# 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.

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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yagombyeR
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.
$$\displaystyle{S}\subseteq{T}\cap{U}$$.