195 reputation
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visits member for 2 years, 7 months
seen 4 hours ago

Jul
2
awarded  Curious
Mar
15
awarded  Notable Question
Feb
11
awarded  Notable Question
Jan
9
awarded  Citizen Patrol
Sep
25
awarded  Popular Question
Sep
2
accepted A proof by contradiction
Sep
2
accepted Is $3^n - 2^n$ composite for all integers $n \geq 6$?
Sep
2
accepted Is there a relation which is neither symmetric nor antisymmetric?
Sep
1
accepted Prove that $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \}$, where $D_a = \{ (x,y) \in \mathbb{R}^2\ | \ |x-y| < a \}$
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 $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ \exists z \in \mathbb{R}((x,z) \in D_s \text{ and } (z,y) \in D_r) \}$
Sep
1
asked Prove that $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \}$, where $D_a = \{ (x,y) \in \mathbb{R}^2\ | \ |x-y| < a \}$
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
24
accepted Is it true that $\forall b \forall c \forall x ((x^2 + bx + c \neq 0) \rightarrow b^2 - 4c < 0)$?
Aug
24
asked Is it true that $\forall b \forall c \forall x ((x^2 + bx + c \neq 0) \rightarrow b^2 - 4c < 0)$?
Aug
21
accepted Indicating when $|x + y + z| = |x| + |y| + |z|$ holds
Aug
21
comment Indicating when $|x + y + z| = |x| + |y| + |z|$ holds
Nice approach! Thank you :-)
Aug
21
accepted The set of all finite sequences of members of a countable set is also countable
Aug
21
asked Indicating when $|x + y + z| = |x| + |y| + |z|$ holds