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visits member for 1 year, 11 months
seen Apr 15 at 12:36

Jan
11
comment Will SADMEP always work to evaluate the inverse of a function, and I should not evaluate right to left?
I see that it is false, I was very suspicious of that statement and now I see why I was mistaken.
Jan
8
comment Fixed: Is this set empty? $ S = \{ x \in \mathbb{Z} \mid \sqrt{x} \in \mathbb{Q}, \sqrt{x} \notin \mathbb{Z}, x \notin \mathbb{P}$ }
I want to know if any numbers greater than 1 that are not prime have a root that is not irrational. I will try and make a new question that is better.
Jan
8
comment Fixed: Is this set empty? $ S = \{ x \in \mathbb{Z} \mid \sqrt{x} \in \mathbb{Q}, \sqrt{x} \notin \mathbb{Z}, x \notin \mathbb{P}$ }
How do I rewrite the question to say X is integer and not prime? Is there a notation for the prime numbers?
Jan
8
comment For every set A, the empty set is a subset of A. The empty set is a set. Therefore, the empty set has a cardinality >= 1..
I understand now, thank you.
Jan
8
comment For every set A, the empty set is a subset of A. The empty set is a set. Therefore, the empty set has a cardinality >= 1..
That is only because the empty set is the empty set. But A is not the empty set. Yes the empty set is not a proper subset of itself, I agree. For it would need to have an element of itself which itself does not have, which it has none to begin with.
Jan
8
comment For every set A, the empty set is a subset of A. The empty set is a set. Therefore, the empty set has a cardinality >= 1..
Yes there is no element of $\emptyset$ that is not in $A$ because it has none.
Jan
8
comment For every set A, the empty set is a subset of A. The empty set is a set. Therefore, the empty set has a cardinality >= 1..
That makes a bit more sense too.. according to your set definitions $\emptyset \notin A$, but, $\emptyset \subseteq A$, and also I think $\emptyset \subset A$
Jan
8
comment For every set A, the empty set is a subset of A. The empty set is a set. Therefore, the empty set has a cardinality >= 1..
Yes that makes sense because $\emptyset = \emptyset$
Jan
8
comment For every set A, the empty set is a subset of A. The empty set is a set. Therefore, the empty set has a cardinality >= 1..
So an element is something without the {} curly brackets. And since $\emptyset$ has such brackets, it does not fit.
Jan
8
comment How does one show using algebra or basic mathematical prowess to show that ψ = 1 - φ
Thank you, that does not make me feel dumb at all!
Dec
13
comment Have I expressed this proposition correctly?
Okay, I think my miscalulation occurred when I said "not for all" and then I magically threw in another negation and said "not not for all," thanks for the feedback.
Oct
16
comment What are algebraic steps can I use to solve this equation or get it into quadratic form neatly?
Thank you, that advice was roughly 10 times more effective than what the T.A's tried to do.