Lösung 1.1:7b - Versionsgeschichte

Silke um 14:09, 13. Aug. 2009

Silke — Thu, 13 Aug 2009 14:09:33 GMT

← Nächstältere Version		Version vom 14:09, 13. Aug. 2009
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	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}

-	Jetzt müssen wir die Zahl nur noch als ~~eine Quote zwischen ganzen~~ Zahlen schreiben.	+	Jetzt müssen wir die Zahl nur noch als Bruch zweier ganzer Zahlen schreiben.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
	Wenn wir unsere Zahl als x benennen haben wir		Wenn wir unsere Zahl als x benennen haben wir
Zeile 20:		Zeile 20:
	::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>		::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	Jetzt sieht man, dass 10000''x'' dieselbe Dezimalbruchentwicklung wie ''x'' hat.	+	Jetzt sieht man, dass 10000''x'' nach dem Komma dieselbe Dezimalbruchentwicklung wie ''x'' hat.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
	Also,		Also,

Moni: Ausdruck

Moni — Sat, 08 Aug 2009 13:05:42 GMT

Ausdruck

← Nächstältere Version		Version vom 13:05, 8. Aug. 2009
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	::<math>10000x-x = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots</math>		::<math>10000x-x = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	::<math>\phantom{10000x-x}{}= 31413\quad</math>(die ~~Dezimale kanzellieren einander~~)	+	::<math>\phantom{10000x-x}{}= 31413\quad</math>(die Nachkommastellen verschwinden)
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
	und nachdem <math>10000x-x = 9999x</math> haben wir		und nachdem <math>10000x-x = 9999x</math> haben wir

Moni: Rechtschreibung

Moni — Sat, 08 Aug 2009 13:01:23 GMT

Rechtschreibung

← Nächstältere Version		Version vom 13:01, 8. Aug. 2009
Zeile 20:		Zeile 20:
	::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>		::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	Jetzt sieht man dass 10000''x'' dieselbe Dezimalbruchentwicklung wie ~~10000~~''x'' hat.	+	Jetzt sieht man, dass 10000''x'' dieselbe Dezimalbruchentwicklung wie ''x'' hat.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
	Also,		Also,
Zeile 30:		Zeile 30:
	::<math>9999x = 31413\,\mbox{.}</math>		::<math>9999x = 31413\,\mbox{.}</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	Wenn wir beide seiten mit 9999 dividieren, sehen wir dass	+	Wenn wir beide seiten mit 9999 dividieren, sehen wir, dass
	::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>		::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	<!--<center> [[Image:1_1_7b-1(2).gif]] </center>		<!--<center> [[Image:1_1_7b-1(2).gif]] </center>
	<center> [[Image:1_1_7b-2(2).gif]] </center>-->		<center> [[Image:1_1_7b-2(2).gif]] </center>-->

Tekbot: hat „Solution 1.1:7b“ nach „Lösung 1.1:7b“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 13:24:04 GMT

hat „Solution 1.1:7b“ nach „Lösung 1.1:7b“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 13:24, 22. Okt. 2008

Martin um 13:54, 18. Okt. 2008

Martin — Sat, 18 Oct 2008 13:54:51 GMT

← Nächstältere Version		Version vom 13:54, 18. Okt. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	~~A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically~~.	+	Eine rationale Zahl hat ab einer bestimmten Dezimale eine periodische Dezimalbruchentwicklung.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	In ~~our case, the sequence~~ 1416 ~~is repeated indefinitely~~	+	In diesem Fall ist die periodische Dezimalbruchentwicklung 1416.
	<center><math>3\textrm{.}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>		<center><math>3\textrm{.}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>
-	~~In other words, the number is~~ rational.	+	Also ist unsere Zahl rational.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}

