I can thnk of only one way to tackle this problem, however, it is very tedious. $$ a \oplus b \iff (a \land \neg b) \lor(\neg b \land a)$$ And so now I evaluate, by definition, $(a \oplus b) \oplus c$ and $a \oplus(b\oplus c)$ And then, using the associativity of OR and AND, I get the same expression for arrangements of parentheses.  However, is there an easier way to solve this problem?
123850
Through the use of Boolean algebra, show that the XOR operator ⊕ is both commutative and associative. I know I can show using a truth table. But using boolean algebra? How do I show? I totally have no clue. Any help please?
56312
Can someone help me? I can't even understand what the question is asking for.     Let $S = \{x_1,x_2,x_3,\ldots,x_n\}$. Show that ${\sim}[\forall x\in S,P(x)\lor Q(x)]\iff \exists x\in S,[{\sim}P(x)\land {\sim}Q(x)]$
49785
I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:  $$ x * y = \frac{xy}{x+y+1}$$  The book solution:  $$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}$$  $$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right) = \frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}$$  However the $(x+y+1)$ and $(y+z+1)$ terms appear to actually cancel out:  $$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} $$  $$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right)  =\frac{x\left(\frac{yz}{y+z+1}\right)}{x+\frac{yz}{y+z+1} + 1} $$  So the denominators cancel out, the two groupings are equivalent, the operator is associative.  This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something.  Sorry for bothering people with such a basic question.
7251
Suppose we have a set  $S=\{a,b\}.$ Obviously, the total number of binary operations on $S$ is the number of all the pairs such that $$*(a,b)=a \, \text{or}  \, b,$$ which gives  $ 2 \cdot 2 \cdot 2 \cdot 2=2^4$ possible ones. Is there a way to find which of these $16$ binary operations are associative without writing all the operation matrices and checking it by hand?
124057
I am working on a problem where I need to show the logical equivalence of two propositions.  One is a biconditional: $p \leftrightarrow q$.  And the other is this: $(p \land q) \lor (\lnot p \land \lnot q)$  I can get the biconditional down to this: $(\lnot p \lor q) \land (\lnot q \lor p)$  And I don't know how to change the second proposition.  Can someone give me a place to start?
122834
This question comes from from my algebra paper:  $(p \rightarrow q)$ is logically equivalent to ... (then four options are given).   The module states that the correct option is $(\sim p \lor q)$. That is:      $$(p\rightarrow q) \iff (\sim p \lor q )$$   but I could not understand this problem or the solution. Could anybody help me?
25606
Can someone give me a hint how to evaluate $$ \binom{n}{0}+\binom{n}{2}+\cdots+\binom{n}{o(n)},$$ where $o(n)$ is $n$ if $n$ is even and $n-1$ otherwise ?
42719
Could someone tell me a simple proof to If a|b and b|a iff a=+-b.
69960
If $A,B,$ and $C$ are sets, then $A\times(B-C)$ = $(A \times B)$ $-$ $(A \times C)$.  Proof. Observe the following sequence of equalities.  $$\begin{align} A\times(B-C) &amp;= \{(x,y)\} : (x \in A) \wedge (y \in (B-C))\} \, (\text{Definition of Cartesian Product}) \\ &amp;=\{(x,y) : (x \in A) \wedge \big((y \in B) \wedge (y\notin C)\big)\} \,  (\text{Definition of } -) \\  &amp;=\{(x,y) : (x \in A) \wedge (x \in A) \wedge \big((y \in B) \wedge (y\notin C)\big)\} \, (P=P \wedge P) \\ &amp;=\{(x,y) : \big((x \in A) \wedge (y \in B)\big) \wedge \big((x \in A) \wedge (y\notin C)\big)\} \, (\text{Rearrange}) \\ &amp;=\{(x,y) : \big((x \in A) \wedge (y \in B)\big) \wedge \big((x \in A) \wedge (y\notin C)\big)\} \, (\text{Definition of }\cap) \\ \end{align}$$  I'm stuck on the last part -- $(x \in A) \wedge (y\notin C)$ translates to $(A-C)$ but I need it to be $(A \times C)$. I can't quite figure out how to reach that.
987
How does one show that $A\cap B = A\setminus (A\setminus B)$?  My thinking is that if $x\in A\setminus (A\setminus B)$ then this means that $x\in A$ and $x\notin A\setminus B$, which means that $x\in A$ and ($x\notin A$ and $x\in B$), but I don't see why the expression in parentheses should simplify to just $x\in B$. Any insight would be helpful.  Maybe $(x\in A^c$ and $x\in B)$, i.e, $x\in A^c\cap B$ is a possible simplification of the expression in parentheses, but still don't see why this simplifies to just $x\in B$.
77327
Edit: I was trying to find a general form of the question but I will just put the real question.  My textbook says |a-b|&lt;|b|/2 implies |a|>|b|/2. I don't see how and the text doesn't provide a proof.
87766
Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then $$ $$ (i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$) $$ $$ (ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$; $$ $$ (iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.  Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column and the reason for the step in the second column.  My book is really vague. I'm not really sure what to do..  First I was thinking something like this: $a\mid b\Rightarrow b=as$ and  $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$ $$ $$ $b + c + c = b + 2c$  $$ $$ $=as + 2(bt+au)$  Yea then I just got lost.
1896
This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give me an understandable explanation of the proof and let me know how I'd go about finding such a bijection.
93154
I'm having problems with this; the strategy is to show the left side is in the right side, and vice versa. Thus far I can only show that both sides of the equality are in A\B. Can I use "the empty set = A and not A" as a fact? Because then it works. Or just show me how to do it!
4402
I'm trying to prove $A \cap (A \cup B) = A$. I'm stuck on the last part of my proof, not sure how to show next:  $$x \in A \cap (A \cup B)$$ $$\iff x \in A \;\;\text{and}\;\; x \in A \cup B$$ $$\iff x \in A \;\;\text{and}\;\; (\text{either}\;\; x \in A \;\;\text{or}\;\; x \in B).$$
18423
So I took a calculus course and my first homework has this one. I am stuck on this one for 2 days straight. Please help me; I will appreciate some hints.   So far we learned the axioms of a field and the axioms for inequality (trichotomy, transitivity, etc.) Do I actually need to divide it into cases where both $x,y$ are with the same sign and another case where they are with different signs?  Is there an elegant and short way for this?
23744
I may be wrong, but I was thinking of:  $\operatorname{lcm}(a, b)$ as $\min(a, b)$ and $\gcd (a, b)$ as $\max(a, b)$ and $a &lt;b &lt;c$   I know I'm wrong, but I think I can do it