tag:blogger.com,1999:blog-36828909040915714822024-02-20T18:21:12.131-08:00Elementary Number TheoryLife is all about trying and fighting. If two wrongs don't make a right, try three.Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.comBlogger14125tag:blogger.com,1999:blog-3682890904091571482.post-55905356889485718442009-04-09T08:44:00.000-07:002009-04-09T09:49:26.957-07:00Representation Of IntegersGiven an integer b>1,any positive integer N can be written uniquely in tems of powers of b as<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=N=a_{m}b^{m}@plus;a_{m-1}b^{m@plus;1}@plus;....@plus;a_{2}b^{2}@plus;a_{1}b@plus;a_{0}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?N=a_{m}b^{m}+a_{m-1}b^{m+1}+....+a_{2}b^{2}+a_{1}b+a_{0}" title="N=a_{m}b^{m}+a_{m-1}b^{m+1}+....+a_{2}b^{2}+a_{1}b+a_{0}" /></a><br /><br />where the coefficient <a href="http://www.codecogs.com/eqnedit.php?latex=0\leq a_{k}\leq b-1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?0\leq a_{k}\leq b-1" title="0\leq a_{k}\leq b-1" /></a>.<br /><br /><span style="font-weight:bold;">Example 1</span><br /><br />N=10,b=2<br /><br />First step.<br />10=5•2+0<br /> 5=2•2+1<br /> 2=1•2+0<br /><br />Second step.<br />Let b=2.<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=10=5\cdot b@plus;0=(2\cdot b@plus;1)\cdot b=2\cdot b^{2}@plus;1\cdot b=(1\cdot b)\cdot b^{2}@plus;1\cdot b=1\cdot b^{3}@plus;1\cdot b=1\cdot 2^{3}@plus;1\cdot 2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?10=5\cdot b+0=(2\cdot b+1)\cdot b=2\cdot b^{2}+1\cdot b=(1\cdot b)\cdot b^{2}+1\cdot b=1\cdot b^{3}+1\cdot b=1\cdot 2^{3}+1\cdot 2" title="10=5\cdot b+0=(2\cdot b+1)\cdot b=2\cdot b^{2}+1\cdot b=(1\cdot b)\cdot b^{2}+1\cdot b=1\cdot b^{3}+1\cdot b=1\cdot 2^{3}+1\cdot 2" /></a><br /><br /><span style="font-weight:bold;">Example 2</span><br /><br />Let N=105,b=3.<br />105=35•3+0<br /> 35=11•3+2<br /> 11=3•3+2<br /> 3=1•3+0<br /><br />Let b=3.<br /><a href="http://www.codecogs.com/eqnedit.php?latex=105=35\cdot b@plus;0=(11\cdot b@plus;2)\cdot b=11\cdot b^{2}@plus;2\cdot b=(3\cdot b@plus;2)\cdot b^{2}@plus;2\cdot b=3\cdot b^{3}@plus;2\cdot b^{2}@plus;2\cdot b=(1\cdot b)\cdot b^{3}@plus;2\cdot b^{2}@plus;2\cdot b=1\cdot b^{4}@plus;2\cdot b^{2}@plus;2\cdot b" target="_blank"><img src="http://latex.codecogs.com/gif.latex?105=35\cdot b+0=(11\cdot b+2)\cdot b=11\cdot b^{2}+2\cdot b=(3\cdot b+2)\cdot b^{2}+2\cdot b=3\cdot b^{3}+2\cdot b^{2}+2\cdot b=(1\cdot b)\cdot b^{3}+2\cdot b^{2}+2\cdot b=1\cdot b^{4}+2\cdot b^{2}+2\cdot b" title="105=35\cdot b+0=(11\cdot b+2)\cdot b=11\cdot b^{2}+2\cdot b=(3\cdot b+2)\cdot b^{2}+2\cdot b=3\cdot b^{3}+2\cdot b^{2}+2\cdot b=(1\cdot b)\cdot b^{3}+2\cdot b^{2}+2\cdot b=1\cdot b^{4}+2\cdot b^{2}+2\cdot b" /></a><br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=\therefore 105=1\cdot 3^{4}@plus;2\cdot 3^{2}@plus;2\cdot 3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\therefore 105=1\cdot 3^{4}+2\cdot 3^{2}+2\cdot 3" title="\therefore 105=1\cdot 3^{4}+2\cdot 3^{2}+2\cdot 3" /></a><br /><br /><p align="center"><a href="http://www.anylove.com"><br /><img src="http://i64.photobucket.com/albums/h191/curion123/myspacej/quotes/2.jpg" border="0"></a><br><font size="2"><a href="http://www.myspacejunks.com">Myspace <br />Graphics</font></a>- At Myspacejunks.com</p>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-4183315196964894352009-04-09T07:53:00.000-07:002009-04-09T08:04:17.304-07:00new on our page...!!!salam to all readers...