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Jul
2
asked Prove A Group is Not Simple
Jun
24
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.
Jun
24
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.
Jun
24
answered Difference between 'true' and 'provable'
Jun
19
reviewed Leave Open Convergence radius and is a series convergent in the ends of that radius
Jun
19
revised two points with same tangent line
added 41 characters in body; edited title
Jun
18
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
Jun
14
comment Integrals of Pullbacks
A homotopy gives you a map $f:X\times I\rightarrow Y$ where the boundary is $X\sqcup X$.
Jun
3
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 $\neg b$.
May
24
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 $S$.
May
23
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 $0$.
May
23
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 $a-(b+1)$.
May
23
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 $a<b+1$, $a<b$.
May
23
reviewed Close Is $\mathbb{Z}[2\sqrt{2}]$ a PID?
May
22
answered Determine whether $f(x)$ is increasing or decreasing
May
21
comment Homotopy of a pair
@user135988 It is not the case that $\pi_1(X,A)$ is trivial for all $X$ and subspaces $A$. Just let $A=x_0$.
May
21
comment Homotopy of a pair
@user135988 Because $I^n$ is a contractible space. Given any space, $Y$, any map $I^n\rightarrow Y$ is homotopic to a constant map, as long as you have no restrictions on the homotopy. In your case as long as the map is contained entirely in $A$, you have no restrictions on the homotopy.
May
20
revised Homotopy of a pair
added 138 characters in body
May
20
answered Homotopy of a pair
May
19
comment Can a nonempty set ever equal its Cartesian product with another set?
If $T$ is the one element set, yes. I suppose they might not be $\textit{equal}$, depending on your definitions.