Currying and Contrapositive in Practice - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-21T01:33:56Z http://mathoverflow.net/feeds/question/48189 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48189/currying-and-contrapositive-in-practice Currying and Contrapositive in Practice turnersr 2010-12-03T16:05:58Z 2010-12-03T16:10:58Z <p>I am studying for a final by reproving theorems using logically equivalent forms. But I am having a little trouble negating. Consider the following,</p> <p>Suppose that V is an n-dimensional vector space over a field F and that \${v_1, . . ., v_m}\$ is a linearly independent set in V. Then \$m &lt;= n \$ and there exists vectors \$v_{m+1}, . . ., v_n\$ such that \${v_1, . . ., v_m, v_{m+1}, . . . v_n}\$ is a basis for V.</p> <p>The theorem is stated using this logical flow: A_1 AND A_2 => B_1 AND B_2 where</p> <p>A_1 = "Suppose that V is an n-dimensional vector space over a field F" </p> <p>A_2 = "\${v_1, . . ., v_m}\$ is a linearly independent set in V"</p> <p>B_1 = "\$m&lt;=n\$"</p> <p>B_2 = "there exists vectors \$v_{m+1}, . . ., v_n\$ such that \${v_1, ... ,v_m, v_{m+1}, ... v_n}\$ is a basis for V"</p> <p>We reformulate this as the following</p> <p>A_1 AND A_2 => B_1 AND B_2 is the same as,</p> <p>By currying, A_1 => (A_2 => B_1 AND B_2) this is the same as, taking the contrapositive, A_1 => (NOT B_1 AND NOT B_2 => NOT A_2)</p> <p>Where: </p> <p>A_1 = "Suppose that V is an n-dimensional vector space over a field F"</p> <p>NOT A_2 = "\${v_1, . . ., v_m}\$ is a linearly dependent set in V"</p> <p>NOT B_1 "\$m > n\$"</p> <p>NOT B_2 "for all vectors \$v_{m+1}, . . ., v_n\$ such that \${v_1, . . ., v_m, v_{m+1}, . . . v_n}\$ is a NOT a basis for V"</p> <p>So we get:</p> <p>Suppose that V is an n-dimensional vector space over a field F, Then \$m > n\$ OR for all vectors \$v_{m+1}, . . ., v_n\$ such that \${v_1, ..., v_m, v_m+1, . . . v_n}\$ is a NOT a basis for V Implies \${v_1, . . ., v_m}\$ is a linearly dependent set in V.</p> <p>So for example let V be a 2-dimensional vector space over the field \$R^2\$,</p> <p>Then 3 > 2 OR for all vectors \$(x_4,y_4), (x_3,y_3), (x_2, y_2)\$ such</p> <p>that \${(x_1, y_2), (x_2, y_2) , (x_3,y_3), (x_4,y_4), (x_3,y_3), (x_2, y_2) }\$ is a NOT a basis for V ( Is this the correct way to interpret \$v_{m+1}, . . ., v_n\$, since now \$m > n\$ ?? ) Implies \${(x_1, y_2), (x_2, y_2) , (x_3,y_3)}\$ is a linearly dependent set in V.</p> <p>So in this case: \$3 > 2\$ OR for all vectors \$(x_4,y_4), (x_3,y_3), (x_2, y_2)\$ such that</p> <p>\${(x_1, y_2), (x_2, y_2) , (x_3,y_3), (x_4,y_4), (x_3,y_3), (x_2,y_2) }\$ IS A BASIS for V</p> <p>Implies \${(x_1, y_2), (x_2, y_2) , (x_3,y_3)}\$ is a linearly dependent set in V?</p> <p>I am not sure what is the correct mathematical usage of the OR in this case.</p> <p>Am I on the right track? Thank you for all your help.</p>