Looking at the truth table, they're equivalent:  A   B      (A+B)  (A+A'B) ------------------------------- 1   1       1       1 1   0       1       1 0   1       1       1 0   0       0       0   But what manipulation can one do using basic identifies and laws to show that they're the same?
77653
For me, it is pretty clear that $a + \overline{a}*b = a + b$, because the first $a$ in the or will make sure that if the second term must be 'evaluated', $a$ will always be false, and therefore won't matter in the and - probably because I'm a programmer and it is very common to see this unnecessary cluttering in long if-else structures.  I can easily prove it with the truth table as well, just make both expressions', and turns out they are identical.  But I was looking for a 'algebraic' way to prove it. That is, something using the basic operations properties like the fact that they are transitive, commutative, associative, etc. For example, if I was to prove this other simple identity $a + a*b = a$ I would simply do something like:  $a + a*b = a*(1 + b) = a*1 = a$.  Is there a way to also prove the former using this algebraic operations, without resorting to the truth table?
56767
I have a set identity: $(A \cap B) \cup C = A \cap (B \cup C)$ if and only if $C \subset A$.  I started with Venn diagrams and here is the result:   It is evident that set identity is correct. So I started to prove it algebraic:  1) According to distributive law: $(A \cap B) \cup C = (A \cup C) \cap (B \cup C)$  2) ...  I stuck a little. Because $C$ is a subset of $A$. I thought of pulling out: $(B \cup C)$ but it seems wrong step to me.  How to prove this identity having in mind that $C \subset A$?  Updated  Venn diagram for $C ⊈ A$
57696
I am having trouble proving this directly. I know that it is easy to prove by contradiction by assuming both $a$ and $b$ to be odd, but how should I start to try to prove this? This is a homework problem and I am just looking for help getting started, not the full answer.
101417
Can anyone help me to understand what is suppose to do in this question?  "A matrix that is similar to the identity matrix" I should say something about this but I do not understand what is meant to do.  I know that two n x n matrices A and B are similar if $B=P^{-1}AP $.   They also have some properties such as:  The same:  Rank; Characteristic equation;  Determinant; Trace; Eigenvalues etc.   However I do not understand the question and consequently I do not know how to start.  Can anyone help me on this?  Thanks
58551
Form the assumption, we can say $b=ak$ ,$k$ integer, $a=bm$, $m$ integer.  Intuitively, this conjecture makes sense. But I can't make further step.
24150
Question:  Show that if $ \ A \cup B = A$ and $ \ A\cap  B = A$ then $ \ A = B$  My attempt:  Proof by contradiction:  Assume $ \ A \cup B = A$ and $ \ A\cap  B = A$ and $ \ A \neq B$   Case 1: $ \exists \ x \in A, x\notin B$  If $ x \in A \implies x\in A \cap B \implies x \in A \ and \  x\in B  \implies x\in B$, a contradiction.   Case 2: $ \exists \ x \in B, x\notin A$  If $ \ x \in B \implies x\in A \cup B \implies x\in A$, since $ \ A \cup B = A$. Contradiction.   Is this approach correct? Could someone please show me how to do a direct proof?
99172
The simple proof goes:  Let B be the left inverse of A, C the right inverse.  C = (BA)C = B(AC) = B  This proof relies on associativity yet does not give any insight as to why this surprising fact about matrices is true.  AC means a bunch of linear combinations of of columns of A. CA means a bunch of linear combinations of rows of A. Completely different numbers get multiplied in each case.  The proof above is just a bunch of syntactical steps not having much to do with matrices directly, I cannot see how CA=AC=I.  Can anyone shed some light on this?
117409
Let $A$, and $B$ be commuting $n\times n$ matrices, i.e. $A.B = B.A$. Let \begin{equation} \exp(A) = \sum_{i=0}^\infty\frac{1}{i!} A^i \end{equation} show that $\exp(A+B) = \exp(A).\exp(B)$. Edit: To those who choose to vote for closing of this question and similar ones as being a duplicate. I think the relative timing of the two questions are actually important! When there are two duplicate questions, I think it makes highly more sense to check the relative timings of the two questions and in general, tend to vote to closing of the question which is asked after (Interestingly, for this question, the duplicate is asked 5 years later than this one). Please note that if their research was done properly or the StackExchenge's duplicate detection worked perfectly at the time, they would not have been even able to post. Now by unnecessarily closing well attended questions of earlier contributors (here me) I am wondering what behavior is actually being rewarded?