-	~~The next problem is to rewrite the number as a fraction, for which we use the fact that multiplication by 10 moves the decimal point one place to the right~~.	+	Jetzt müssen wir die Zahl nur noch als eine Quote zwischen ganzen Zahlen schreiben.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~If we write~~	+	Wenn wir unsere Zahl als x benennen haben wir
	::<math>\insteadof[right]{10000x}{x}{} = 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>		::<math>\insteadof[right]{10000x}{x}{} = 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>
-	~~then~~	+	und
	::<math>\insteadof[right]{10000x}{10x}{} = 31\,\textrm{.}\,4161\ 4161\ 4161\,\ldots</math>		::<math>\insteadof[right]{10000x}{10x}{} = 31\,\textrm{.}\,4161\ 4161\ 4161\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
Zeile 20:		Zeile 20:
	::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>		::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~Note that, in~~ 10000''x'' ~~we have moved the decimal point sufficiently many places so that the decimal expansion of~~ 10000''x'' ~~is in phase with the decimal expansion of ''x'', i.e. they have the same~~	+	Jetzt sieht man dass 10000''x'' dieselbe Dezimalbruchentwicklung wie 10000''x'' hat.
-	~~decimal expansion~~.	+
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~Therefore~~,	+	Also,
	::<math>10000x-x = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots</math>		::<math>10000x-x = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	::<math>\phantom{10000x-x}{}= 31413\quad</math>(~~The decimal parts cancel out each other~~)	+	::<math>\phantom{10000x-x}{}= 31413\quad</math>(die Dezimale kanzellieren einander)
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~and as~~ <math>10000x-x = 9999x</math> ~~we get that~~	+	und nachdem <math>10000x-x = 9999x</math> haben wir
	::<math>9999x = 31413\,\mbox{.}</math>		::<math>9999x = 31413\,\mbox{.}</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~Solving for ''x'' in this relationship we find ''x'' as a quotient between two integers~~	+	Wenn wir beide seiten mit 9999 dividieren, sehen wir dass
	::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>		::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	<!--<center> [[Image:1_1_7b-1(2).gif]] </center>		<!--<center> [[Image:1_1_7b-1(2).gif]] </center>
	<center> [[Image:1_1_7b-2(2).gif]] </center>-->		<center> [[Image:1_1_7b-2(2).gif]] </center>-->

Tek: decimal comma --> decimal point

Tek — Fri, 19 Sep 2008 07:13:55 GMT

decimal comma --> decimal point

← Nächstältere Version		Version vom 07:13, 19. Sep. 2008
Zeile 2:		Zeile 2:
	A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically.		A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	In our case, the sequence is repeated indefinitely.	+	In our case, the sequence 1416 is repeated indefinitely
-	<center><math>3{,}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>	+	<center><math>3\textrm{.}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>
	In other words, the number is rational.		In other words, the number is rational.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
Zeile 10:		Zeile 10:
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
	If we write		If we write
-	::<math>\insteadof[right]{10000x}{x}{} = 3\,\~~color~~{~~red}{~~}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>	+	::<math>\insteadof[right]{10000x}{x}{} = 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>
	then		then
-	::<math>\insteadof[right]{10000x}{10x}{} = 31\,\~~color~~{~~red}{~~}\,4161\ 4161\ 4161\,\ldots</math>	+	::<math>\insteadof[right]{10000x}{10x}{} = 31\,\textrm{.}\,4161\ 4161\ 4161\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	::<math>\insteadof[right]{10000x}{100x}{} = 314\,\~~color~~{~~red}{~~}\,1614\ 1614\ 161\,\ldots</math>	+	::<math>\insteadof[right]{10000x}{100x}{} = 314\,\textrm{.}\,1614\ 1614\ 161\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	::<math>\insteadof[right]{10000x}{1000x}{} = 3141\,\~~color~~{~~red}{~~}\,6141\ 6141\ 61\,\ldots</math>	+	::<math>\insteadof[right]{10000x}{1000x}{} = 3141\,\textrm{.}\,6141\ 6141\ 61\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\~~color~~{~~red}{~~}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>	+	::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	Note that, in 10000''x'' we have moved the decimal point sufficiently many places so that the decimal	+	Note that, in 10000''x'' we have moved the decimal point sufficiently many places so that the decimal expansion of 10000''x'' is in phase with the decimal expansion of ''x'', i.e. they have the same
-	expansion of	+
-	10000''x'' ~~has come~~ in phase with the decimal expansion of ''x'', i.e. they have the same	+
	decimal expansion.		decimal expansion.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
	Therefore,		Therefore,
-	::<math>10000x-x = 31416\,{,}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,{,}\,\underline{1416}\ \underline{1416}\,\ldots</math>	+	::<math>10000x-x = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
	::<math>\phantom{10000x-x}{}= 31413\quad</math>(The decimal parts cancel out each other)		::<math>\phantom{10000x-x}{}= 31413\quad</math>(The decimal parts cancel out each other)

