[Merged by Bors] - feat(Rat): Numerator and denominator of q ^ n
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Prove `(q ^ n).num = q.num ^ n` and `(q ^ n).den = q.den ^ n` by making those defeq. It is somewhat painful. `(q ^ n).den = q.den ^ n` is also defeq for `NNRat`, but not `(q ^ n).num = q.num ^ n` due to the `Int.natAbs` in the definition of `NNRat.num`.
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Prove `(q ^ n).num = q.num ^ n` and `(q ^ n).den = q.den ^ n` by making those defeq. It is somewhat painful. `(q ^ n).den = q.den ^ n` is also defeq for `NNRat`, but not `(q ^ n).num = q.num ^ n` due to the `Int.natAbs` in the definition of `NNRat.num`.
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Build failed: |
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bors merge |
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I think the build is still failing? bors r- |
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Canceled. |
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bors merge |
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Sorry, the mobile app is very limiting in what I can see 🙈 |
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Canceled. |
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bors merge |
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Prove `(q ^ n).num = q.num ^ n` and `(q ^ n).den = q.den ^ n` by making those defeq. It is somewhat painful. `(q ^ n).den = q.den ^ n` is also defeq for `NNRat`, but not `(q ^ n).num = q.num ^ n` due to the `Int.natAbs` in the definition of `NNRat.num`. Co-authored-by: Eric Wieser <wieser.eric@gmail.com>
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Prove `(q ^ n).num = q.num ^ n` and `(q ^ n).den = q.den ^ n` by making those defeq. It is somewhat painful. `(q ^ n).den = q.den ^ n` is also defeq for `NNRat`, but not `(q ^ n).num = q.num ^ n` due to the `Int.natAbs` in the definition of `NNRat.num`. Co-authored-by: Eric Wieser <wieser.eric@gmail.com>
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Pull request successfully merged into master. Build succeeded: |
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q ^ nq ^ n
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Prove `(q ^ n).num = q.num ^ n` and `(q ^ n).den = q.den ^ n` by making those defeq. It is somewhat painful. `(q ^ n).den = q.den ^ n` is also defeq for `NNRat`, but not `(q ^ n).num = q.num ^ n` due to the `Int.natAbs` in the definition of `NNRat.num`. Co-authored-by: Eric Wieser <wieser.eric@gmail.com>
callesonne
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Prove `(q ^ n).num = q.num ^ n` and `(q ^ n).den = q.den ^ n` by making those defeq. It is somewhat painful. `(q ^ n).den = q.den ^ n` is also defeq for `NNRat`, but not `(q ^ n).num = q.num ^ n` due to the `Int.natAbs` in the definition of `NNRat.num`. Co-authored-by: Eric Wieser <wieser.eric@gmail.com>
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Prove
(q ^ n).num = q.num ^ nand(q ^ n).den = q.den ^ nby making those defeq. It is somewhat painful.(q ^ n).den = q.den ^ nis also defeq forNNRat, but not(q ^ n).num = q.num ^ ndue to theInt.natAbsin the definition ofNNRat.num.