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Metamath Proof Explorer

Theorem densq

Description: Square commutes with canonical denominator. (Contributed by Stefan O'Rear, 15-Sep-2014)

		Ref	Expression
	Assertion	densq	$⊢ A \in ℚ \to denom (A^{2}) = {denom (A)}^{2}$

Step	Hyp	Ref	Expression
1		numdensq	$⊢ A \in ℚ \to (numer (A^{2}) = {numer (A)}^{2} \land denom (A^{2}) = {denom (A)}^{2})$
2	1	simprd	$⊢ A \in ℚ \to denom (A^{2}) = {denom (A)}^{2}$