<br /><br />have you all heard about the documentary about THE ARRIVAL? If not, now you can watch it from our page! Its a documentary video that have been done by a muslim just for da'wah. Its very interesting and you can learn a lot from it. let me briefly give you the introduction about the documentary. This documentary revealed about our world today that have been ruled long time ago by the companion of DAJJAL (the arrival). Its tell us how they work,for who they work,what are their goals,why they make all this and the most important thing is are we well prepared to face all this obstacles to fight them..?? If you all want to know more about this documentary, check out their website at www.wakeupproject.com.Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-67982660787280033032009-04-09T07:15:00.000-07:002009-04-09T09:51:40.711-07:00The Theory Of Congruences.Definition.<br /><br />Let n be a fixed positive integer. Two integer are said to be congruent modulo n,symbolized by <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b (mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b (mod n)" title="a\equiv b (mod n)" /></a> (pronounced a congruent to b modulo n)if n|a-b. If it's not,hence a is said to be not congruent to b modulo n.<br /><br />Example:<br />Let say a=7,b=10 and n=3.<br /><br />Since 7-10=-3 and -3|3,hence <a href="http://www.codecogs.com/eqnedit.php?latex=7\equiv 10(mod 3)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?7\equiv 10(mod 3)" title="7\equiv 10(mod 3)" /></a>.<br /><br /><span style="font-weight:bold;">Theorem 14</span> - For any integers a and b,<a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a> if and only if a and b leave the same nonnegative remainder when divided by n.<br /><br /><span style="font-weight:bold;">Theorem 15</span> - Len n be fixed and a,b,c and d can be any integers. The following properties hold<br /><br />a) <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv a(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv a(mod n)" title="a\equiv a(mod n)" /></a><br /><br />b) If <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a>,then <a href="http://www.codecogs.com/eqnedit.php?latex=b\equiv a(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?b\equiv a(mod n)" title="b\equiv a(mod n)" /></a><br /><br />c) If <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a>,and <a href="http://www.codecogs.com/eqnedit.php?latex=b\equiv c(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?b\equiv c(mod n)" title="b\equiv c(mod n)" /></a>,then <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv c(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv c(mod n)" title="a\equiv c(mod n)" /></a>.<br /><br />d) If <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a>,and <a href="http://www.codecogs.com/eqnedit.php?latex=c\equiv d(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?c\equiv d(mod n)" title="c\equiv d(mod n)" /></a>,then <a href="http://www.codecogs.com/eqnedit.php?latex=a@plus;c\equiv b@plus;d(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a+c\equiv b+d(mod n)" title="a+c\equiv b+d(mod n)" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=ac\equiv bd(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?ac\equiv bd(mod n)" title="ac\equiv bd(mod n)" /></a>.<br /><br />e) If <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a>,then <a href="http://www.codecogs.com/eqnedit.php?latex=a@plus;c\equiv b@plus;c(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a+c\equiv b+c(mod n)" title="a+c\equiv b+c(mod n)" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=ac\equiv bc(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?ac\equiv bc(mod n)" title="ac\equiv bc(mod n)" /></a>.<br /><br />f) If <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a>,then <a href="http://www.codecogs.com/eqnedit.php?latex=a^{k}\equiv b^{k}(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a^{k}\equiv b^{k}(mod n)" title="a^{k}\equiv b^{k}(mod n)" /></a> for any k>0.