A DIRECT PROOF OF THE POWER RULE FOR RATIONAL
NUMBERS
PETER F. MCLOUGHLIN
The orthodox proof of the power rule for rational numbers follows by first obm
serving that y = x n if and only if y n = xm and then using implicit differentiation
and the power rule for integers. In this paper we provide a direct proof of the power
rule for rational numbers.
proof of power rule for positive rational numbers
Let m and n 6= 0 be whole numbers. The following two formulas can be derived
using long division:
1
1
1− 1
1− 2 1
1− 3 2
1− 1
(1)
y1 − y2 = (y1n − y2n )(y1 n + y1 n y2n + y1 n y2n + . . . + y2 n )
m−1
m
xm
+ xm−2
x2 + xm−3
x22 + . . . + x2m−1 )
1 − x2 = (x1 − x2 )(x1
1
1
1
1
It follows limy2 →y1 (
y1n −y2n
y1 −y2
1
1− n
) = limy2 →y1 (y1
2
1− n
+y1
1
3
1− n
y2n +y1
(2)
2
1
1− n
−1
y2n +. . .+y2
)
=
1
n −1
1
n y1
.
m
m
1
1
1
1
Next note: a n −b n = (a n )m −(b n )m . Letting x1 = a n and x2 = b n in formula (2)
m
n
m
n
−b
we get limb→a ( (a (a−b)
1
(a n
1
−b n
)
) = limb→a (a
m−1
m−2
1
n
1
−b n )(a
m−1
n
m−3
1
2
+a
m−2
n
1
b n +a
(a−b)
m−3
n
2
b n +...+a
m−1
m−1
n
1
)
(limb→a (a−b) ) )(limb→a (a n +a n b n +a n b n +. . .+a n )) = ( n1 a n −1 )(ma
m m
n −1 . This proves the power rule for positive rational numbers.
na
=
m−1
n
)=
proof of power rule for negative rational numbers
limb→a ( (a
m
n −1
−m
n
−m
−b n )
)
(a−b)
−2m
n
m
= limb→a
m
b n −a n
m
a n (a−b)
m
bn
−m
n −1
( −m
) · (a
) = −m
.
n a
n a
This proves the power rule for rational numbers.
E-mail address: [email protected]
1
m
n
−b
= (limb→a −(a(a−b)
m
n
)
)·(limb→a
1
m m
bnan
)=