Answer:
To prove that N is a proper subset of M, show that every element of N is an element of M but there exists some elements of M which are not in N.
M= {4, 8, 12, 16, 20, 24...}
N= {12, 24, 36, 48, 60...}
Similar to example 6, 12 is a multiple of 4.
If x∈N,
then x= 12m for some positive integer m.
∴ x= 4(3m)
Hence, x is a multiple of 4.
i.e. x∈N.
Thus, every element of N is an element of M, but there exists some elements of M which are not elements of N. For example, 4, 8 and 16 are elements of M but are not elements of N. Thus, N is a proper subset of M. (shown)
Explanation:
∈ is the symbol of 'element of'. Thus, x∈N means that x is an element of N.
☆ Think of subsets like subtopics:
If differentiation is a subtopic of calculus, differentiation is a part of calculus but calculus is not only about differentiation since it includes integration too. Thus, there are numbers, which we call elements, in M that are not in N but all elements in N are in M.
