Question #35f5d

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Question #35f5d

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Answer 1 · 2016-10-22T08:10:50

No

Explanation:

A ring must contain a multiplicative identity. As there is only one such element $1_{R}$ $1_R$ in $R$ $R$ , and $T$ $T$ is a restriction of $R$ $R$ with the same operations, then the only possible multiplicative identity for $T$ $T$ is $1_{R}$ $1_R$ .

However, as $n > 0$ $n > 0$ , we have $n \cdot 1_{R} = n \neq 0$ $n*1_R = n != 0$ . Thus $1_{R} \notin T$ $1_R !inT$ , meaning $T$ $T$ has no multiplicative identity and thus cannot be a ring (or subring).

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Question #35f5d

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