I was wondering if my proof is right and if there are better methods of solving this proof?   My proof:   Assume  $x \equiv_k y$ and  $y \equiv_k z$ is true. This means that :   \begin{align}x &amp;= y + kq,&amp;&amp;\text{where }\;q ∈\Bbb Z,\tag{1} \\ y &amp;= z + ka,&amp;&amp; \text{where } a ∈ \Bbb Z. \tag{2} \end{align}  By substitution of $(2)$ into $(1)$ we get:   \begin{align} x &amp;= z + ka + kq \\   &amp;= z + k(a + q),&amp;&amp;\qquad \text{where }\: a ∈\Bbb Z \text{ and } q ∈ \Bbb Z \end{align}  This proves $x \equiv_k z$ is true. $\blacksquare$
57102
Let $m \in \mathbb{N}$. Show that $\,a \text{ R } b \Leftrightarrow a \equiv b \text{ }(\text{mod } m)\,$ is an equivalence realtion.   We need to show 3 things:  (i) reflexive: $\forall a \in \mathbb{Z}: a \equiv a \text{ mod } m$  (ii) symmetric: $\forall a,b \in \mathbb{Z}: a \equiv b \text{ mod } m \text{ } \Rightarrow b \equiv a \text{ mod } m$  (iii) transitive: $\forall a,b,c \in \mathbb{Z}: a \equiv b \text{ mod } m \text{ } \wedge b \equiv c \text{ mod } m \Rightarrow a \equiv c \text{ mod } m$  So I tried it like this:  (i) $m \text{ | } 0 \Rightarrow m \text{ | } a-a \Rightarrow a \equiv a \text{ mod } m$  (ii) $a \equiv b \text{ mod } m \Rightarrow a \text{ mod } m = b \text{ mod } m \Rightarrow b\text{ mod } m = a \text{ mod } m \Rightarrow b \equiv a \text{ mod } m$  (iii) $a \equiv b \text{ mod } m \text{ } \wedge \text{ }b \equiv c \text{ mod } m \Rightarrow m \text{ | } (a-b) \text{ } \wedge \text{ } m \text{ | } (b-c) \Rightarrow m \text{ | } (a-b) + (b-c) \Rightarrow m \text{ | } a-c \Rightarrow a \equiv c \text{ mod } m$  Because every condition is fulfilled, we are done (?)  Please help me and tell me if it would be correct like this? Maybe there is better way to prove it?
69512
I am trying to give a non-algebraic proof for this equality:  $$\dbinom{a}{b}\dbinom{b}{c}=\dbinom{a}{a-c}\dbinom{a-c}{a-b}$$  So far, I could only use the identity $\dbinom{x}{y}=\dbinom{x}{x-y}$.  What I got is $\dbinom{a}{a-c}\dbinom{a-c}{a-b}=\dbinom{a}{c}\dbinom{a-c}{b-c}$. But I don't think I have made any progress from there. It seems reasonable if we think about the identity, but I just find it difficult to give a proper proof for it.  Thanks for the help.
86926
Let $X,Y,Z$ be subspaces of $V$ so that $X$ is a subspace of $Y$. Prove that $Y\cap (X+Z)=X+(Y\cap Z)$  I know that I need to prove that $Y\cap (X+Z)\subseteq X+(Y\cap Z)$ and $X+(Y\cap Z)\subseteq Y\cap (X+Z)$ but I´m having a hard time.  Can you lend me a hand please? I would really appreciate it
52713
The problem statement is: for $a,b$ coprime, prove that there exists positive integers $x,y$ such that $ax-by=1$.  The question is form Arthur Egnels' problem solving text, and from the chapter on the pigeonhole principle.    My question: His proof begins by generating a list $a,\ldots,a(b-1)$, where he then points out that each element in this list(which proceeds sequentially) doesn't have remainder $0\pmod{b}$.  He then shows that we arrive at a contradiction if we assume that we also don't get remainder $1$ in this list $\bmod b$.  To show the contradiction in this last statement, he first states that we would have positive integers $p,q $ where $ 0&lt;p&lt;q&lt;b$ so that $pa\equiv qa \pmod{b}$.  He then goes on to point out that since we $a$ and $b$ are coprime, we then have that $b| q-p$.  I understand the last part here, but don't see why we get $pa\equiv qa \pmod{b}$.  Thanks in advance
6701
If $A|B$ and $B|A$ then prove $A=\pm B$   So far I have $A|B \iff AX=B$ and $B|A \iff BY=A$ with $X,Y \in \mathbb{Z}$  Not sure how to finish, any help.
74091
Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]  This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$  But how should I prove it using that hint or at least a similar proof.
116030
I know how to prove this using an element and subset proof but that's not what I'm being asked to do. I'm suppose to prove it using the distributive law and the associative law but I'm just not seeing how to do this. Can you give me some advice on how to start this. I figure the first thing would be $$((A \cup B) \cap A) \cup ((A \cup B) \cap B^c))$$ But I'm not sure where to go from there.
96450
Just wanted to confirm that this is a correct solution:    Proof: Suppose $f(x_0) &gt; 0$ for some $x_0 \in [a,b]$. Then, by continuity of $f$, for $\epsilon &lt; f(x_0)$, there exists $\delta &gt; 0$ so that, if $|x-x_0| &lt; \delta$, we have $|f(x) - f(x_0)| &lt; \epsilon$. That is to say, $f(x)$ is non-zero in a neighborhood $B_\delta (x_0)$. Now,   \begin{eqnarray*} U(f,P) &amp;=&amp; \sum_{P([a,b])} M_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} M_i \Delta_i \\ &amp;=&amp; \sup_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \sup_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \sup_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;&gt;&amp; 0 \end{eqnarray*}  In the same manner,   \begin{eqnarray*} L(f,P) &amp;=&amp; \sum_{P([a,b])} m_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} m_i \Delta_i \\ &amp;=&amp; \inf_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \inf_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \inf_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;\geq&amp; 0 \end{eqnarray*}    We can conclude from here that $\int_{a}^{b} f(x) dx &gt; 0$.  Is this a valid proof?