18157
I've done a truth table after reducing it to this and it seems to be equal to $\neg Q$:  $$\lnot Q \lor (\lnot Q \land R) = \lnot Q$$  But when I try to show it without a truth table (with just transformations), I end up in a loop between that and:  $$\lnot Q \land (\lnot Q \lor R)$$  Is there a way to show this is true without using a truth table? What am I missing?!  Thanks in advance!
38840
I have gone through various sources... But I am not quite sure what it is.I want an and gate and the logical equivalent is two inputs feeding to one gate and for Y=AB' the logical equivalent is feeding to one not gate and one and gate. But it is the same LUT for both AND and Y=AB'. I think we store the values as desired in the LUT. Someone elaborate on this
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!
38627
Let $a$ and $b$ be integers. Show that if $(a,b) = 1$ then $(a,a + b) = 1$.   I'm stuck on how to do this proof. I tried using the fact that $ax+by=1$ but that didn't seem to help.
47928
Determine if this is true or not:  For any natural numbers $a$, $b$, $c$, and $d$, if $a | c$ and $b | d$ then $ab|cd$.  My answer was True, I went:  $a = 2$, $b = 4$, $c = 6$, and $d = 8$. If $2|6$ and $4|8$, then $(2 * 4) | (6 * 8)$.  Is there another way to do it using, for example, a contradiction proof?
62597
I know that this is valid because by thinking about it, you can see it's true, but I'm just not sure how I would go about actually proving it mathematically. I've kind of tried proof by contradiction, since that's what I use most of the time, but it doesn't seem to work.    In my mind, A ∪ B ⊆ C kind of just implies that A ⊆ C and B ⊆ C.
44977
Please don't sign this as a duplicate of , they aren't. I am interested about the actual algorithm, not a proof of the existence.  Does a such mapping exists? I think, it must, because both of them have the same cardinality ($\aleph_1$).  My actual suggestion for a such mapping were simply merge the digits of the real numbers. For example, from $\pi$ and $e$ we could got  $3.1415...$ and $2.7182...$ would lead to $32.17411852...$ .  Although I am not sure it were ok. Maybe a better, clearer solution also exists?
114311
Assume that my definition of a Boolean algebras is the following one: I have a set $B$ with two binary operations $\vee$ and $\wedge$ which both satisfy the commutative, associative and distributive laws. Both the operations have an identity element ($0$ and $1$) and for each element $b\in B$ there exists a complement (i.e. an element $b{'}\in B$ such that $b\vee b{'}=1$ and $b\wedge b{'}=0$).  Now I found the following definition of Boolean Algebra. Let $B$ be a set on which we define a binary operation $\cap$ and an unary operation ${'}$. We also require that those operations satisfy the following rules:  1) $x\cap y=y\cap x$ for each $x,\, y\in B$  2) $(x\cap y)\cap z=x\cap (y\cap z)$  3) $x\cap y{'}=z\cap z{'}$ iff $x\cap y=x$  Are those equivalent definitions? I was trying to work the thing out in the following way.  a) First we may put $0:=z\cap z'$ by property 3).  b) We may define $x\cup y:=(x{'}\cap y{'}){'}$ and put $0{'}=:1$ and we may prove the commutative law for $\cup$.  Anyway I cannot prove the associativity of $\cup$ or either the distributivity or even the fact the $0$ and $1$ are identity elements.
1802
I`m trying to prove this claim and I need some advice how to continue, $$(A \setminus B) \cup (A \setminus C) = B \Leftrightarrow A=B , C\cap B=\varnothing$$ what I did is: $$(A \setminus B) \cup (A \setminus C) = (A \cap B') \cup (A \cap C') = A \cup (B' \cap C')$$ thanks!