<br /><br /><span style="font-weight:bold;">Proof</span><br /><br />a) Since a-a=0,and n|0,hence <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv a(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv a(mod n)" title="a\equiv a(mod n)" /></a>.<br /><br />b) Suppose that <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a>. So from the definition,n|a-b. It follows that a-b=nk for some k. Since b-a=-nk,we multiply both side by -1,we'll ontain -b+a=a-b=nk. So,n|a-b,thus <a href="http://www.codecogs.com/eqnedit.php?latex=b\equiv a(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?b\equiv a(mod n)" title="b\equiv a(mod n)" /></a>.<br /><br />c) Suppose that <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=b\equiv c(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?b\equiv c(mod n)" title="b\equiv c(mod n)" /></a>. So from the definition,n|a-b and n|b-c. It follows that a-b=nk for some k and b-c=nj for some j. By substituting b=a-nk into b-c=nj,we'll obtain a-nk-c=nj. So a-c=nj+nk. From that we get a-c|n and thus <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv c(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv c(mod n)" title="a\equiv c(mod n)" /></a>.<br /><br />d) Suppose that <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=c\equiv d(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?c\equiv d(mod n)" title="c\equiv d(mod n)" /></a>. From the definition,n|a-b and n|c-d. It follows that a-b=nk for some k and c-d=nj for some j.<br /><br />a-b + (c-d) = nk + nj<br />a-b + c-d = n (k+j)<br />a+c - b-d = n (k+j)<br />a+c - (b+d) = n (k+j)<br /><br />Because a+c - (b+d)|n,hence <a href="http://www.codecogs.com/eqnedit.php?latex=a@plus;c\equiv b@plus;d(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a+c\equiv b+d(mod n)" title="a+c\equiv b+d(mod n)" /></a>.<br /><br />ac-bd=ac-bc+bc-bd<br /> =c(a-b)+b(c-d)<br /> =c(nk)+b(nj)<br /> =n(ck-bj)<br /><br />Because n|ac-bd,therefore <a href="http://www.codecogs.com/eqnedit.php?latex=ac\equiv bd(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?ac\equiv bd(mod n)" title="ac\equiv bd(mod n)" /></a>.<br /><br />e) Since <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a>,we know n|a-b.<br /><br />a-b=(a+c)-(b+c)<br /><br />Hence n|(a+c)-(b+c). So,<a href="http://www.codecogs.com/eqnedit.php?latex=a@plus;c\equiv b@plus;c(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a+c\equiv b+c(mod n)" title="a+c\equiv b+c(mod n)" /></a>.<br /><br />Since <a href="http://www.codecogs.com/eqnedit.php?latex=a\equiv b(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\equiv b(mod n)" title="a\equiv b(mod n)" /></a>,we know n|a-b.<br /><br />ac-bc = c(a-b)<br /><br />Hence n|c(a-b). So,<a href="http://www.codecogs.com/eqnedit.php?latex=ac\equiv bc(mod n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?ac\equiv bc(mod n)" title="ac\equiv bc(mod n)" /></a>.<br /><br /><p align="center"><a href="http://www.anylove.com"><br /><img src="http://i64.photobucket.com/albums/h191/curion123/myspacej/quotes/7.jpg" border="0"></a><br><font size="2"><a href="http://www.myspacejunks.com">Myspace <br />Graphics</font></a>- At Myspacejunks.com</p>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-90366191909828697482009-04-09T04:25:00.000-07:002009-04-09T04:38:30.717-07:00FINAL EXAM!!!!!!!To all our beloved classmates (082's CTS) and everybody who's gonna sit for the final exam,<br /><br />Number Theory Brothers wish u good luck in your final exam. Pray for our success.