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## Homework Statement

In the process of trying to prove something else I found it would be helpful if r

^{n}[itex]\in[/itex]I, where I is an ideal, n[itex]\in[/itex]N, and r[itex]\in[/itex]R and R is a ring, then r is in I.

## Homework Equations

I is an ideal if a[itex]\in[/itex]I and b[itex]\in[/itex]I then a+b[itex]\in[/itex]I, a[itex]\in[/itex]I and r[itex]\in[/itex]R then ar[itex]\in[/itex]I, and I is not the empty set.

## The Attempt at a Solution

Base Case: Assume r

^{1}[itex]\in[/itex]I. Then r[itex]\in[/itex]I.

Inductive Case: Assume that if r

^{n}[itex]\in[/itex]I then r[itex]\in[/itex]I for all n<n+1.

Assume r

^{n+1}[itex]\in[/itex]I. Since r

^{n+1}= r

^{n}*r then either r

^{n}or r is in I. We only need to show the first case works since the second is trivial. If r

^{n}[itex]\in[/itex]I then r[itex]\in[/itex]I by the inductive hypothesis. (QED)

Is this correct?

thank you for your time.