30520
I haven't found any useful method to solve the following problem: Prove that if $x,y,z\in\mathbb{Z}$ and $x^3+y^3=3z^3$ then $xyz=0$.  Source:
98111
Prove that $x^2+y^2=z^n$ has a solution $(x, y, z)$ in $\mathbb{N}$ for all $n\in\mathbb{N}$  I tried to prove this by induction, but couldn't. ( This was probably because the solution for some $n$ isn't necessarily related to the solution for $n+1$)  I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :)
22425
Prove $a \equiv b \pmod{c} \implies a^n \equiv b^n \pmod{c}$.   Here is my proof, which I'm slightly doubtful I've done correctly:  Suppose  $a \equiv b \pmod{c}$ and $d \equiv e \pmod{c}$  We have:  $a-b=ck$ and $d-e=cs$  $a=ck+b$ and $d=cs+e$  And hence:  $$ad=(cs+e)(ck+b)=c^2sk+cab+eck+eb$$  $$=c(csk+ab+eck)+eb$$  Therefore,   $$ad-eb=cn$$  $$ad \equiv eb \pmod{c}$$  The special case $d=a^{n-1}$ and $e=b^{n-1}$ through induction is reduced to:  $$a^n \equiv b^n \pmod{c}$$  For $n \geq 1$  Induction:  If $n=1$, then by the proof above  $a=b \pmod{c}$  implies $a^2=b^2 \pmod{c}$. Assume that $a^{n-1}=b^{n-1} \pmod {c}$ is true, then we have  by the proof above: $a^{n}=b^{n} \pmod{c}$. Hence because we proved $a^{2}=b^{2} \pmod{c}$,  $a^{3}=b^{3} \pmod{c}$. And because of this $a^{4}=b^{4} \pmod{c}$. And because of this $a^{5}=b^{5} \pmod{c}$. Etc..  Is my proof valid? Because this question is more of a yes or no one, a different proof would also a nice supplement to the answer (or just  a nice answer) if you wish to provide one.
77659
Let $a,b,c\in \mathbb{Z}$. If $ac&gt;bc$ and $c&gt;0$, then $a&gt;b$   I think I would use contrapositive to prove this. But I am not sure proper contrapositive statement.   Is it If $a&lt;b$, then $ac&lt;bc$, and $c&lt;0$.? If so, contrapositive is not a good choice, isn't it?
57864
Can somebody please check if my proof is okay?  Prove that $||z_1|-|z_2|| \leq |z_1-z_2|$.  Attempt: This proof seems similar to the triangle inequality proof.  $$\begin{split} |z_1-z_2|^2 &amp;= (z_1-z_2) \overline{(z_1-z_2)} \\ &amp;= (z_1-z_2)(\bar z_1 - \bar z_2) \\ &amp;= z_1\bar z_1 - z_1 \bar z_2 - \bar z_1 z_2 + z_2 \bar z_2\\ &amp;=|z_1|^2 - 2Re(z_1 \bar z_2) + |z_2|^2\\ &amp; \geq |z_1|^2 -2|z_1 \bar z_2| + |z_2|^2 \\ &amp;= |z_1|^2 - 2|z_1||z_2| + |z_2|^2 \\ &amp;= (|z_1| - |z_2|)^2. \end{split}$$  Taking the square root of both sides gives $|z_1|-|z_2| \leq |z_1-z_2|$  Can I just take the norm of both sides so that $||z_1|-|z_2|| \leq ||z_1-z_2|| = |z_1-z_2|$?  Thank you.
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.
2419
Prove $||a|-|b|| \leq |a-b|$  I do it by making 4 cases :  Case 1 : $ a \geq 0$, $ b\geq 0$  So we have   $||a||-||b||  = |a - b|$ . hence proved  Case 2 :   $a &lt; 0 , b&lt; 0$  $|-a+b| = |-(a-b)|=|a-b|$  Case 3  $a &gt;0, b&lt;0$  $||a||-||b|| =|a+b| \leq |a|+|b| \leq a -b \leq |a-b| $  Case 4:  $a &lt;0 ,b&gt;0$  $|-a+b| = |-(a-b)| = |a-b|$  Is this OK ? Thanks
96573
I recently came across this identity:  $$E \left[ E \left(Y|X,Z \right) |X \right] =E \left[Y | X \right]$$  I am of course familiar with the simpler version of that rule, namely that $E \left[ E \left(Y|X \right) \right]=E \left(Y\right) $ but I was not able to find justification for its generalization.   I would be grateful if someone could point me to a not-so-technical reference for that fact or, even better, if someone could lay out a simple proof for this important result.
114930
This is a question the prof asked in my university math proofs course. I've tried this for a bit over an hour and I am stuck.   The method I personally chose was proof by cases:  Case 1: n is even:    Thus $n = 2k$ for some k ∈ Z. So $n = 2k = 5a + 2b$  Here I can choose $a=0$, thus $n = 2k = 2b$ Therefore, if n is even there always exists some a,b∈ Z such that $n = 5a + 2b$ and case 1 is true.  Case 2: n is odd:  Thus $n = 2k+1$ for some k ∈ Z. So $n = 2k+1 = 5a + 2b$  This is where I'm stuck, I cant seem to show that the case is always true for when n is odd.  If you could give me a hint in the right direction or complete my method of proof by cases that would be great, or if you had another method of proving it that would be helpful too.  Thanks, John