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Continuing in the same pattern, we get
Observe that negative exponents
1 1 Note: becomes positive when it is in the
–1
10 = =
10 10 1 denominator.
1 1 1 1
10 = ÷ 10 = × = Quick Check
–2
10 10 10 100
1 Fill in the following table.
= Exponential Expanded Value Positive
10 2 form form exponential form

–3
10 = 1 ÷ 10 = 1 × 1 = 1 3 3 … … …
100 100 10 10 3
1 1 1 3
Thus, in general, ×××… up to n 3 0 1 1
a a a 3
n
1
1
times =    == () n == () -- n 3 –3 … … 3 1 3
a
 
a
a
where, a is the base and (–n) is the 3 –4 … … …
exponent, which is a negative integer.
1
Or we can say that, for any non-zero integer, a, a − n = , where n is any integer.
a n
Laws of Exponents for Negative Exponents

n
m
We know that for any non-zero integer a, a × a = a m + n holds when m and n are whole numbers.
But what happens when the exponents are negative? Can we apply the same rule? Let us investigate.
Consider 3 and 3 .
–2
–4
1 1  − n 1 
–2
–4
Here, 3 = and 3 =  Q a () = n 
3 4 3 2  a 
1 1 1 1
–2
–4
\ 3 × 3 = × = = = 3 − 6
3 4 3 2 3 42 3 6 Maths Talk
+
Now, –6 = (–4) + (–2) Check whether the following laws of
exponents are true.
–2
–4
So, 3 × 3 = 3 (–4) + (–2) m
m n
1. a = a mn 2. (a ) = a mn
−
2
–3
Similarly for 9 and 9 , we have a n

a
1 9 2 1 1 3. a × b = (ab) 4. a m =   m
m
m
m
2
–3
2
9 × 9 = × 9 = = = b m  
b
−
9 3 9 3 9 32 9 1
= 9 –1 where a and b are non-zero integers
and m, n are any integers.
–3
\ 9 × 9 = 9 (–3) + 2 [Q –1 = (–3) + 2] Discuss your observation in the class.
2
–5
2
Again, for (–4) and (–4) , we have
1 (−4 ) 2 1
–5
2
2
(–4) × (–4) = (−4 ) × = = 2 = (–4) –3
5
(−4 ) 5 (−4 ) 5 (− 4) × (− 4) −
–5
2
\ (–4) × (–4) = (–4) 2 + (–5) [Q –3 = 2 + (–5)]
In general, for any non-zero integer a, a × a = a m + n , where m and n are integers.
m
n
Mathematics-8 266