Today is our topic of discussion Rational and Irrational Exponents .
Rational and Irrational Exponents
We recall some notations: R denotes the set of real number
N denotes the set of natural numbers
Z denotes the set of integers
Q denotes the set of rational numbers
If a is any real number and n is any natural number, when a is multiplied n times, the product is written as aaaa… (n times) and a” is called n th power of a. In such cases, a is called the base and n is called the exponent or index. in 3×4 the base is 3 and exponent is 4.
Again, in 2/3×4 the base is 2/3 and exponent is 4.
Definition: For all a € R
1. a¹ = a
2. axn=a.a.a……a (n times factor of a), where nЄN,n> 1
Irrational Exponent
When an exponent x is irrational, we fix the value of ax(a > 0) so that for some rational value p approximate to x, the value of a” is close to the value of a*. For example, we consider the number 3√5. We know, √5 is an irrational number and √5 = 2.236067977… (We have obtained this value using calculator and … indicates that the decimal expression is infinite).
As the approximate values of √5 considering
P₁ = 2.23
P2 = 2.236
P3 = 2.2360
P4 = 2.23606
P5 =2.236067
P6 = 2.2360679
P7 = 2.23606797
we obtain the following approximate values of 35
q₁ = 32.23 = 11.5872505 q2 = 32.236 = 11.6638822 q3 = 32.2360 94 = 32.2360611.66465109 95 = 32.236067 11.6647407 =
11.6638822 96 = 32.2360679 = 11.6647523 q7 = 32.23606797 = 11.6647532
This values have also been obtained using calculator.
Actually, 35 11.6647533…
Laws of Exponents
Formula 1. If a Є R and nЄN, a¹ = a, an+1 = a”.a.
Proof:
By definition a¹ a and for all nЄN: an+1 = n times
Note: N is the set of all natural numbers.
Formula 2. For every a Є R and m, n Є N, a™ · axm.axn = a(m+n)
Proof:
Let us consider the statement axm = axm+n… (1) fixing some value for mЄ N and taking n as the variable.
Putting n = 1 in (1) we get,
L.H.S = axm . a¹=a . ma am+1 = R.H.S [Formula 1]
(1) is true for n = 1.
Now, let (1) is true for n k. i.e. axm . axk =axm+k
Then, axm . axk+1= axm (axk . a) [Formula 1]
= (axm . axk). a
= axm+k . a
= a(m+k+1) [Formula 1]
i.e. (1) is true for n = k +1.
Hence, by mathematical induction for all n € N (1) is true.
for every m, nЄ N axm . axn=axm+n
Formula 3
For every a ∉ R, a 0 and m,n ∉ N,m ≠ n,
Proof:
1. Suppose, m>n. Then m-nЄ N.
axm-n . axn = a(m-n)+n=axm [Formula 2]
axm/axm-n+n” [definition of division]
2. Suppose, m<n. Then n – mЄN
axn-m .axm= a(n-m)+ma” [Formula 2]
axm/an= 1/an-m = [definition of division]
Note: Prove the above formula by mathematical induction. [like formula 2]
Formula 4. If a Є R and m, nЄ N, then (am)” = amn
Formula 5. For all a, bЄ R and n Є N, (a – b)” = a”. br
[Prove the above formulae by mathematical induction]
Zero and negative integer exponents:
Definition: For some a ∉R, ifa ≠0,
3. aº = 1
4. a “= 1 an where n Є N
Remark: While expanding the concept of exponent, the validity of the functional law of exponents ama” = am+” is assured carefully.
If the formula is true for m = 0, then aº. a” = aº+” i.e, aº held. = an 1 must be if the formula is true for m = 0, then aº a” aº+ i.e, aº: = held.
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