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Sep
1
comment Prove that $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \}$, where $D_a = \{ (x,y) \in \mathbb{R}^2\ | \ |x-y| < a \}$
Nice answer! Any tip on how you came up with this $z$?
Sep
1
comment Prove that $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \}$, where $D_a = \{ (x,y) \in \mathbb{R}^2\ | \ |x-y| < a \}$
Suppose $D_r$ and $D_s$ are both relations on $\mathbb{R}$. Then $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ \exists z \in \mathbb{R}((x,z) \in D_s \text{ and } (z,y) \in D_r) \}$
Aug
24
comment Is it true that $\forall b \forall c \forall x ((x^2 + bx + c \neq 0) \rightarrow b^2 - 4c < 0)$?
Great answers! Thanks, I understand now.
Aug
24
comment Is it true that $\forall b \forall c \forall x ((x^2 + bx + c \neq 0) \rightarrow b^2 - 4c < 0)$?
Yes, your answer helped a lot. Is it true that if $\forall b,c,x.\ P(b,c,x) \to Q(b,c)$ then $\forall b,c.\ (\forall x.\ P(b,c,x)) \to Q(b,c)$? How can we prove that they are not equivalent? Any hint?
Aug
21
comment Indicating when $|x + y + z| = |x| + |y| + |z|$ holds
Nice approach! Thank you :-)
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
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
5
comment If $\forall \mathcal F(\bigcup \mathcal F = A \Rightarrow A \in \mathcal F)$ then A has exactly one element
Hey! Kalispera :)
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 $A \setminus \bigcap S = \bigcup_{X \in S}(A \setminus X)$.
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
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
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?
May
22
comment If $x^2\equiv 1 \pmod{n}$ and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of n
Thanks! I'm new to number theory, so I have some questions. Why the latter assumption implies that $\gcd$s are $<n$? And why every nontrivial, proper divisor of n is prime?
May
22
comment If $x^2\equiv 1 \pmod{n}$ and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of n
@JyrkiLahtonen: Yes, you're right! I'm studying an RSA problem and trying to understand the proof of the Fact 1 on page 205 from this paper.
May
22
comment If $x^2\equiv 1 \pmod{n}$ and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of n
Oops, sorry. I'm reading this article and may have misinterpreted the problem.
Jun
30
comment Is $3^n - 2^n$ composite for all integers $n \geq 6$?
@Mavris: Thank you! Do we know whether or not there are infinitely many primes of that form?
Jun
30
comment Is $3^n - 2^n$ composite for all integers $n \geq 6$?
Cameron Buie: Important remark! Byron Schmuland: Nice webpage! Thank you guys.