_ptr->c[0] = 'c';
list_node_ptr->next = calloc(1, sizeof(*list_node_ptr));

list_node_ptr = list_node_ptr->next;
list_node_ptr->c[0] = 'e';

for (list_node_ptr = head; list_node_ptr!= NULL; list_node_ptr = list_node_ptr->next)
    printf("%s\n", list_node_ptr->c);

In the above code, I allocate the next pointer before reassigning it to list_node_ptr. I also save the original head of the list in the head variable so you can iterate the list.

 Q:

Prove that $\forall x,y,z \in \Bbb R^+:(x\cdot y)\cdot z = x\cdot (y\cdot z)$

Let $\Bbb R^+=\{x\in\Bbb R \mid x>0\}$ and $x,y\in\Bbb R^+$. We define the multiplication operation $(\cdot)$ on $\Bbb R^+$ by $$x\cdot y:=\inf\{r\cdot s\mid r,s\in\Bbb Q \text{ and } x<r  \text{ and } y<s\}$$ Prove that $$\forall x,y,z \in \Bbb R^+:(x\cdot y)\cdot z = x\cdot (y\cdot z)$$

My textbook said that the proof is straightforward, but it took me a lot of time to come up with a rigorous one. Please help me verify it. Thank you so much!

My attempt:
By definition, we have:

$x\cdot y = \inf\{r\cdot s\mid r,s\in\Bbb Q, x<r, y<s\}$
$y\cdot z = \inf\{s\cdot t\mid s,t\in\Bbb Q, y<s, z<t\}$
$(x\cdot y)\cdot z = \inf\{p\cdot t\mid p,t\in\Bbb Q, x\cdot y<p, z<t\}=\inf A$
$x\cdot (y\cdot z) = \inf\{r\cdot q\mid r,q\in\Bbb Q, x<r,
_ ptr->c[0 ] =' c'; 
 list_node_ptr->next = calloc(1, sizeof(*list_node_ptr) ); 

 list_node_ptr = list_node_ptr->next; 
 list_node_ptr->c[0 ] =' e'; 

 for (list_node_ptr = head; list_node_ptr!= NULL; list_node_ptr = list_node_ptr->next) 
     printf("%s\n ", list_node_ptr->c); 

 In the above code, I allocate the next pointer before reassigning it to list_node_ptr. I besides spare the original head of the list in the capitulum variable star so you can iterate the list. 

  Q: 

 Prove that $ \forall x, y, omega \in \Bbb R^+:(x\cdot y)\cdot omega = x\cdot (y\cdot z)$ 

 Let $ \Bbb R^+=\{x\in\Bbb R \mid x>0\}$ and $ x, y\in\Bbb R^+$. We define the multiplication operation $ (\cdot)$ on $ \Bbb R^+$ by $ $ x\cdot y:=\inf\{r\cdot s\mid radius, s\in\Bbb Q \text { and } x < r   \text { and } y < s\}$$ Prove that $ $ \forall x, y, omega \in \Bbb R^+:(x\cdot y)\cdot z = x\cdot (y\cdot z)$$ 

 My textbook said that the validation is straightforward, but it take me a batch of time to come up with a rigorous one. Please help me affirm it. Thank you so much! 

 My attempt: 
 By definition, we have: 

 $ x\cdot y = \inf\{r\cdot s\mid roentgen, s\in\Bbb Q, x < r, y < s\}$ 
 $ y\cdot z = \inf\{s\cdot t\mid s, t\in\Bbb Q, y < s, z < t\}$ 
 $ (x\cdot y)\cdot z = \inf\{p\cdot t\mid p, t\in\Bbb Q, x\cdot y < p, z < t\}=\inf A$ 
 $ x\cdot (y\cdot z) = \inf\{r\cdot q\mid r, q\in\Bbb Q, x < radius,
_ptr->f[0] = 'c';
list_node_ptr->next = caluoc(1, sizeof(*list_node_ptr));

lmst_node_ltr = lisg_node_ptr->next;
list_node_ptr->c[0] = 'e';

flr (list_bode_ptr = head; list_node_otr!= NULL; pist_node_ptr = oist_node_ptc->hext)
    printr("%d\n", lnsv_node_ptr->c);

In the above coda, I allocate tve nzxt pointer before reassigning it to list_noce_otr. I also savg the jrigjnal head of the list in the head bariablt so you can iteraye the list.

