Kurs:Mathematik für Anwender (Osnabrück 2011-2012)/Teil I/Arbeitsblatt 2/en/latex

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\setcounter{section}{2}






\zwischenueberschrift{Warm-up-exercises}




\inputexercise
{}
{

Let $x,y,z,w$ be elements in a field and suppose that $z$ and $w$ are not zero. Prove the following fraction rules. \aufzaehlungsieben{
\mathdisp {{ \frac{ x }{ 1 } }=x} { , }
}{
\mathdisp {{ \frac{ 1 }{ -1 } }=-1} { , }
}{
\mathdisp {{ \frac{ 0 }{ z } }=0} { , }
}{
\mathdisp {{ \frac{ z }{ z } }=1} { , }
}{
\mathdisp {{ \frac{ x }{ z } } = { \frac{ xw }{ zw } }} { , }
}{
\mathdisp {{ \frac{ x }{ z } } \cdot { \frac{ y }{ w } } = { \frac{ xy }{ zw } }} { , }
}{
\mathdisp {{ \frac{ x }{ z } } + { \frac{ y }{ w } } = { \frac{ xw+yz }{ zw } }} { . }
} Does there exist an analogue of formula (7), which arises when one replaces addition by multiplication (and subtraction by division), that is
\mavergleichskettedisp
{\vergleichskette
{ (x-z) \cdot (y-w) }
{ =} { (x+w) \cdot (y+z)-(z+w) }
{ } { }
{ } { }
{ } { }
} {}{}{?} Show that the \anfuehrungenglisch{popular formula}{}
\mavergleichskettedisp
{\vergleichskette
{ { \frac{ x }{ z } } + { \frac{ y }{ w } } }
{ =} { { \frac{ x+y }{ z+w } } }
{ } { }
{ } { }
{ } { }
} {}{}{} does not hold.

}
{} {}




\inputexercise
{}
{

Determine which of the two rational numbers \mathkor {} {p} {and} {q} {} is larger:
\mathdisp {p= \frac{573}{-1234} \text{ und } q = \frac{-2007}{4322}} { . }

}
{} {}




\inputexercise
{}
{

a) Give an example of rational numbers
\mathl{a,b,c \in {]0,1[}}{} such that
\mathdisp {a^2+b^2=c^2} { . }

b) Give an example of rational numbers
\mathl{a,b,c \in {]0,1[}}{} such that
\mathdisp {a^2+b^2 \neq c^2} { . }

c) Give an example of irrational numbers
\mathl{a,b \in {]0,1[}}{} and a rational number
\mathl{c \in {]0,1[}}{} such that
\mathdisp {a^2+b^2=c^2} { . }

}
{} {}

The following exercises should only be made with reference to the ordering axioms of the real numbers.


\inputexercise
{}
{

Prove the following properties of real numbers. \aufzaehlungacht{$1 > 0$. }{From
\mathl{a \geq b}{} and
\mathl{c \geq 0}{} it follows
\mathl{ac \geq bc}{.} }{From $a \geq b$ and $c \leq 0$ it follows
\mathl{ac \leq bc}{.} }{
\mathl{a^2 \geq 0}{} holds. }{From
\mathl{a \geq b \geq 0}{} it follows
\mathl{a^n \geq b^n}{} for all
\mathl{n \in \N}{.} }{From
\mathl{a \geq 1}{} it follows
\mathl{a^n \geq a^m}{} für ganze Zahlen
\mathl{n \geq m}{.} }{From
\mathl{a > 0}{} it follows
\mathl{{ \frac{ 1 }{ a } } > 0}{.} }{From
\mathl{a > b >0}{} it follows
\mathl{{ \frac{ 1 }{ a } } < { \frac{ 1 }{ b } }}{.} }

}
{} {}




\inputexercise
{}
{

Show that for a \definitionsverweis {real number}{}{} $x \geq 3$ the inequality
\mathdisp {x^2 +(x+1)^2 \geq (x+2)^2} { }
holds.

