حمل (رياضيات)

الجمع (+)
الطرح (−)
الضرب (×)
القسمة (÷)
الحمل لأس
th root (√)
لوغاريتم (log)

Graphs of

y = b x

for various bases b: base 10, , base 2, and base 1/2. Each curve passes through the point

(0, 1)

because any nonzero number raised to the power of 0 is 1. At

x = 1

, the value of y equals the base because any number raised to the power of 1 is the number itself.

الأُسُّ في الحساب Power ناتج عدد مضروب في نفسه عددًا محدَّدًا من المرَّات. على سبيل المثال ثلاثة × ثلاثة × ثلاثة × ثلاثة × ثلاثة يسمى الأس الخامس للعدد ثلاثة ويخط ¹3. وفيما يتعلق بـ ¹3 يسمى العدد الأساس والعددخمسة الدليل الأسي. أما الأُسَّين الثاني والثالث لعدد ما فيسميان التربيع ؛ التكعيب . والأس الأول لعدد ما العدد ذاته، أما أس الصفر لعدد ما فهوواحد، بمعنى ثلاثة ¥هوثلاثة و3¤قد يكون 1. وينطبق مفهوم الأس أيضًا على الأعداد والكسور السالبة

Exponentiation is a mathematical operation, written as $b n$ , involving two numbers, the base $b$ and the exponent or power $n$ . When $n$ is a positive integer, exponentiation corresponds to repeated multiplication of the base: that is, $b n$ is the product of multiplying $n$ bases:

{\displaystyle b^{n =\underbrace {b\times \dots \times b _{n\,{\textrm {times .

From the associativity of multiplication, it follows that for any positive integers $m$ and $n$ ,

{\displaystyle b^{m+n =b^{m \cdot b^{n .

Zero exponent

Any nonzero number raised to the $0$ power is $1$ :

{\displaystyle b^{0 =1.

One interpretation of such a power is as an empty product.

The case of $00$ is more complicated, and the choice of whether to assign it a value and what value to assign may depend on context. نطقب:Crossref

Negative exponents

The following identity holds for an arbitrary integer $n$ and nonzero $b$ :

{\displaystyle b^{-n ={\frac {1 {b^{n .

Raising 0 to a negative exponent is undefined, but in some circumstances, it may be interpreted as infinity ( $\infty$ ).

The identity above may be derived through a definition aimed at extending the range of exponents to negative integers.

For non-zero $b$ and positive $n$ , the recurrence relation above can be rewritten as

{\displaystyle b^{n ={\frac {b^{n+1 {b ,\quad n\geq 1.

By defining this relation as valid for all integer $n$ and nonzero $b$ , it follows that

{\displaystyle {\begin{aligned b^{0 &={\frac {b^{1 {b =1,\\[3pt]b^{-1 &={\frac {b^{0 {b ={\frac {1 {b ,\end{aligned

and more generally for any nonzero $b$ and any nonnegative integer $n$ ,

{\displaystyle b^{-n ={\frac {1 {b^{n .

This is then readily shown to be true for every integer $n$ .

Identities and properties

The following identities hold for all integer exponents, provided that the base is non-zero:

{\displaystyle {\begin{aligned b^{m+n &=b^{m \cdot b^{n \\\left(b^{m \right)^{n &=b^{m\cdot n \\(b\cdot c)^{n &=b^{n \cdot c^{n \end{aligned

Unlike addition and multiplication:

Exponentiation is not commutative. For example, $23 =ثمانية \neq 3 2 = 9$ .
Exponentiation is not associative. For example, $(2 3) 4 = 8 4 = 4096$ , whereas $2 (3 4) = 2 81 = 7024241785163922925♠ 2417851639229258349412352$ . Without parentheses, the conventional order of operations in superscript notation is top-down (or right-associative), not bottom-up (or left-associative). That is,
${\displaystyle b^{p^{q =b^{\left(p^{q \right) ,$

which, in general, is different from

${\displaystyle \left(b^{p \right)^{q =b^{pq .$

Combinatorial interpretation

For nonnegative integers $n$ and $m$ , the value of $n m$ is the number of functions from a set of $m$ elements to a set of $n$ elements (see cardinal exponentiation). Such functions can be represented as $m$ -tuples from an $n$ -element set (or as $m$ -letter words from an $n$ -letter alphabet). Some examples for particular values of $m$ and $n$ are given in the following table:

$n m$	The $n m$ possible $m$ -tuples of elements from the set $نطقب:Mset$
${\displaystyle 0^{5 =0$	none
${\displaystyle 1^{4 =1$	${\displaystyle (1,1,1,1)$
${\displaystyle 2^{3 =8$	${\displaystyle (1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)$
${\displaystyle 3^{2 =9$	${\displaystyle (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)$
${\displaystyle 4^{1 =4$	${\displaystyle (1),(2),(3),(4)$
${\displaystyle 5^{0 =1$	${\displaystyle ()$

