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Metamath Proof Explorer

Theorem numexp

Description: Elevating to a nonnegative power commutes with canonical numerator. Similar to numsq , extended to nonnegative exponents. (Contributed by Steven Nguyen, 5-Apr-2023)

		Ref	Expression
	Assertion	numexp	$⊢ (A \in ℚ \land N \in ℕ_{0}) \to numer (A^{N}) = {numer (A)}^{N}$

Proof

Step	Hyp	Ref	Expression
1		numdenexp	$⊢ (A \in ℚ \land N \in ℕ_{0}) \to (numer (A^{N}) = {numer (A)}^{N} \land denom (A^{N}) = {denom (A)}^{N})$
2	1	simpld	$⊢ (A \in ℚ \land N \in ℕ_{0}) \to numer (A^{N}) = {numer (A)}^{N}$