Joe Johnson 126
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 Jul2 asked Prove A Group is Not Simple Jun24 comment Difference between 'true' and 'provable' @user2520938 You are losing sight of the intent of what you read. The original statement is just to show that truth and proof are different. You can have some facts that you want to be true, but have no way to prove them if you don't have the right deduction system. You are correct, we can add anything we want as an axiom and consider it proven. But, if you add some statement $P$ and the negation of $P$ is provable, you tend to run into problems. Again, this points out the difference between "truth" and "proof", at least as defined in mathematical systems. Jun24 comment Difference between 'true' and 'provable' @user2520938 The reason we know that $17$ is prime is because there exist rules of deduction from which we can conclude it is prime. The point is that we can state a lot of things in math, but cannot prove anything without a system of deduction. We can state "$17$ is prime", but have no way of showing it without such a system. But, either $17$ is prime or it is not, regardless of whether we can prove it. Thankfully we have a system of deduction (several really) that can show it is indeed prime. Jun24 answered Difference between 'true' and 'provable' Jun19 reviewed Leave Open Convergence radius and is a series convergent in the ends of that radius Jun19 revised two points with same tangent line added 41 characters in body; edited title Jun18 comment The topology on $X / G$ where $G$ acts on $X$ @ThePortakal This is the Sierpinski Space: en.wikipedia.org/wiki/Sierpi%C5%84ski_space Jun14 comment Integrals of Pullbacks A homotopy gives you a map $f:X\times I\rightarrow Y$ where the boundary is $X\sqcup X$. Jun3 comment Natural deduction proof for : p → ( c ∨ b) , b → s ⊢ ( p ∧ ¬s)→ c From $\neg s$ you can derive $\neg b$. Since you have $c\vee b$ and $\neg b$, you can derive $c$ from or elimination and $\neg b$. May24 comment Basis and dimension of the span of the vectors (0, 0, 0), (9, 0, 0), (8, 1, 0), (1, 8, 9) It is asking you to find a linearly independent subset of $S$ that has the same span as $S$. May23 comment Proving that $a \dot{-} (b+1) = (a \dot{-} b) \dot{-} 1$ @Nagase You prove it using the distributive and commutative laws. Look at $(a-b)-1-(a-(b+1))$ and show it equals $0$. May23 comment Proving that $a \dot{-} (b+1) = (a \dot{-} b) \dot{-} 1$ If $a\geq b+1$, then $a\dot{-}(b+1)=a-(b+1)$ and $(a\dot{-}b)\dot{-}1=(a-b)\dot{-}1$. But, $a-b\geq 1$. So, $(a-b)\dot{-}1=(a-b)-1$, which is the same as $a-(b+1)$. May23 comment Proving that $a \dot{-} (b+1) = (a \dot{-} b) \dot{-} 1$ I wouldn't think induction works. You could try cases: $a\geq b+1$, $a\geq b$ but \$a