Ian um 13:54, 14. Sep. 2008

Ian — Sun, 14 Sep 2008 13:54:13 GMT

← Nächstältere Version		Version vom 13:54, 14. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	~~Ett rationellt tal har alltid en decimalutveckling som från och med en viss~~ decimal ~~upprepar sig periodiskt~~.	+	A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~I vårt fall så upprepas sekvensen 1416 i all oändlighet~~	+	In our case, the sequence is repeated indefinitely.
	<center><math>3{,}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>		<center><math>3{,}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>
-	~~Med andra ord är talet rationellt~~.	+	In other words, the number is rational.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}

-	~~Nästa~~ problem ~~är att skriva om talet som ett bråktal och då utnyttjar vi att multiplikation med~~ 10 ~~flyttar decimalkommat ett steg åt höger~~.	+	The next problem is to rewrite the number as a fraction, for which we use the fact that multiplication by 10 moves the decimal point one place to the right.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~Om vi skriver~~	+	If we write
	::<math>\insteadof[right]{10000x}{x}{} = 3\,\color{red}{}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>		::<math>\insteadof[right]{10000x}{x}{} = 3\,\color{red}{}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>
-	~~så är därför~~	+	then
	::<math>\insteadof[right]{10000x}{10x}{} = 31\,\color{red}{}\,4161\ 4161\ 4161\,\ldots</math>		::<math>\insteadof[right]{10000x}{10x}{} = 31\,\color{red}{}\,4161\ 4161\ 4161\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
Zeile 20:		Zeile 20:
	::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\color{red}{}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>		::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\color{red}{}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~Notera att i~~ 10000''x'' ~~har vi flyttat decimalkommat tillräckligt många steg så att decimalutvecklingen av~~ 10000''x'' ~~har kommit i fas med decimalutvecklingen av~~ ''x'', ~~dvs~~. ~~de har samma decimalutveckling~~.	+	Note that, in 10000''x'' we have moved the decimal point sufficiently many places so that the decimal
		+	expansion of
		+	10000''x'' has come in phase with the decimal expansion of ''x'', i.e. they have the same
		+	decimal expansion.
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~Därför är~~	+	Therefore,
	::<math>10000x-x = 31416\,{,}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,{,}\,\underline{1416}\ \underline{1416}\,\ldots</math>		::<math>10000x-x = 31416\,{,}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,{,}\,\underline{1416}\ \underline{1416}\,\ldots</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	::<math>\phantom{10000x-x}{}= 31413\quad</math>(~~decimalerna tar ut varandra~~)	+	::<math>\phantom{10000x-x}{}= 31413\quad</math>(The decimal parts cancel out each other)
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~och eftersom~~ <math>10000x-x = 9999x</math> ~~så har vi alltså sambandet~~	+	and as <math>10000x-x = 9999x</math> we get that
	::<math>9999x = 31413\,\mbox{.}</math>		::<math>9999x = 31413\,\mbox{.}</math>
	{{NAVCONTENT_STEP}}		{{NAVCONTENT_STEP}}
-	~~Löser vi ut~~ ''x'' ~~ur detta samband får vi~~ ''x'' ~~som en kvot mellan två heltal~~	+	Solving for ''x'' in this relationship we find ''x'' as a quotient between two integers
	::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>		::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	<!--<center> [[Image:1_1_7b-1(2).gif]] </center>		<!--<center> [[Image:1_1_7b-1(2).gif]] </center>
	<center> [[Image:1_1_7b-2(2).gif]] </center>-->		<center> [[Image:1_1_7b-2(2).gif]] </center>-->

