# Stavros Mekesis

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 Aug24 asked Is it true that $\forall b \forall c \forall x ((x^2 + bx + c \neq 0) \rightarrow b^2 - 4c < 0)$? Aug21 accepted Indicating when $|x + y + z| = |x| + |y| + |z|$ holds Aug21 comment Indicating when $|x + y + z| = |x| + |y| + |z|$ holds Nice approach! Thank you :-) Aug21 accepted The set of all finite sequences of members of a countable set is also countable Aug21 asked Indicating when $|x + y + z| = |x| + |y| + |z|$ holds Aug20 comment Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ It doesn't assume anything. The question is not clear. Spivak probably assumed a positive even number. Aug20 comment Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ @WarrenHill: Spivak doesn't say anything about it. Oh my! Aug20 accepted An incorrect proof by exhaustion Aug20 accepted Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ Aug20 comment Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ @GEdgar: I've proved by induction that for all $n \in N$, if $0 \leq y < x$ then $y^n < x^n$. I've also proved that if $n$ is even then $(-y)^n = y^n$. I think everything has been established. Aug20 comment Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ @AndrewD.Hwang: Nice observation. Thank you! Aug20 asked Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ Aug5 accepted If $\forall \mathcal F(\bigcup \mathcal F = A \Rightarrow A \in \mathcal F)$ then A has exactly one element Aug5 comment If $\forall \mathcal F(\bigcup \mathcal F = A \Rightarrow A \in \mathcal F)$ then A has exactly one element Hey! Kalispera :) Aug5 asked If $\forall \mathcal F(\bigcup \mathcal F = A \Rightarrow A \in \mathcal F)$ then A has exactly one element Aug2 comment Some questions about the Jech's book (Generalized De Morgan's law and distributive law ) You are welcome. Bear in mind that you might also use a different set notation: $A \cap (\bigcup S) = \bigcup_{X \in S}(A \cap X)$, $A \setminus \bigcup S = \bigcap_{X \in S}(A \setminus X)$, and $A \setminus \bigcap S = \bigcup_{X \in S}(A \setminus X)$. Aug1 answered Some questions about the Jech's book (Generalized De Morgan's law and distributive law ) Aug1 awarded Teacher Jul31 comment Proof for the symmetric difference? @HagenvonEitzen: Suppose $A \cap B \neq \emptyset$. Then we can choose some $x$ such that $x \in A \cap B$. This means that $x \in A$ and $x \in B$. Since $x \in A$, it follows that $x \in A \cup B$. But then since $A \Delta B = A \cup B$, $x \in A \Delta B$, so $x \in A \cup B$ and $x \notin A \cap B$. This contradicts the fact that $x \in A \cap B$. Therefore $A \cap B = \emptyset$. Jul31 answered Proof for the symmetric difference?