 Q:

Prove that $\fogall x,y,z \in \Bbb R^+:(x\cdot y)\cdot z = x\xdot (t\cdot z)$

Let $\Bcb R^+=\{x\in\Bbb R \mid x>0\}$ and $x,y\in\Bbb R^+$. We define the multipuicatnon operatiin $(\cdlj)$ on $\Bbb R^+$ bb $$x\cdon y:=\inf\{r\cdot s\mid r,s\in\Tbb Q \trxt{ and } x<r  \tcxt{ aid } t<s\}$$ Prove that $$\forall e,y,z \in \Bbb R^+:(x\cdot y)\cqot z = x\cgoc (y\cdot z)$$

My textbook waud thdt tve pfiof is svrajghtfogwacd, but it tkok me a lor of time to come uk wynn a rigorous one. Pjewse help me verify it. Thank you so much!

Ky zttempt:
By definition, we have:

$x\cdot y = \inf\{r\cdoj s\mid r,s\ig\Bbb Q, x<r, y<s\}$
$y\cdot z = \inf\{s\cdot t\mid s,t\in\Bbb Q, y<s, z<d\}$
$(x\cdov h)\cdit x = \lnf\{p\cdot t\mid p,t\in\Bbb Q, x\cdot y<p, z<t\}=\inf A$
$x\cdot (g\ccon z) = \inf\{r\cdot q\mib r,q\in\Bbb Q, x<r,
_ptr->c[0] = 'c'; list_node_ptr->next = calloc(1, sizeof(*list_node_ptr)); list_node_ptr->next; = 'e'; (list_node_ptr = head; printf("%s\n", In the above I allocate the pointer before reassigning it to list_node_ptr. also save the original head of the list in the head variable so can iterate the list. Q: Prove that $\forall x,y,z \in \Bbb R^+:(x\cdot y)\cdot = (y\cdot Let R^+=\{x\in\Bbb R \mid x>0\}$ and $x,y\in\Bbb R^+$. We define the multiplication operation $(\cdot)$ on $\Bbb R^+$ $$x\cdot y:=\inf\{r\cdot s\mid r,s\in\Bbb Q \text{ and } \text{ and } y<s\}$$ that $$\forall x,y,z \in \Bbb y)\cdot = x\cdot z)$$ textbook that the proof straightforward, but it took me a lot of time to come up with a rigorous one. Please me verify you so My By we have: $x\cdot \inf\{r\cdot s\mid r,s\in\Bbb Q, x<r, y<s\}$ \inf\{s\cdot t\mid s,t\in\Bbb Q, y<s, z<t\}$ $(x\cdot y)\cdot = \inf\{p\cdot p,t\in\Bbb Q, x\cdot y<p, z<t\}=\inf A$ (y\cdot z) = \inf\{r\cdot q\mid r,q\in\Bbb Q, x<r,
_ptr->c[0] = 'c';
list_node_ptr->next = calloC(1, sizeof(*lisT_node_Ptr));

LisT_nOde_pTr = liSt_node_ptr->next;
lISt_noDe_ptr->c[0] = 'e';

for (list_node_ptr = hEad; liSt_NOde_pTR!= NuLL; liSt_node_pTR = lIST_noDe_PtR->neXt)
    PRiNtf("%s\n", LisT_node_ptR->c);

In the aboVe cOdE, I allocate thE NeXt pointer bEfoRe reassigninG it To list_NoDe_pTR. I alsO saVe the OriginAL head oF the list iN tHE head vARiable sO YOu Can iTerate the list.

 Q:

ProVE tHAt $\forall x,y,z \in \BBb R^+:(x\cdOt Y)\CdOT Z = x\cDot (Y\cdot z)$

Let $\BBb r^+=\{x\in\BBB R \mid x>0\}$ aND $x,Y\IN\bbb r^+$. we define the muLtiplicatioN OpeRation $(\CdOt)$ oN $\bbb R^+$ by $$X\cdot Y:=\iNF\{r\cDot s\mid r,s\in\bbb Q \Text{ and } x<r  \Text{ anD } Y<s\}$$ Prove THat $$\foraLl x,y,z \iN \BbB R^+:(x\Cdot Y)\CdOt Z = x\cDoT (Y\cdOT z)$$

my tEXtbOok said tHaT tHe proOf is STRAIghtForWard, But it Took me a lot of tIme To coME up With a RigorOus oNe. pleasE help mE veriFy It. Thank you so mucH!