List of whole-number powers

n	n²	n³	n⁴	n⁵	n⁶	n⁷	n⁸	n⁹	n¹⁰
2	4	8	16	32	64	128	256	512	1,024
3	9	27	81	243	729	2,187	6,561	19,683	59,049
4	16	64	256	1,024	4,096	16,384	65,536	262,144	1,048,576
5	25	125	625	3,125	15,625	78,125	390,625	1,953,125	9,765,625
6	36	216	1,296	7,776	46,656	279,936	1,679,616	10,077,696	60,466,176
7	49	343	2,401	16,807	117,649	823,543	5,764,801	40,353,607	282,475,249
8	64	512	4,096	32,768	262,144	2,097,152	16,777,216	134,217,728	1,073,741,824
9	81	729	6,561	59,049	531,441	4,782,969	43,046,721	387,420,489	3,486,784,401
10	100	1,000	10,000	100,000	1,000,000	10,000,000	100,000,000	1,000,000,000	10,000,000,000

Rational exponents

From top to bottom: x^1/8, x^1/4, x^1/2, x¹, x², x⁴, x⁸.

An nth root of a number b is a number x such that $x n = b$ .

If b is a positive real number and n is a positive integer, then there is exactly one positive real solution to $x n = b$ . This solution is called the principal th root of b. It is denoted ^ⁿ√b, where √ is the radical symbol; alternatively, the principal root may be written b^1/n. For example: $9 1/2 = \sqrt 9 = 3$ and $8 1/3 = 3 \sqrt 8 = 2$ .

The fact that ${\displaystyle x=b^{\frac {1 {n$ solves ${\displaystyle x^{n =b$ follows from noting that

{\displaystyle {\begin{aligned x^{n &=\left(b^{\frac {1 {n \right)^{n =\underbrace {b^{\frac {1 {n \times b^{\frac {1 {n \times \cdots \times b^{\frac {1 {n _{n\,{\textrm {times \\&=b^{\underbrace {\left({\frac {1 {n +{\frac {1 {n +\cdots +{\frac {1 {n \right) _{n\,{\textrm {times =b^{\frac {n {n =b^{1 =b.\end{aligned

If n is even and b is positive, then $x n = b$ has two real solutions, which are the positive and negative nth roots of b, that is, $b 1/ n > 0$ and $-(b 1/ n) < 0.$

If n is even and b is negative, the equation has no solution in real numbers.

If n is odd, then $x n = b$ has exactly one real solution, which is positive if b is positive ( $b 1/ n > 0$ ) and negative if b is negative ( $b 1/ n < 0$ ).

Taking a positive real number b to a rational exponent u/v, where u is an integer and v is a positive integer, and considering principal roots only, yields

{\displaystyle b^{\frac {u {v =\left(b^{u \right)^{\frac {1 {v ={\sqrt[{v ]{b^{u =\left(b^{\frac {1 {v \right)^{u =\left({\sqrt[{v ]{b \right)^{u .

Taking a negative real number b to a rational power u/v, where u/v is in lowest terms, yields a positive real result if u is even, and hence v is odd, because then b^u is positive; and yields a negative real result, if u and v are both odd, because then b^u is negative. The case of even v (and, hence, odd u) cannot be treated this way within the reals, since there is no real number x such that $x 2 k = -1$ , the value of b^u/v in this case must use the imaginary unit i, as described more fully in the section § Powers of complex numbers.

Thus we have $(-27) 1/3 = -3$ and $(-27) 2/3 = 9$ . The number أربعة has two 3/2 powers, namelyثمانية and −8; however, by convention the notation 4^3/2 employs the principal root, and results in 8. For employing the v-th root the u/v-th power is also called the v/u-th root, and for even v the term principal root denotes also the positive result.

This sign ambiguity needs to be taken care of when applying the power identities. For instance:

{\displaystyle -27=(-27)^{\left(\left({\frac {2 {3 \right)\left({\frac {3 {2 \right)\right) =\left((-27)^{\frac {2 {3 \right)^{\frac {3 {2 =9^{\frac {3 {2 =27

is clearly wrong. The problem starts already in the first equality by introducing a standard notation for an inherently ambiguous situation –asking for an even root– and simply relying wrongly on only one, the conventional or principal interpretation. The same problem occurs also with an inappropriately introduced surd-notation, inherently enforcing a positive result:

{\displaystyle \left((-27)^{\frac {2 {3 \right)^{\frac {3 {2 ={\sqrt {\left({\sqrt[{3 ]{(-27)^{2 \right)^{3 ={\sqrt {(-27)^{2 \neq -27

instead of

{\displaystyle \left((-27)^{\frac {2 {3 \right)^{\frac {3 {2 =-{\sqrt {\left({\sqrt[{3 ]{(-27)^{2 \right)^{3 =-{\sqrt {(-27)^{2 =-27.

In general the same sort of problems occur for complex numbers as described in the section § Failure of power and logarithm identities.

انظر أيضاً

Double exponential function
Exponential decay
Exponential growth
List of exponential topics
Modular exponentiation
Scientific notation
Unicode subscripts and superscripts
Zero to the power of zero

المراجع

^ Achatz, Thomas (2005). (3rd ed.). Industrial Press. p. 101. ISBN .
^ Raphael M. Robinson (1958). + 1 and on factors of Fermat numbers" (PDF). Proc. Amer. Math. Soc. 9 (5): 677. doi:10.1090/s0002-9939-1958-0096614-7.