This was an exam question that I got totally wrong and am a bit question. Prove $x^3 + x + 1 = 0$ has no solutions. Prove by contradiction. Assume: $x^3 +x +1 =0$ has at least one rational root. ...
Prove that if $(m - 1)! + 1$ is divisible by $m$, $m$ is a prime with $(m - 1)! = 1.2.3…(m - 2)(m - 1)$
$m$ is a positive integer, and $ m > 1$, Prove that if $(m - 1)! + 1$ is divisible by $m$, $m$ is a prime. Solve this by making a contradiction. My english isn't so well. Please help and thank you ...
so basically i have a project about 3 valued logic ie truth=1 false = 0, unknown = 1/2 in a previous project I had to come up with formulae for 2 valued logic as follows: ...