- May 2021
-
math.libretexts.org math.libretexts.org
-
The hand is a straight (consecutive ranks, as in 5, 6, 7, 8, 9, but not all from the same suit).
The back of the book gives the answer as 10(4^5-1), which is incorrect. There are 10 4^5 "straights" where the suit doesn't matter, the sequence can even all be the same suit among the 10 4^5 such hands. So you want to subtract out the "straights" of the same suit, there are 40 such hands. There are 10 possible sequences 1)A2345 2)23456,3)34567,...10)10JQKA. Each sequence can be one of the 4 suits so thats 4 10=40 total. A straight of all the same suit is a straight flush, which is part h of the question. So the final answer to the number of straights but not from the same suit is 104^5-40.
-
-
math.libretexts.org math.libretexts.org
-
m+nm∼n⇔m+nm\sim n \,\Leftrightarrow\, m+n is even
The answer in the back of the book states that there are 3 equivalence classes(ECs): [0], [1], & [2]. But I think it is incorrect and there are only 2, one for evens and one for odds because if you add any 2 even numbers you get an even sum, same if you add 2 odd numbers. Only way you get an odd sum is if you add one even and one odd #. So all the even numbers would belong to the even EC and all the odd numbers would belong to the odd EC.
-
|m−3|=|n−3
The answer in the back of the book states that the equivalence classes(ECs) are of the form {3−k, 3+k} but I dont think that is correct. I believe the correct answer is that the ECs are [a] where a \(\in\) Z, so an equivalence class for each integer in Z.
-
-
math.libretexts.org math.libretexts.org
-
math.libretexts.org math.libretexts.org
-
4725
Answer in the back of the book is incorrect. Correct answer is (3^3)(5^2)(7^1)
-
-
math.libretexts.org math.libretexts.org
-
[−1,1]⋂x∈(1,2)(1−2x,x2)=[−1,1]\bigcap_{x\in(1,2)} (1-2x,x^2) = [-1,1], ⋃x∈(1,2)(1−2x,x2)=(−3.4)
Correction: Part a) (-1,1) Part b) (-3,4)
-
En=E0={0}
Correction: Since 0 is not a natural number and 1 is the smallest natural number, the correct answer to part a,the intersection question is {-1,0,1}.
-
{(−2,−3),(−2,0),(−2,3),(−2,−3),(−2,0),(−2,3)}{(−2,−3),(−2,0),(−2,3),(−2,−3),(−2,0),(−2,3)}\{(-2,-3), (-2,0), (-2,3), (-2,-3), (-2,0), (-2,3)\}
Correction, should be: {(−2,−3),(−2,0),(−2,3),(2,−3),(2,0),(2,3)}
-
true
2 is not a member of the interval (2,3) because (2,3)={x | 2<x<3}. It does not include 2. So the answer should be false.
-
PQRS is a parallelogram
Should read: "PQRS is not a parallelogram."
-
For all integers nnn, if nnn is prime and nnn is even, then n≤2n≤2n\leq2.
There's a typo, it should read: For all integers n, if n is prime and n>2, then n is not even.
-
true
I think there's an error in this solution. The left hand side of the equation in question 2.5.9, is equivalent to the truth value True. Thus, it is not equivalent to Not(P).
-