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>> No.11640839 [View]
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11640839

>>11640512

>> No.11615629 [View]
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11615629

Good morning /mg/!

>>11614799
Some comments:
(12) You claim you changed the set, but that is actually not true. The set is the same. What you are trying to say is that you use the reduced form p/q of n/m, where the greatest common divisor for p & q is 1.
(13) What you are trying to say is that 2 and 3 are prime numbers. The reasoning is correct.
(14) Sufficient and necessary condition can be easily remembered if you look at the truth table of [math]p \rightarrow q[/math]. If we assume that the implication itself is true, then p being true is a sufficient condition for q to be true. Conversely, if the implication is true, then q being true is a necessary condition for p to be true.
(17) & (18) There is always the identity. There are actually m! bijections for a set of m elements. If m=2, you also have f(1)=2, f(2)=1. Note that if there are m! > 1 bijections from S to itself, then there must be more than 1 function from S to itself!
For the last one, I can't think of anything quickly. Keep up the good work!

>>11614814
I'm doing a PhD. I have a bf, but we are not in the same country.

>>11614756
Suppose [math]g^n = g^m[/math], for [math]m<n[/math], so that [math]g^m = g^m\cdot g^{n-m}[/math]. Inverting gives us [math]e \neq g^{n-m}=e[/math], a result most problematic.

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