If n = number of digits , x is a number in the given radix

Now , if radix = p , then radix complement ie p's complement = p^{n} - x

if radix = q , then diminished radix complement ie. (q-1)'s complement = (q^{n}-1) - x

Now , if we equate both , p^{n} - x = (q^{n}-1) - x

p^{n }= q^{n}-1

Now, suppose , if n = 1 , p=2 ,q= 3...then both will give same result

but if n = 2 then both will not give the same result...

So, we can say radix complement for base 'r' will not always be same as diminished radix complement for base 's'

for example , 2's complement for base 2 = 2^{n }- x but 2's complement for base '3' , it will be (3^{n}-1) - x

Here , We have to find 4's complement of 7_{10} in base 4

So, 7_{10 }= (13)_{4}

Now , (4^{2}-1)_{10} - (13)_{4} + 1

(15)_{10} - (13)_{4} + 1

(33)_{4} - (13)_{4 }+ 1

(21)_{4} = (9)_{10}

So, 4's complement of 7_{10} in base 4 will be 9_{10}

Now , We will find 4's complement of 7_{10} in base 5

So, 7_{10 }= (12)_{5}

(5^{2}-1)_{10} - (12)_{5}

(24)_{10} - (12)_{5}

(44)_{5} - (12)_{5}

(32)_{5} = (17)_{10}

So, both are different because radix are different in both cases.