<br /><br /><a href="http://www.trippytext.com"><img src="http://www.trippytext.com/gen/47/331/06b64640c6c7ed28449f16af648d6c5c.gif" border="0" alt="http://www.trippytext.com/ - Trippy Text"></a><br>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-20402613891733239702009-04-07T06:14:00.000-07:002009-04-09T04:54:12.856-07:00Typical Kampong Problem..blergh!During school holiday,Pak Hamid offers Hakimi to do a part time jobs to get an extra pocket money. He’ll pay RM 14 per hour for the painting his house and RM21 per hours to take care of his farm. By the end of the school holiday,Hakimi get a well paid RM147 from Pak Hamid. How many hours possible did Hakimi spend for each of the respective job?<br /><br /><p align="center"><a href="http://www.myspacejunks.com"><br /><img src="http://i172.photobucket.com/albums/w31/graphichv/funnyanimated/29.gif" border="0"><br><font size="1">Myspace Graphics</font></a></p><br /><br />Solution.<br /><br />The equation derived is 21x + 14y = 147<br /><br />Let x represent the hours spent for taking care of Pak Hamid’s farm.<br />Let y represent the hours spent for the painting job.<br /><br />Using Euclidean Algorithm we evaluate gcd(21,14)<br />21 = 1•14 + 7<br />14 = 2•7 + 0<br /><br />Hence,the gcd(21,14) equals to 7.<br /><br />Since gcd(21,14) equals to 7 and 7|147,hence the equation 21x+14y=147 has a solution.<br /><br />To obtain the integer 7 as a linear combination of 21 and 14,we work backward from the previous computation as follows:<br />7 = 21 - 1•14<br /><br />Hence,7=1•21- 1•14<br /><br /><br />Upon multiplying this equation by 21,we obtain<br />147 = 21•21+(-21)•14<br /><br />Thus,<a href="http://www.codecogs.com/eqnedit.php?latex=x_{0}=21" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x_{0}=21" title="x_{0}=21" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=y_{0}=-21" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y_{0}=-21" title="y_{0}=-21" /></a>.<br />The general solution of 21x+14y=147 is given by the equation <a href="http://www.codecogs.com/eqnedit.php?latex=x=21@plus;\frac{14}{7}t=21@plus;2t" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x=21+\frac{14}{7}t=21+2t" title="x=21+\frac{14}{7}t=21+2t" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=y=-21-\frac{21}{7}t=-21-3t" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=-21-\frac{21}{7}t=-21-3t" title="y=-21-\frac{21}{7}t=-21-3t" /></a><br /><br />Since x and y represent time,hence it cannot be negative.<br /><br />21+2t>0 and -21-3t>0<br />Hence,<br />t>-10.5 and t<-7.<br /><br />The overlapped real value of t is -8,-9 and -10.<br />Substitute the value of t obtained to the general solution of x and y,we’ll get;<br />When t=-8,x=5 and y=3<br />When t=-9,x=3 and y=6<br />When t=-10,x=1 and y=9.<br /><br /><p align="center"><a href="http://www.myspacejunks.com"><br /><img src="http://i87.photobucket.com/albums/k156/jnds/funnygif/84.gif" border="0"><br><font size="1">Myspace Graphics</font></a></p>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com2tag:blogger.com,1999:blog-3682890904091571482.post-41470595840153074582009-04-06T02:17:00.001-07:002009-04-06T02:17:48.797-07:00lets get to the basic..<object width="445" height="364"><param name="movie" value="http://www.youtube.com/v/4GxP9EVKCjo&hl=en&fs=1&color1=0x5d1719&color2=0xcd311b&border=1"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/4GxP9EVKCjo&hl=en&fs=1&color1=0x5d1719&color2=0xcd311b&border=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="445" height="364"></embed></object>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-38163685171639564582009-04-06T01:54:00.000-07:002009-04-06T02:01:53.890-07:00fun math : exciting..!!!!<object width="445" height="364"><param name="movie" value="http://www.youtube.com/v/wm_T-FiXkmY&hl=en&fs=1&color1=0x2b405b&color2=0x6b8ab6&border=1"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/wm_T-FiXkmY&hl=en&fs=1&color1=0x2b405b&color2=0x6b8ab6&border=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="445" height="364"></embed></object>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-24770233683777308592009-04-02T06:06:00.000-07:002009-04-09T04:55:03.317-07:002 = 1? Explain please,Prof. Hafiz..