Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

To finite sets have m and n elements. The total no. of subsets of first set is 60 more than the total number of subsets of second sets. Find the value of m

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Given m and n are integers. They are cardinalities of two sets.

The total number of sub sets of first set: 2^m (I hope the null set is a subset of any set.). total number of subsets of second set: 2^n. Hence, 2^m = 2^n + 60

The above equation is not satisfied by m =0 or 1, or n = 0 or 1. So m and n are >= 2. Hence, 2^(m-2) and 2^(n-2) are also integers.

Now divide the above equation by 4. we get: 2^(m-2) = 2^(n-2) + 15

LHS is even as it is a power of 2. On RHS 15 is an odd number. So only way this equation can be solved is if 2^(n-2) is an odd number. The only possibility is n-2 = 0 for which it is equal to 1 and becomes an odd number. Hence, n = 2. So m -2 = 4 and so m = 6

=================================== if you want a simple solution,

2^m = 2^n + 60 try n =0, 1, 2, 3,.. try to get RHS as a power of 2.

for n =2, RHS = 64 , it is a power of 2. and 2^6. Hence, m = 6.