I was given the following task  Define a bijection between ${(A^B)}^C$ and $A^{B \times C}$  I have no clue where to start and I don't understand the task. Could you please explain what I am required to show and how should I start the proof ?
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One of the basic (and frequently used) properties of cardinal exponentiation is that $(a^b)^c=a^{bc}$.  What is the proof of this fact?  As Arturo pointed out in his comment, in computer science this is called .
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I'm well aware of the standard proof based on cardinality arithmetic to show that these two sets have the same cardinality and the question of defining a bijection between the two sets came up. I worked on it a little while and was wondering if there were any gaps, or a simpler method of coming up with such a bijection.   The main idea I had was to take a family of sequences indexed by the real numbers. So that each one of these sequences corresponds to a unique real number. Then I can map each one of these sequences to another number, by applying the binary function from $\mathbb R$ to $\{0,1\}$ and mapping the infinite binary sequence to a real number, use one of the standard bijections.  First let $E$ be a family of sequences indexed by the real numbers. Such that $E_{xi}\neq E_{yj}$ for all $x,y \in \mathbb R$ and $i,j \in \mathbb N$ also with the property that  $\mathbb R= \bigcup_{x \in \mathbb R, i \in \mathbb N} E_{xi}$.   What sort of theorems would I need to prove such a family existed, is it enough to note that the reals can be represented as a uncountable union of countably infinite sets?  Also let $P: 2^{\mathbb N} \rightarrow \mathbb R$ be you favorite bijection.  Now consider an element of $2^{\mathbb R}$ as a function $f: \mathbb R \rightarrow \{0,1\}$ and $g \in \mathbb{ R^R}$ in a similar manner. Then our bijection between the two sets will be the function $\Phi$ which maps $f$ to a function $g$, such that $g(x)=P(F(E_x))$ where $F$ applies $f$ to each element of the sequence $E_x$.   I'm fairly sure this is an onto and one-to-one function. The one-to-one property should follow because the sequences are disjoint, cover $\mathbb R$ and $P$ is a bijection. The onto argument is a little hazier, but it should be possible to recreate the function $f$, by first inverting each element with $P$ and then defining $f$ based on these sequences.
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Let $f = aX^2 + bX + c \in \mathbb{R}[X]$. I want to show that:   $\mathbb{R}[X]/(f) \cong \mathbb{C}$ if $b^2 - 4ac &lt; 0$,  $\mathbb{R}[X]/(f) \cong \mathbb{R}[\epsilon]$ if $b^2 - 4ac = 0$,  $\mathbb{R}[X]/(f) \cong \mathbb{R} \times \mathbb{R}$ if $b^2 - 4ac &gt; 0$.  My idea so far is to prove that in the first case, $(f) = (X^2 + 1)$, in the second case $(f) = (X^2)$ , in the third case $(f) = (X^2 - 1)$, which would lead to the desired result. However, I don't know how to prove this or if I'm on the correct road at all. Does anyone have any ideas?
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Under what conditions on the sets $A$ and $B$ will it be true that $A\times B= B\times A$?   I think it should be set $A$ $=$ set $B$ but don't know how to prove it or is there any other way for those two cross product to be true?