Tekbot: Lösning 1.1:7b moved to Solution 1.1:7b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:13:43 GMT

Lösning 1.1:7b moved to Solution 1.1:7b: Robot: moved page

← Nächstältere Version

Version vom 08:13, 9. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 06:22:29 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

← Nächstältere Version		Version vom 06:22, 21. Aug. 2008
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	::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>		::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
-	<!--<center> [[~~Bild~~:1_1_7b-1(2).gif]] </center>	+	<!--<center> [[Image:1_1_7b-1(2).gif]] </center>
-	<center> [[~~Bild~~:1_1_7b-2(2).gif]] </center>-->	+	<center> [[Image:1_1_7b-2(2).gif]] </center>-->

Tek um 08:14, 29. Apr. 2008

Tek — Tue, 29 Apr 2008 08:14:02 GMT

← Nächstältere Version		Version vom 08:14, 29. Apr. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:~~1_1_7b-1(2).gif~~]] </~~center~~>	+	Ett rationellt tal har alltid en decimalutveckling som från och med en viss decimal upprepar sig periodiskt.
-	<~~center~~> [~~[Bild~~:~~1_1_7b~~-2(2).~~gif]]~~ </~~center~~>	+	{{NAVCONTENT_STEP}}
		+	I vårt fall så upprepas sekvensen 1416 i all oändlighet
		+	<center><math>3{,}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>
		+	Med andra ord är talet rationellt.
		+	{{NAVCONTENT_STEP}}
		+
		+	Nästa problem är att skriva om talet som ett bråktal och då utnyttjar vi att multiplikation med 10 flyttar decimalkommat ett steg åt höger.
		+	{{NAVCONTENT_STEP}}
		+	Om vi skriver
		+	::<math>\insteadof[right]{10000x}{x}{} = 3\,\color{red}{}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>
		+	så är därför
		+	::<math>\insteadof[right]{10000x}{10x}{} = 31\,\color{red}{}\,4161\ 4161\ 4161\,\ldots</math>
		+	{{NAVCONTENT_STEP}}
		+	::<math>\insteadof[right]{10000x}{100x}{} = 314\,\color{red}{}\,1614\ 1614\ 161\,\ldots</math>
		+	{{NAVCONTENT_STEP}}
		+	::<math>\insteadof[right]{10000x}{1000x}{} = 3141\,\color{red}{}\,6141\ 6141\ 61\,\ldots</math>
		+	{{NAVCONTENT_STEP}}
		+	::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\color{red}{}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>
		+	{{NAVCONTENT_STEP}}
		+	Notera att i 10000''x'' har vi flyttat decimalkommat tillräckligt många steg så att decimalutvecklingen av 10000''x'' har kommit i fas med decimalutvecklingen av ''x'', dvs. de har samma decimalutveckling.
		+	{{NAVCONTENT_STEP}}
		+	Därför är
		+	::<math>10000x-x = 31416\,{,}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,{,}\,\underline{1416}\ \underline{1416}\,\ldots</math>
		+	{{NAVCONTENT_STEP}}
		+	::<math>\phantom{10000x-x}{}= 31413\quad</math>(decimalerna tar ut varandra)
		+	{{NAVCONTENT_STEP}}
		+	och eftersom <math>10000x-x = 9999x</math> så har vi alltså sambandet
		+	::<math>9999x = 31413\,\mbox{.}</math>
		+	{{NAVCONTENT_STEP}}
		+	Löser vi ut ''x'' ur detta samband får vi ''x'' som en kvot mellan två heltal
		+	::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+	<!--<center> [[Bild:1_1_7b-1(2).gif]] </center>
		+	<center> [[Bild:1_1_7b-2(2).gif]] </center>-->