<a href="http://www.codecogs.com/eqnedit.php?latex=x=y" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x=y" title="x=y" /></a> Given<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=x^{2}=xy" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x^{2}=xy" title="x^{2}=xy" /></a> Multiply both sides by X<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=x^{2}-y^{2}=xy-y^2{}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x^{2}-y^{2}=xy-y^2{}" title="x^{2}-y^{2}=xy-y^2{}" /></a> Subtract Y^2 from both sides<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=(x@plus;y)(x-y)=y(x-y)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(x+y)(x-y)=y(x-y)" title="(x+y)(x-y)=y(x-y)" /></a> Factor both sides<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=(x@plus;y)=y" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(x+y)=y" title="(x+y)=y" /></a> Cancel out common factors<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=y@plus;y=y" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y+y=y" title="y+y=y" /></a> Substitute in from line (1)<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=2y=y" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2y=y" title="2y=y" /></a>Collect the Y's<br /><br /><a href="http://www.codecogs.com/eqnedit.php?latex=2=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2=1" title="2=1" /></a> Divide both sides by Y<br /><br />Now obviously, 2 doesn't equal 1. So what's wrong with this proof?<br /><br /><br /><br /><br />Answer<br />'Cancelling' the common factors from line (4) to (5) means dividing by the factor (X-Y). Since X=Y, this is a division by 0, the results of which are undefined.<br /><br /><p align="center"><a href="http://www.myspacejunks.com"><br /><img src="http://i87.photobucket.com/albums/k156/jnds/funnygif/77.gif" border="0"><br><font size="1">Myspace Graphics</font></a></p>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-75737879872459007562009-03-30T01:03:00.000-07:002009-03-30T01:10:56.530-07:00Take 5 - The Beauty Of Numbers.1 x 8 + 1 = 9<br />12 x 8 + 2 = 98<br />123 x 8 + 3 = 987<br />1234 x 8 + 4 = 9876<br />12345 x 8 + 5 = 98765<br />123456 x 8 + 6 = 987654<br />1234567 x 8 + 7 = 9876543<br />12345678 x 8 + 8 = 98765432<br />123456789 x 8 + 9 = 987654321<br /><br />1 x 9 + 2 = 11<br />12 x 9 + 3 = 111<br />123 x 9 + 4 = 1111<br />1234 x 9 + 5 = 11111<br />12345 x 9 + 6 = 111111<br />123456 x 9 + 7 = 1111111<br />1234567 x 9 + 8 = 11111111<br />12345678 x 9 + 9 = 111111111<br />123456789 x 9 +10= 1111111111<br /><br />9 x 9 + 7 = 88<br />98 x 9 + 6 = 888<br />987 x 9 + 5 = 8888<br />9876 x 9 + 4 = 88888<br />98765 x 9 + 3 = 888888<br />987654 x 9 + 2 = 8888888<br />9876543 x 9 + 1 = 88888888<br />98765432 x 9 + 0 = 888888888<br /><br />Brilliant, isn't it?<br /><br />And finally, take a look at this symmetry:<br /><br />1 x 1 = 1<br />11 x 11 = 121<br />111 x 111 = 12321<br />1111 x 1111 = 1234321<br />11111 x 11111 = 123454321<br />111111 x 111111 = 12345654321<br />1111111 x 1111111 = 1234567654321<br />11111111 x 11111111 = 123456787654321<br />111111111 x 111111111=12345678987654321Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com9tag:blogger.com,1999:blog-3682890904091571482.post-52826590986097934562009-03-29T23:11:00.000-07:002009-04-09T05:01:29.285-07:00Application Of Euclidean AlgorithmDefinition<br /><br />Two integers a and b,not both of which are zero,are said to be relatively prime whenever gcd(a,b)=1.