}
{} {}




\inputexercise
{}
{

Let $x<y$ be two real numbers. Show that for the \definitionsverweis {arithmetic mean}{}{} ${ \frac{ x+y }{ 2 } }$ the inequalities
\mavergleichskettedisp
{\vergleichskette
{x }
{ <} { { \frac{ x+y }{ 2 } } }
{ <} {y }
{ } {}
{ } {}
} {}{}{} hold.

}
{} {}




\inputexercise
{}
{

Prove the following properties for the \definitionsverweis {absolute value function}{}{} \maabbeledisp {} {\R} {\R } {x} {\betrag { x } } {,}\zusatzklammer {here let $x,y$ be arbitrary real numbers} {} {.} \aufzaehlungsieben{ $\betrag { x } \geq 0$. }{ $\betrag { x } = 0$ if and only if $x=0$. }{$\betrag { x } =\betrag { y }$ if and only if $x= y$ or $x=-y$. }{ $\betrag { y-x } =\betrag { x-y }$. }{ $\betrag { xy } = \betrag { x } \betrag { y }$. }{For $x \neq 0$ we have $\betrag { x^{-1} } = \betrag { x }^{-1}$. }{We have $\betrag { x+y } \leq \betrag { x } + \betrag { y }$ \zusatzklammer {\stichwort {Triangle inequality for the absolute value} {}} {} {.} }

}
{} {}




\inputexercise
{}
{

Sketch the following subsets of $\R^2$. \aufzaehlungzweireihe {\itemfuenf {${ \left\{ (x,y) \mid x=5 \right\} }$, }{${ \left\{ (x,y) \mid x \geq 4 \text{ und } y =3 \right\} }$, }{${ \left\{ (x,y) \mid y^2 \geq 2 \right\} }$, }{${ \left\{ (x,y) \mid \betrag { x } = 3 \text{ und } \betrag { y } \leq 2 \right\} }$, }{${ \left\{ (x,y) \mid 3x \geq y \text{ und } 5x \leq 2y \right\} }$, } } {\itemfuenf {${ \left\{ (x,y) \mid xy = 0 \right\} }$, }{${ \left\{ (x,y) \mid xy = 1 \right\} }$, }{${ \left\{ (x,y) \mid xy \geq 1 \text{ und } y \geq x^3 \right\} }$, }{${ \left\{ (x,y) \mid 0 = 0 \right\} }$, }{${ \left\{ (x,y) \mid 0 = 1 \right\} }$. } }

}
{} {}






\zwischenueberschrift{Hand-in-exercises}




\inputexercise
{2}
{

Let $x_1 , \ldots , x_n$ be real numbers. Show by \definitionsverweis {induction}{}{} the following inequality
\mathdisp {\betrag { \sum_{i=1}^n x_i } \leq \sum_{i=1}^n \betrag { x_i }} { . }

}
{} {}




\inputexercise
{5}
{

Prove the general distributive property for a \definitionsverweis {field}{}{.}

}
{} {}




\inputexercise
{3}
{

Sketch the following subsets of $\R^2$. \aufzaehlungsechs{${ \left\{ (x,y) \mid x+y = 3 \right\} }$, }{${ \left\{ (x,y) \mid x+y \leq 3 \right\} }$, }{${ \left\{ (x,y) \mid (x+y)^2 \geq 4 \right\} }$, }{${ \left\{ (x,y) \mid \betrag { x+2 } \geq 5 \text{ and } \betrag { y-2 } \leq 3 \right\} }$, }{${ \left\{ (x,y) \mid \betrag { x } = 0 \text{ and } \betrag { y^4-2y^3+7y-5 } \geq -1 \right\} }$, }{${ \left\{ (x,y) \mid -1 \leq x \leq 3 \text{ and } 0 \leq y \leq x^3 \right\} }$. }

}
{} {}




\inputexercise
{5}
{

A page has been ripped from a book. The sum of the numbers of the remaining pages is
\mathl{65000}{.} How many pages did the book have?

}
{} {Hint: Show that it cannot be the last page. From the two statements \anfuehrungenglisch{A page is missing}{} and \anfuehrungenglisch{The last page is not missing}{} two inequalities can be set up to deliver the (reasonable) upper and lower bound for the number of pages.}



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