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I need to give an explicit bijection between $(0, 1]$ and $[0,1]$ and I'm wondering if my bijection/proof is correct. Using the hint that was given, I constructed the following function $f: (0, 1] \to [0,1]$: $$ x \mapsto \left\{ \begin{array}{ll} 2 - 2^{-i} - 2^{-i-1} - x &amp; \text{if } x \in (1-2^{-i}, 1-2^{-i-1}]\text{ for an } i \in \mathbb{N}_0 \\ 1 &amp; \text{if } x = 1 \end{array}\right. $$  It's easy to see that for every $x \in (0, 1)$, there exists such an $i$.  Now define $\tilde{f}: [0,1] \mapsto (0,1]$ with  $$ x \mapsto \left\{ \begin{array}{ll} 2 - 2^{-i} - 2^{-i-1} - x &amp; \text{if } x \in [1-2^{-i}, 1-2^{-i-1})\text{ for an } i \in \mathbb{N}_0 \\ 1 &amp; \text{if } x = 1 \end{array}\right. $$  I want to prove that $\tilde{f}(f(x)) = f(\tilde{f}(x)) =  x$, so it has an inverse and therefore is a bijection. The case $x=1$ is trivial, so assume that $x \in (0,1)$ with $x \in (1-2^{-i}, 1-2^{-i-1}]$ for some $i \in \mathbb{N}_0$. This interval has length $1-2^{-i-1} - (1-2^{-i}) = 2^{-i-1}$, so we can write $x = 1-2^{-i} + \epsilon\cdot 2^{-i-1}$ for some $\epsilon \in (0, 1]$. We now calculate $f(x)$: \begin{align*} f(x) &amp;= 2 - 2^{-i} - 2^{-i-1} - x\\ &amp;= 2 - 2^{-i} - 2^{-i-1} - (1-2^{-i} + \epsilon\cdot 2^{-i-1})\\ &amp;= 1 - 2^{-i-1}(1+\epsilon). \end{align*} We conclude that $f(x) \in [1-2^{-i}, 1-2^{-i-1})$. We now use the definition of $\tilde{f}$, so if we calculate $\tilde{f}(f(x))$, we get \begin{align*} \tilde{f}(f(x))  &amp;= 2 - 2^{-i} - 2^{-i-1} - f(x) \\ &amp;= 2 - 2^{-i} - 2^{-i-1} - (2-2^{-i} - 2^{-i-1} - x) \\ &amp;= x. \end{align*}  We conclude that $f$ has an inverse. Using exactly the same reasoning, we get that $f(\tilde{f}(x)) = x$ for all $x \in [0,1]$. Therefore it's inverse exists and it has to be a bijection.  I know there are less cumbersome methods of proving this fact, but as of now this is the only thing I can come up with.
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Why does $a^{b}·a^{c}=a^{b+c}$ ?  I want proof for it, I asked professor and he replied: "I don't know, it's a property anyway that is true and that's all what you need to  know."
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The  lists various combinatorial objects that are described by them. I posit that there might be bijections between these various combinatorial objects. For some of them (like Dyck paths, correctly matched parentheses and paths from the bottom left to top right of a $2n \times 2n$ grid) they are quite obvious.   I was then trying to find a bijection between the number of Dyck words and the number of ways of associating $n$ applications of a binary operator to $n+1$ items (third one on the list). I attempted to do this for a simple case ($n=3$ which is the example provided in the Wikipedia article). However, couldn't find one after multiple hours. Is it reasonable to expect such a bijection will exist? If so, how do we go about finding it?    EDIT: In addition to the very nice answer by @Marc, the following page also helped me see the bijection:   "Let $P$ be the Dyck path and $f(P)$ be the binary tree. If you go up in the Dyck path, create a left child. Otherwise, go up one vertex until creating a new right child is possible and create one."    Here is one of my attempts:   Number of Dyck words with length $2 \times 3$ is $\frac{6 \choose 3}{4} = 5$. They are:  hhhttt; hhthtt; hhttht; hthhtt; hththt  And the number of applications of a binary operator among $3+1=4$ factors is:  ((ab)c)d;     (a(bc))d;     (ab)(cd);     a((bc)d);     a(b(cd))  Both combinatorial objects have been arranged in a way that there is some kind of ordering between them. For the Dyck words for example, if h equals +1 and t equals -1, then the order is lexicographical from the left to the right of the cumulative score along the sequence.   Now, the first and last characters of the Dyck words are always h and t respectively. So, we can ignore them. We are left with:  hhtt; htht; htth; thht; thth  I tried to start from the left of the sequence abcd and if I see 'h', merge the character with the one to its right. This approach didn't produce a valid mapping from the third Dyck word to the third binary operator precedence order.
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I'm so puzzled about this:   $$a^{b^c} = a^{(b^c)} \neq (a^b)^c = a^{(bc)}.$$  Why isn't $a^{b^c}$ equal to $a^{(bc)}$? Why is $a^{b^c}$ instead equal to $a^{(b^c)}$? And how is it possible that $(a^b)^c = a^{(bc)}$?  My mind is pretty much exploding from trying to understand this.
