I found this one more challenging, I think the length of it confused me the most.    Given Statement: $\forall x \in \Bbb N \left[\left(\exists y \in \Bbb N \left(2 \le y \land y \lt x \land y \mid x \right)\right) \to \left(\exists z \in \Bbb N \left(2 \le z \land z \le \sqrt x \land z \mid x\right)\right)\right]$  So far I have written it as: For all $x$ in the natural numbers and there exists $y$ for all natural numbers where $z$ is smaller or equal to $y$ if and only if $y$ is smaller than $x$ if and only if $y$ is a divisible of $x$ then, $\cdot \cdot \cdot$  I only wrote it up to the $\rightarrow$, but not sure if it makes sense and furthermore if it's correct. Thus, can I continue like this for the next part.  Any edits of my existing statement and any help in converting to plain English is welcomed. Thank you in advance.
120749
Please help me to prove this statement. I'm very confused, Thank you! I Have tried prove it using examples but it needs to be proved using mathematical method
108689
I have several of these types of problems, and it would be great if I can get some help on one so I have a guide on how I can solve these.  The question is:      Prove $A \setminus (B\setminus C) \neq  (A\setminus B) \setminus C$   I know I must prove both sides are not equivalent to each other to complete this proof. Here's my shot: We start with left side.   if $x$ is in $A$, then $x$ is not in $B$, not not in $C$ so $x$ is in $A$ and $C$ if $x$ is in $A$, then it's not in $B$ $\rightarrow$ $(A\setminus B)$ but if $x$ is in $A\setminus B$, then it must not be in $C$ however, earlier we stated $x$ is in both $A$ and $C$ we see that the two sides are not equal   Is this the right idea?  Should I then reverse the proof to prove it the other way around, or is that unnecessary?  Should it be more formal?  Thanks!
50635
Question    Suppose we are dealing with real valued functions $f(x)$ of one real variable $x$. We say that the limit of $f(x)$ at point $a$ exists and equals to $L$ if and only if  $$\exists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)$$  and show this by the symbolism  $$\mathop {\lim f(x)}\limits_{x \to a}  = L$$  Now, what do we say when we want to state that the limit of function $f(x)$ does not exist at $a$? In fact, what is the negation of the above statement? I am interested to obtain the negation with a step by step approach using tautologies in logic. To be specific, I want to start from  $$\neg \left[ {\exists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)} \right]$$  and then go through it to get the final form of negation (See the example below).    My Thought  I just wrote down the two following negations without going through a step by step approach.  $$\forall L,\exists \varepsilon  &gt; 0:\left( {\forall \delta  &gt; 0,\exists x:0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| \ge \varepsilon } \right)$$  or  $$\nexists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)$$  I want to know weather these are true or not.       Example    I will give an example of what I mean by a step by step approach. Consider the following statement  $${P}\implies R$$  I want to take a step by step approach to obtain the negation of the above statement. Here is what they usually do in logic  \begin{align} \, \neg \left( {P \implies R} \right) &amp;\iff \neg \left( {\neg P \vee R} \right) &amp; \text{Conditional  Disjunction} \\ \qquad \qquad \quad &amp;\iff \neg \neg P \wedge \neg R &amp; \text{Demorgan's  Law} \\ \qquad \qquad \quad &amp;\iff P \wedge \neg R &amp; \text{Double  Negation} \end{align}
50394
Denote by $\mathbb{N}$ the set of natural numbers and by $2^{\mathbb N}$ the set of all subsets of $\mathbb N$. Let $E$ be some subset of $2^{\mathbb N}$ such that for  every pair of elements in $E$, one is a subset of the other. Clearly, $E$ could be countable by considering $E=\big\{\{1,\cdots, n\}, n\in\mathbb N\big\}$. My question is the following: could we find an example that $E$ is uncountable? Many thanks for the answer!
123907
Question:  Part a:  Prove that for any $ b\in \Bbb N,$ if $ 2^{3b -1} + 5 . 3^{b}$ is divisible by $ 11$, then $ 2^{3(b+2) -1} + 5 . 3^{b+2}$ is divisible by $11$.  Part b:  Is statement 1 or statement 2 true? Explain answer.   For any odd number $a\in \Bbb N $, $ 2^{3a -1} + 5 . 3^{a}$ is divisible by $ 11$ For any even number $a\in \Bbb N $, $ 2^{3a -1} + 5 . 3^{a}$ is divisible by $ 11$   My attempt:  Part a:  I am not sure what the base case should be.  Induction hypothesis: Assume $ 2^{3k -1} + 5 . 3^{k}$ is divisible by $ 11$, for some $k$ natural number.   I am not sure how to prove true for $ 2^{3(k+2) -1} + 5 . 3^{k+2}$.  Part b:  Would statement 2 be correct since the expression is divisible by $11$ when $ a=2$
55029
Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally:     $$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$   If this statement is true, what is the proof ?
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?
14753
I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides.
