Дополнение к статье "Рiвняння"
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{\displaystyle ~t\vdash \quad \mathrm {f} =\{\langle x,y\rangle |\ \ \langle x,y\rangle \in X_{\mathrm {f} }\times Y_{\mathrm {f} }\ \land \ y=f(x)\}\quad \land \quad \mathrm {g} =\{\langle x,y\rangle |\ \ \langle x,y\rangle \in X_{\mathrm {g} }\times Y_{\mathrm {g} }\ \land \ y=g(x)\}}
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{\displaystyle ~\to \quad \mathrm {f} \cap \mathrm {g} =\{\langle x,y\rangle |\ \ x\in X_{\mathrm {f} }\cap X_{\mathrm {g} }\ \land \ y\in Y_{\mathrm {f} }\cap Y_{\mathrm {g} }\ \ \land \ \ y=f(x)\ \land \ y=g(x)\ \land \ f(x)=g(x)\}}
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{\displaystyle ~d\vdash \quad \mathrm {To\ solve\ an\ equation} \ f(x)=g(x)\quad \mathrm {is\quad to\ find\ the\ set} \ \{x|\ \ x\in X_{\mathrm {f} }\cap X_{\mathrm {g} }\ \land \ f(x)=g(x)\}.}
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{\displaystyle ~\vdash \ X_{\mathrm {f} }\cap X_{\mathrm {g} }=\varnothing \quad \lor \quad \forall x\ (x\in X_{\mathrm {f} }\cap X_{\mathrm {g} }\to f(x)\neq g(x))\ \ \leftrightarrow \ \ \{x|\ x\in X_{\mathrm {f} }\cap X_{\mathrm {g} }\ \land \ f(x)=g(x)\}=\varnothing }
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{\displaystyle ~\vdash \ \{x|\ x\in X_{\mathrm {f} }\cap X_{\mathrm {g} }\ \land \ f(x)=g(x)\}=\varnothing \ \ \leftrightarrow \ \ \mathrm {The\ equation} \ f(x)=g(x)\ \mathrm {has\ no\ root} .}
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{\displaystyle ~\vdash \ X_{\mathrm {f} }\cap X_{\mathrm {g} }\neq \varnothing \quad \land \quad \exists x\ (x\in X_{\mathrm {f} }\cap X_{\mathrm {g} }\ \land \ f(x)=g(x))\ \ \leftrightarrow \ \ \{x|\ x\in X_{\mathrm {f} }\cap X_{\mathrm {g} }\ \land \ f(x)=g(x)\}\neq \varnothing }
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{\displaystyle ~\vdash \ \{x|\ x\in X_{\mathrm {f} }\cap X_{\mathrm {g} }\ \land \ f(x)=g(x)\}\neq \varnothing \ \ \leftrightarrow \ \ \mathrm {The\ equation} \ f(x)=g(x)\ \mathrm {has\ at\ least\ one\ root.} }
Галактион 11:42, 27 серпня 2009 (UTC) Відповісти