All variable are $\in \mathbb{Z}$. $k$ divides $n$ and $k$ divides $m$ therefore because $k$ is a divisor of $n$ and $m$ it must be also a divisor for the greatest common divisor of $n$ and $m$. Assume there is no number $w\in \mathbb{Z}$ such that $kw$ = gcd$(n,m):=z$ then there would be a contradiction since there exists $n',n'',m',m''$ such that $n+m = k(n'+m') = z(n''+m'') \Rightarrow \frac{k(n'+m')}{k(n'+m')} = \frac{z(n''+m'')}{k(n'+m')} = 1 $   That's my thought process so far, Maybe it's the wrong Approach- Maybe I am in the Right direction if the latter is the case please tell me how I should continue otherwise it would be Kind if you could give me the proof.
Let $S = \{ax + by: x,y \in \mathbb{Z}\}$  and let $e &gt; 0$ be the smallest element in $S$. Prove that $e \mid a$, and hence prove that $e = \gcd(a,b)$  I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.  I suppose that to show $e\mid a$ I would have to show that $eg = a$ for some $g \in \mathbb{Z}$. But I have no idea how I would do this.
C Scanf suddenly stopped reading in values Why would I want to use Interfaces? I understand that they force you to implement methods and such but what I cant understand is why you would want to use them. Can anybody give me a good example or explanation on why I would want to implement this. I am trying to assign an image texture to an Image Texture node in 2.80. Very simple script - I'm just starting to learn how this stuff works.  This is what I have. It creates an image and and IT node, but it doesn't assign the one to other. Can anybody help? I've searched around, but can't find it in another question.  import bpy  #create blank image bpy.ops.image.new(name="TestImg", width=1024, height=1024)  #add image texture to object, to hold baked lighting mat = bpy.context.view_layer.objects.active.active_material image_node = mat.node_tree.nodes.new('ShaderNodeTexImage')  #assign image to image texture node image_node.image = "TestImg"   Thanks in advance. According to "Moving Around Other Creatures" (PH p. 191 / SRD p. 92):     You can move through a nonhostile creature’s space. [...] another creature’s space is difficult terrain for you.    Whether a creature is a friend or an enemy, you can’t willingly end your move in its space.   So it's possible for Alice to , the only penalty being that the ally's space counts as difficult terrain to enter.  If an enemy adjacent to the ally's space has readied a grapple against Alice and is successful, then Alice (unwillingly) ends her move in her ally's space.  Do allied characters in the same space suffer any negative effect?  Now the rest of Alice's allies do the same thing.   Is there a limit to the number of characters that can be stacked on a space?   What if Alice's ally was the one with the readied grapple?  Can you stack an ally on your space by readying a grapple to prevent them from leaving? Can I visit UK with Italian stay permit? I'm an Egyptian citizen, who holds Schengen visa and stay permit in Italy. Can I go to UK for some days with my stay permit and Schengen visa? In what order should you do linear regression diagnostics? No indent in the first paragraph in a section? Are the Night's Watch still required? How do I fix my USB drive to get its original 8GB size back? My USB drive used to be originally 8GB when I bought it.  I'm trying to reformatted in Windows 7 by right clicking on the drive and selecting Format.... But the capacity only shows 250MB.  Is there something I can do to get the original size back? Maybe it got partitioned in a weird way?   The  flash drive is a SanDisk Cruzer Micro 8GB. I got it from Wal-Mart but this is the same drive. How to find range $a_{75}$ of the term of the series $a_n=a_{n-1}+ {1 \over {a_{n-1}}} $ If $a_1=1$ and for n&gt;1$$a_n=a_{n-1}+ {1 \over {a_{n-1}}} $$ $a_{75}$ lies between (a) (12,15) (b) (11,12) (c) (15,18)  Now , in this question, I rewrote, $a_n-a_{n-1} = {1 \over {a_{n-1}}}$, to make a telescopic type and summed the terms to get $$a_n-a_{1}= {1 \over {a_{n-1}}}+ {1 \over {a_{n-2}}}...{1 \over {a_{1}}}$$ And as ${1 \over {a_{r}}} &lt;1$ for all r , I got $a_{75} &lt;76$ , but it's of no help as , the upper bound is much higher. I got no further ideas , on how to solve it, please help. How can I pause a render in progress and then resume it at a later time? Error 1052 Field is Ambiguous using Inner Join Function How to add days to current Date using JavaScript. Does JavaScript have a built in function like .Net's AddDay? I would like to typeset the top part of Pascal's triangle. To get the triangle with the names of the binomial coefficients, i.e., {n \choose k}, I used the following code  \begin{tikzpicture} \foreach \n in {0,...,4} {   \foreach \k in {0,...,\n} {     \node at (\k-\n/2,-\n) {${\n \choose \k}$};   } } \end{tikzpicture}   The result is this   Now I want to be equally lazy and do something like this for the values of the binomial coefficients, i.e., replace {\n \choose \k} in the node label with \CalculateBinomialCoefficient{\n}{\k} where \CalculateBinomialCoefficient is a hypothetical macro that calculates the binomial coefficient. Has anyone done something like that?  The result should look like this: Ideals in valuation rings How do I prove the fact that for any valuation ring $V$ the ideals are totally ordered under inclusion? Why do we need the trilogy instead of just one site? Concatenate multiple rows Range of A and null space of the transpose of A So I'm complete stuck with something. I know it the following statements are true (or at least the seem to be from the results that I got from messing around with it a bit on MATLAB), but I don't understand why they are true or how to show so. Let $A$ be and $m$X$n$ matrix. Show that:  a) if $x \in N(A^TA)$ then $Ax$ is in both $R(A)$ and $N(A^T)$.   For this one I messed around with it with my own examples and I got $Ax=0$, therefore satisfing the statement, but I don't understand what's actually going on.   b) $N(A^TA)=N(A)$  again, makes sense when I see the results in MATLAB, but don't undestand why it works.  c) $A$ and $A^TA$ have the same rank  d) If $A$ has linearly independent columns, the $A^TA$ is nonsingular.  For the last two I have no idea on how to even start showing the relationship. I feel like I'm missing some crucial relationship between $A$ and $A^TA$, but I'm just not seeing it.   I would greatly appreciate any help of sugestions on how to show that these statements are true.   Thank you very much. Converting Raw HTTP Request into HTTPWebRequest Object Resistor in series We know that when a resistor is connected in series , the current flowing through that resistor will be constant.But how do we use a resistor to limit the current flowing through a circuit ,when resistor is connected in series with the circuit ?