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# Let U= {$x:x$ is a letter in English alphabet}, A= {$X:X$ is a vowel in English alphabet}, find A’ and (A’)’.

Last updated date: 22nd Jul 2024
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Hint:To find a complement of a set, we subtract or remove its element from the universal set. The given sets are written in set builder notation; write it in roster notation to get clearer image in solving the question.

Complete step by step solution:
To solve this question, we will first elaborate the given data in the question. U is the universal set given in which all the alphabets of English are present and A is a set of all vowels in the English alphabets, in this manner we can say that A is a subset or a proper subset of the universal set U.
We can write this all in roster notation as follows
${\text{U}} = \{ {\text{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z}}\} \\ {\text{A}} = \{ {\text{a,}}\;{\text{e,}}\;{\text{i,}}\;{\text{o,}}\;{\text{u}}\} \\$
Now coming to the question, we have to find A’ and (A’)’, that is complement of A and A’ respectively.
Complement of a set is the set resulting from removal of the given set from the universal set.
According to this, A’ will be given as
${\text{A'}} = {\text{U}} - {\text{A}} = \{ {\text{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z}}\} - \{ {\text{a,}}\;{\text{e,}}\;{\text{i,}}\;{\text{o,}}\;{\text{u}}\} \\ {\text{A'}} = \{ {\text{b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}}\} \\$
And (A’)’ will be given as
${\text{(A')'}} = {\text{U}} - {\text{A'}} = \{ {\text{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z}}\} - \\ \{ {\text{b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}}\} \\ {\text{(A')'}} = \{ {\text{a,}}\;{\text{e,}}\;{\text{i,}}\;{\text{o,}}\;{\text{u}}\} \\$
Additional information: Complement of the universal set gives null set and also vice versa.

Note: U is denoted as the universal set. If we find a complement of a set then we will eventually get the set itself whose complement we have taken, it works just as negative of a negative equals positive. This question can also be solved with the help of Venn diagrams, try to solve it by yourself.