forall x \in \hspace{0.05em} \mathbb{R}^+, $$
$$ x^{\frac{1}{2}} = \sqrt{x} $$
By applying the same process:
$$ \forall a \in \hspace{0.05em} \mathbb{Z}^*, \ \Bigl[ \forall x \in \hspace{0.05em} \mathbb{R}_+, \ n \in \{2a\} \Bigr] \lor \Bigl[ \forall x \in \hspace{0.05em} \mathbb{R}, \ n \in \{2a + 1\} \Bigr], $$
$$ \underbrace{\sqrt[n]{x} \ \sqrt[n]{x} \ \sqrt[n]{x} ... } _\text{n fois} \ = x $$
$$ \underbrace{x^p \ x^p \ x^p ... } _\text{n fois} \ = x^{np} $$
And in the same way:
$$ x^1 = x^{np} \Longrightarrow np = 1 \Longrightarrow p = \frac{1}{n} $$
So,
$$ \forall a \in \hspace{0.05em} \mathbb{Z}^*, \ \Bigl[ \forall x \in \hspace{0.05em} \mathbb{R}_+, \ n \in \{2a\} \Bigr] \lor \Bigl[ \forall x \in \hspace{0.05em} \mathbb{R}, \ n \in \{2a + 1\} \Bigr], $$
$$ x^{\frac{1}{n}}= \sqrt[n]{x} $$
$$ with \enspace \mathcal{D}(x^n) = \Biggl[(x = 0) \land (n \in \mathbb{R}) \Biggr] \lor \underbrace{ \overbrace { \Biggl[ \forall x \in \hspace{0.05em} \mathbb{R}^*, \enspace \forall n \in \frac{\mathbb{Z}}{2\mathbb{N} + 1 } \Biggr] } ^\text{even and odd functions} \lor \overbrace { \Biggl[ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace \forall n \in \frac{\mathbb{Z}}{2\mathbb{N}^*} \Biggr] } ^\text{ square root } } _\text{\(n \ \in \ \mathbb{Q}\) under an irreducible form} \lor \underbrace{ \Biggl[ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace \forall n \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \mathbb{Q} \Bigr] \Biggr] } _\text{irrational numbers}$$
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