101103
We are given with  $$S_n=\{(x,y)|x^2+y^2=n, \text{where }x,y,n \in \mathbb Z\text{ and }n\geq0  \}$$ Now observe that $-\sqrt n\leq x\leq\sqrt n $ and similarily, $-\sqrt n\leq y\leq\sqrt n $. This means that $-\left \lfloor{\sqrt n}\right \rfloor\leq x\leq \left \lfloor{\sqrt n}\right \rfloor$ and  $-\left \lfloor{\sqrt n}\right \rfloor\leq y\leq \left \lfloor{\sqrt n}\right \rfloor$. Through this result one can easily write code and figure out the cardinality of set $S_i$ but I was wondering whether an expression can be found that would allow one to directly compute $|S_i|$.
93710
Statement: There are uncountably many surjections from $\mathbb N$ to $\mathbb N$.  Is the following proof valid:  The real numbers are uncountable.   There exists a bijection between the set of real numbers and the set of all functions from $\mathbb N$ to $\{0,1\}$.   Given a function, $f$, from $\mathbb N$ to $\{0,1\}$, construct a surjection from $\mathbb N$ to $\mathbb N$, denoted $g$, as follows:  We map the $n$th composite number to $n$, and the nth prime, to $f(n)$.   Clearly this is a surjection since the composite numbers are countable. Clearly different functions $f$ lead to different surjections $g$. Since there are uncountably many $f$s, there are uncountably many surjections $g$.
100786
Show that the logic $ \ \forall x \exists y \ P(x,y) \ $ is not logically equivalent to $ \ \exists x \forall y \ P(x,y) \ $  in the domain $ \ \{-1,0,1 \} \ $ by giving a counter example .  Answer:  Let   $ x=\{0,1 \} \\ y=\{-1,0 ,1 \} \ $  Let $ P(x,y) \ $ be the property such that $ \ x \geq y \ $   Then , $ \forall x \exists y \ P(x,y) \ $ is true.   But, $ \ \exists x \forall y \ P(x,y) \ $ is False. Because for $ x=0 \ $ , we have $ \ 0 \ngeqslant y=1 \ $.  I need confirmation of my work.
103593
Assume that a,b,n are all natural numbers.  I was going to set it up as:  na = q(1)*n(b) + r(1)     where a>b  and go down the chain:  nb = q2 * r(1) + (r2)  but something seems off. Someone told me that in the first part, it should be n*r(1), but I don't understand why. May I please have some direction?
21599
If I know for two numbers a and b that    $${a &lt; b }$$  Then is it correct to say that    $$ a \leq b $$  I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information.
48176
I'm having trouble with this homework question:  "Let there be a sequence $\left(b_{n}\right)_{n=1}^{\infty}  $ such that $\lim_{n\rightarrow\infty}\left(b_{n}\right)=1$. Let there also be some bounded sequence  $\left(a_{n}\right)_{n=1}^{\infty}$  Prove that the following statement is correct: $\limsup\left(b_{n}a_{n}\right)=\limsup\left(a_{n}\right)  $"  Alright so I sat a good two hours on this question with no major breakthroughs as of yet. I have some semi-proofs all of which use some arbitrary assumptions which I can't make.    Any hints?  Thank you!
116156
Having some issue with some logic. The question is to formally prove;  $$q \land \neg q \vdash r \land \neg r$$  I've never done this before so would appreciate some help with it. No idea really where to start as $r \land \neg r$ doesn't exist on the left hand side of the equation. I guess I break down the left hand side first but then I don't know where to go.   Would really appreciate any help anyone can provide.  Thanks.  Update  Have attempted it myself but still not sure what to do. With the first part of the formula $q \land \neg q$ ... do I declare this as false? Really have no idea where to go with this, other than deducting the $q \land \neg q$ apart.   Would really appreciate some help on this.   Thanks.
7716
Definition: The least common multiple of $a.b\in\Bbb{Z}$ is the smallest  $k\in\Bbb{N}$ that is divisible by both a and b.         Proposition: $\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))\tag{2}$    Proof: We want to show $\text{lcm}(\text{lcm}(a,b),c)\mid\text{lcm}(a,\text{lcm}(b,c))$ and $\text{lcm}(a,\text{lcm}(b,c))\mid \text{lcm}(\text{lcm}(a,b),c)$ such that $\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))$.  Let $k=\text{lcm}(\text{lcm}(a,b),c)$, then $c\mid k$ and $\text{lcm}(a,b)\mid k$, but if we let $l=\text{lcm}(a,b)$, then $a\mid l\,\,\text{and}\,\, b\mid l$. Therefore $c\mid k\,\,,a\mid k\,\,\text{and}\,\, b\mid k$, but then $\text{lcm}(b,c)\mid k$. Thus $\text{lcm}(a,\text{lcm}(b,c))\mid k$      Question: Does this makes sense, ifso can I do the same for $\text{lcm}(\text{lcm}(a,b),c)\mid \text{lcm}(a,\text{lcm}(b,c))$?
29661
Why is "$P \Rightarrow Q$" equivalent to "$\neg Q \Rightarrow \neg P$"?  I am trying to understand why this does always apply, in terms of pure logic. Can you please explain it to me?