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Aug
20
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.
Aug
20
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!
Aug
20
accepted An incorrect proof by exhaustion
Aug
20
accepted Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$
Aug
20
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.
Aug
20
comment Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$
@AndrewD.Hwang: Nice observation. Thank you!
Aug
20
asked Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$
Aug
5
accepted If $\forall \mathcal F(\bigcup \mathcal F = A \Rightarrow A \in \mathcal F)$ then A has exactly one element
Aug
5
comment If $\forall \mathcal F(\bigcup \mathcal F = A \Rightarrow A \in \mathcal F)$ then A has exactly one element
Hey! Kalispera :)
Aug
5
asked If $\forall \mathcal F(\bigcup \mathcal F = A \Rightarrow A \in \mathcal F)$ then A has exactly one element
Aug
2
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 $A \setminus \bigcap S = \bigcup_{X \in S}(A \setminus X)$.
Aug
1
answered Some questions about the Jech's book (Generalized De Morgan's law and distributive law )
Aug
1
awarded  Teacher
Jul
31
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 $A \cap B = \emptyset$.
Jul
31
answered Proof for the symmetric difference?
Jul
31
comment An incorrect proof by exhaustion
In both cases I prove that $\exists y \in R(xy^2 \neq y - x)$. Then I say that since was arbitrary, $\forall x \in R \exists y \in R(xy^2 \neq y - x)$. Is that all right?
Jul
31
comment An incorrect proof by exhaustion
"When you choose cases, you lose the supposition that x is arbitrary." Now consider the statement $\forall x \in R \exists y \in R(xy^2 \neq y - x)$. To prove it, I assume that x is an arbitrary real number. Then I consider the following cases: Case 1. $x = 0$, Case 2. $x \neq 0$. Do I restrict what my x can be in this particular case?
Jul
31
asked An incorrect proof by exhaustion
Jul
24
comment A proof by contradiction
Yes, in this particular case assumption is a better word choice because I assumed that $x \notin A$. Is assumption generally a better word choice than fact?
Jul
24
revised A proof by contradiction
added 28 characters in body