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I am trying to determine if $\Bbb Z_6 \cong   \Bbb Z_3 \times \Bbb Z_2 $. I noticed that $\Bbb Z_6$ has a generator $1$ and $\Bbb Z_3 \times \Bbb Z_2$ has generator $(1,1)$. Now I set up the bijection $f :\Bbb Z_6 \to   \Bbb Z_3 \times \Bbb Z_2 $ where \begin{eqnarray} f(1)&amp;=&amp;(1,1)\\ f(1+1)&amp;=&amp;(1,1)+(1,1) = (2,0)\\ f(2+1)&amp;=&amp;(2,0)+(1,1) = (0,1)\\ f(3+1)&amp;=&amp;(0,1)+(1,1) = (1,0)\\ f(4+1)&amp;=&amp;(1,0)+(1,1) = (2,1)\\ f(0)&amp;=&amp;(1,1)+(2,1) = (0,0)\\ \end{eqnarray} So basically $f(1^n) = (1,1)^n$ where $1^n$ means we add the $1$ to itself $n$ times and take the result mod 6, so as per usual $1^n$ means applying the group operation on 1 to itself for $n$ times. So this is a bijection but I am stuck on showing that $f(ab)=f(a)f(b)$ for all $a,b \in \Bbb Z_6$. I could check all possibilities but there must be a better way to show this. I was thinking something like $f(1^n1^m) = f(1^{n+m})=(1,1)^{n+m}=(1,1)^n(1,1)^m = f(a)f(b) $ because every element in $\Bbb Z_6$ can be expressed as some power of $1$.  Is this right? And is it enough to show $\Bbb Z_6 \cong   \Bbb Z_3 \times \Bbb Z_2 $?
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For all $ a, b \text{ and } c \in \mathbb{R}$ and $a&gt;1$, Prove that $a^b\cdot a^c=a^{b+c}$  I have come across this question and its  bugging me. Its a basic property that we learn in HS  and I was hoping someone can enlighten me
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Prove that the function $f: R\to R$ given $f(x)=x\cdot |x|$ is a bijection  Proof 1: f is injective  Suppose $$f(a) = f(b) \implies a\cdot |a| = b\cdot |b|$$ $$\implies (a\cdot |a|)^2 = (b\cdot |b|)^2$$ $$\implies a^4 =b^4$$ $$a=b$$ Thus, f is injective  Proof 2: f is surjective  Pick $n\in R$, we want to show $n=f(x)$ for some $x\in R$  $$\implies n= a\cdot |a|$$ for some $a\in R$  I dont know how to complete the surjective proof
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If $A$ and $B$ are two $4\times3$ matrices such that $A\mathbb{v}=B\mathbb{v}$ for all $\mathbb{v}\in\mathbb{R}^3$, then $A=B$?  If it's not true, can you give me an example of it?  Thank you.
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Problem: Hi, I'm trying to solve the following proof:  If $a &gt; b \geq 0$, Prove that $a^2 &gt; b^2$.  But I'm a bit new with proofs and I don't manage to write a proper one. I would be very grateful if someone could explain me which are the steps required to demonstrate the proof. Thank you.
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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.
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I'm having some difficulties continuing this problem. I get that $b^2\mid a$ and $c\mid a$ but I am not sure where to go from there.
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For which $a,b,c,d \in \Bbb R$ does $f(x) = \frac {ax+b}{cx+d}$ define a bijection $f: \Bbb R \to \Bbb R$?  I'm guessing I need a system of equations and I know that $cx + d \ne 0$.  Thanks!
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A) Specify a bijection from [0,1] to (0,1]. This shows that  |[0,1]|   =  |(0,1]|    B) The Cantor-Bernstein-Schroeder (CBS) theorem says that if there's an injection from A to B and an injection from B to A, then there's a bijection from A to B (ie, |A| = |B|). Use this to come to again show that   |[0;1]| = |(0;1]|