Theorem qmuldeneqnum

Description: Multiplying a rational by its denominator results in an integer. (Contributed by Stefan O'Rear, 13-Sep-2014)

	Ref	Expression
Assertion	qmuldeneqnum	$$⊢A\in \mathbb{Q}\to A\mathrm{denom}\left(A\right)=\mathrm{numer}\left(A\right)$$

Proof

Step	Hyp	Ref	Expression
1		qeqnumdivden	$$⊢A\in \mathbb{Q}\to A=\frac{\mathrm{numer}\left(A\right)}{\mathrm{denom}\left(A\right)}$$
2	1	oveq1d	$$⊢A\in \mathbb{Q}\to A\mathrm{denom}\left(A\right)=\left(\frac{\mathrm{numer}\left(A\right)}{\mathrm{denom}\left(A\right)}\right)\mathrm{denom}\left(A\right)$$
3		qnumcl	$$⊢A\in \mathbb{Q}\to \mathrm{numer}\left(A\right)\in \mathbb{Z}$$
4	3	zcnd	$$⊢A\in \mathbb{Q}\to \mathrm{numer}\left(A\right)\in ℂ$$
5		qdencl	$$⊢A\in \mathbb{Q}\to \mathrm{denom}\left(A\right)\in \mathbb{N}$$
6	5	nncnd	$$⊢A\in \mathbb{Q}\to \mathrm{denom}\left(A\right)\in ℂ$$
7	5	nnne0d	$$⊢A\in \mathbb{Q}\to \mathrm{denom}\left(A\right)\ne 0$$
8	4 6 7	divcan1d	$$⊢A\in \mathbb{Q}\to \left(\frac{\mathrm{numer}\left(A\right)}{\mathrm{denom}\left(A\right)}\right)\mathrm{denom}\left(A\right)=\mathrm{numer}\left(A\right)$$
9	2 8	eqtrd	$$⊢A\in \mathbb{Q}\to A\mathrm{denom}\left(A\right)=\mathrm{numer}\left(A\right)$$