Let S = {10n - 1 : n \in Z}, T be the set of odd integers, and U = {6m + 4 : m \in Z} Prove that discrete math S \subseteq T \cap U.

pedzenekO 2021-08-02 Answered
Let \(\displaystyle{S}={\left\lbrace{10}{n}-{1}:{n}\in{Z}\right\rbrace}\), T be the set of odd integers, and \(\displaystyle{U}={\left\lbrace{6}{m}+{4}:{m}\in{Z}\right\rbrace}\)
Prove that discrete math \(\displaystyle{S}\subseteq{T}\cap{U}\).

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Expert Answer

yagombyeR
Answered 2021-08-03 Author has 24499 answers
Consider the given information.
Here,
\(\displaystyle{S}={\left\lbrace{10}{n}−{1}:{n}\in{Z}\right\rbrace}\)
And,
\(\displaystyle{U}={\left\lbrace{5}{m}+{4}:{m}\in{Z}\right\rbrace}\)
And, the set T is the set of odd integers.
So, the set is defined as,
\(\displaystyle{T}={\left\lbrace{2}{k}+{11}:{k}\in{Z}\right\rbrace}\)
Now, find the first \(\displaystyle{T}\cap{U}\).
Let k is the arbitrary element of S.
Thus,
\(\displaystyle{k}={10}{n}−{1}{n}\in{Z}\)
= 10n − 1 + 2 − 2
= (10n − 2) + 1
= 2 (5n − 1) + 1
Here, (5n - 1) is belongs to integers.
Let (5n - 1) is equal p. Put the value in the expression.
Thus,
\(\displaystyle{k}={2}{p}+{1}\in{T}\)
Thus, the number k belongs to T.
\(\displaystyle{k}\in{T}\)
Now, define for S.
k = 10n − 1
= 10n − 1 + 5 − 5
= (10n − 5) + 4
= 5(2n − 1) + 4
Let (2n - 1) is equal m. Put the value in the expression.
Thus,
\(\displaystyle{k}={5}{m}+{4}\in{U}\)
Thus, the number k belongs to S.
\(\displaystyle{k}\in{U}\)
From the above both the result, the given condition proved.
\(\displaystyle{S}\subseteq{T}\cap{U}\).
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