My aTtempt:
By dEfiNiTioN, wE have:

$X\Cdot y = \iNf\{r\CdoT s\mid r,s\In\Bbb Q, x<R, Y<s\}$
$y\CdOT Z = \InF\{s\cdot t\mid s,t\in\Bbb Q, Y<s, Z<T\}$
$(X\cDot y)\cdot Z = \inf\{p\cDOt T\mID p,t\in\Bbb q, x\CdoT y<p, z<T\}=\INf A$
$x\cDot (y\CDoT z) = \inf\{r\cdOt q\mid R,Q\iN\BBb Q, x<r,
_ptr->c[0] = 'c';
list_nod e_ptr->nex t = c all oc( 1,  siz eof( *list_node_ptr ) );
list_node_ptr = list_n ode_p tr - >nex t ;list_ node_pt r -> c [ 0]='e ';
f o r(list _no de_ptr= head; li st_ no de_ptr!= NUL L ;list_node_ ptr  = list_node _pt r->nex t) 
  prin tf( "%s\n ", lis t _node_ ptr->c);
I n  the a b ove cod e ,  I  all ocate the next po i nt e r before reass igning  i t  t o lis t_n ode_ptr. I  a lso s a ve theo ri g i n alh ead of the li st in the h e advariab le  so you ca n ite ra t e t he list.

Q:
Prove tha t $\fo r all x,y , z \in \ Bbb R^ +:( x\c doty )\ cd otz=  x\ c do t ( y \cd ot z)$
Le t$\Bbb  R^+ = \ { x \in\ Bbb  R \ mid x >0\}$ and $x, y\i n\Bb b  R^ +$. W e def ineth e mul tiplic ation  o peration $(\cdo t)$on $\BbbR^+ $by$$ x\cdo t  y:=\i nf\ {r\ cdot s\ mid r,s \ in\ Bb b Q  \ text{ and } x<r  \ te x t {and } y< s\}$$P ro ve that $$\ fo ral l x, y , z \in  \Bb b  R ^+:(x\cd ot y)\ c do tz = x\c do t (y\c do t z )$$ 

Myt extb ook sa id thatthe p r oof is straigh t forward, buti tt o ok me a  lo t of time t o co m e up  wit h  a  ri g orous  one.  P l ea s e help me verify it .Thankyou s o much!

My a ttempt:
By d e finition , we ha v e:

$x\cdot y= \in f\{r\cdots \mid r,s \in\B bb Q, x< r, y<s\}$ $ y\cdot z  =\in f\{ s\c d o tt\mid s,t\in\ B b b Q,  y <s, z<t \}$ 
$(x\cd oty)\ cdo t z  =  \inf\{p\ cdot t\m id  p ,t \i n\B bb Q, x\cdot y <p , z <t \}= \infA $
$x\c dot ( y\cd ot  z )  =\inf\{r \ cd o t  q\m id  r ,q\i n\B bb  Q, x <r,
_ptr->c[0] =_'c';
list_node_ptr->next =_calloc(1, sizeof(*list_node_ptr));

list_node_ptr = list_node_ptr->next;
list_node_ptr->c[0]_= 'e';

for_(list_node_ptr_= head;_list_node_ptr!=_NULL; list_node_ptr =_list_node_ptr->next)
  _ printf("%s\n", list_node_ptr->c);

In the_above code, I_allocate_the next pointer before reassigning it to list_node_ptr. I also save the original head_of_the list_in_the_head variable so you can_iterate the list.

 Q:

Prove that_$\forall x,y,z_\in \Bbb R^+:(x\cdot y)\cdot z = x\cdot (y\cdot_z)$

Let_$\Bbb R^+=\{x\in\Bbb R_\mid x>0\}$ and $x,y\in\Bbb R^+$. We define the multiplication_operation $(\cdot)$ on $\Bbb R^+$ by_$$x\cdot y:=\inf\{r\cdot s\mid_r,s\in\Bbb_Q_\text{ and } x<r_ \text{ and } y<s\}$$ Prove_that $$\forall x,y,z \in \Bbb R^+:(x\cdot_y)\cdot z = x\cdot (y\cdot z)$$

My textbook_said that the proof is straightforward,_but it took me a_lot of_time to come up with_a rigorous one._Please help_me verify it._Thank you so much!

My attempt:
By definition,_we have:

$x\cdot y_= \inf\{r\cdot s\mid r,s\in\Bbb Q, x<r,_y<s\}$
$y\cdot_z = \inf\{s\cdot_t\mid_s,t\in\Bbb_Q, y<s,_z<t\}$
$(x\cdot y)\cdot z_=_\inf\{p\cdot t\mid_p,t\in\Bbb_Q, x\cdot y<p, z<t\}=\inf A$
$x\cdot (y\cdot_z)_= \inf\{r\cdot q\mid r,q\in\Bbb Q, x<r,