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seen Oct 10 at 3:57

Jan
11
comment Will SADMEP always work to evaluate the inverse of a function, and I should not evaluate right to left?
Yes I am taking it as an algorithmic method to evaluating, where it uses the reverse steps as PEMDAS. Thank you for clarifying why this SADMEP is not "literally" the opposite in the sense that I have considered it.
Jan
11
comment Will SADMEP always work to evaluate the inverse of a function, and I should not evaluate right to left?
So there is no reason to use SADMEP, and infact it is not a correct method? I think maybe if there is a correct SADMEP, it is very ambiguous and apparently beyond me at this point. That is, starting with subtraction and ending with exponents/parentheses is itself not correct.
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
11
asked Will SADMEP always work to evaluate the inverse of a function, and I should not evaluate right to left?
Jan
8
accepted 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}$ }
Jan
8
revised 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}$ }
changed question again
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
awarded  Commentator
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
asked 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}$ }
Jan
8
revised 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..
added 48 characters in body
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
accepted 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..
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
asked 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..
Jan
8
accepted How does one show using algebra or basic mathematical prowess to show that ψ = 1 - φ