<br /><br />Theorem 5 – Let a and b be integers,not both zero. Then a and b is relatively prime if and only if there exist integers x and y such that 1 = ax +by. (in other words,gcd(a,b)=1)<br /><br />Theorem 6 – If gcd (a,b) = d,then <a href="http://www.codecogs.com/eqnedit.php?latex=gcd(\frac{a}{d},\frac{b}{d})=1" target="_blank"><img title="gcd(\frac{a}{d},\frac{b}{d})=" src="http://latex.codecogs.com/gif.latex?gcd(\frac{a}{d},\frac{b}{d})=1" /></a><br />Theorem 7 – If a|c and b|c,with gcd (a,b)=1,then a|c.<br /><br />Theorem 8 – If ab|c,with gcd (a,b)=1,then a|c.<br /><br />Theorem 9 – Let a,b be integers,not both are zero. For a positive integer d,d=gcd(a,b) if and only if<br />a) d|a and d|b<br />b) whenever c|a and c|b,then c|d<br /><br />Theorem 10 – If a=bq+r,then gcd(a,b)=gcd(b,r)<br /><br />Theorem 11 – If k>0,then gcd (ka,kb) = k•gcd(a,b)<br /><br />Theorem 12 – The linear Diophantine equation ax+by=c has a solution if and only if dc where d=gcd(a,b). If x0 and y0 is any particular solution of this equation,then all other solution are given by x = x0 + (b/d)t and y = y0 – (a/d)t where t is any integer.<br /><br />Theorem 13 – If gcd(a,b)=1 and if x0 and y0 is a particular solution of the linear Diophantine equation ax+by=c ,then all solution are given by x= x0 + bt and y= y0 – at for any t.<br /><br />Exercise<br /><br />a) Given integers a,b,c,and d. Prove that if a|b and a|c,then <a href="http://www.codecogs.com/eqnedit.php?latex=a^{2}\mid bc" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a^{2}\mid bc" title="a^{2}\mid bc" /></a><br /># Since ab and ac,then exist d and e such that b=ad and c=ae. Because bc= ad•ae = <a href="http://www.codecogs.com/eqnedit.php?latex=a^{2" target="_blank"><img title="a^{2}" src="http://latex.codecogs.com/gif.latex?a^{2" /></a>•de. Thus,<a href="http://www.codecogs.com/eqnedit.php?latex=a^{2}\mid bc" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a^{2}\mid bc" title="a^{2}\mid bc" /></a><br /><br />b) Given integers a,b,c,and d. Prove that if a|b and c|d,then ac|bd.<br /># Since a|b and c|d,then exist e and f such that b=ae and d=cf. Because bd=ae•cf=ac•ef,then ac|bd.<br /><br />c) If gcd(a,b)=1 and c|a,then gcd(b,c)=1.<br /># Since gcd (a,b)is equal to 1,then exist x and y such that 1=ax +by and since ca,then exist d such that a=cd. Because 1=ax + by=cdx + by,thus gcd(b,c)=1.<br /><br />Example Of Problem Which Apply The Use Of Euclidean Algorithm.<br /><br />Problems<br />Consider the problem of purchasing RM 510 of travellers cheque using only RM 20 and RM 50. How many each type of cheque should be used.<br /><br />Let x represent the amount of cheque for RM 50.<br />Let y represent the amount of cheque for RM 20.<br /><br />Hence,the equation could be represent as follows;<br />50x + 20y = 510.<br /><br />Using Euclidean Algorithm,we evaluate the gcd (50,20)<br />50 = 2•20 + 10<br />20 = 2•10 + 0<br /><br />Since gcd(50,20)=10 and 10|510,the equation has a solution.<br /><br />To obtain the integer 10 as a linear combination of 20 and 50,we work backward through the previous computations as follows<br />10 = 50 – 2•20 = 1•50 – 2•20<br />So 10 = 1•50 + (-2)•20<br /><br />Upon multiplying this relation by 51,we obtain<br />510 = 51•50 + (-102)•20.<br /><br />Thus x0 = 51 and y0=-102 is a solution for 50x + 20y=510.<br />The other solution is given by the equation x = 51+2t and y = -102 – 5t.<br />Since x and y are always positive (because it involve money),hence only few value of t is applicable.<br /><br />51+2t>0 and -102-5t>0<br />hence,<br />t>-25.5 and t<-20.4<br /><br />Hence,overlapped real value of t is -21,-22,-23,-24,-25. Thus the value of x and y which is applicable are ;<br /><br />When t=-25, x=1 and y=23.<br />When t=-24, x=3 and y=18.<br />When t=-23, x=5 and y=13.<br />When t=-22, x=7 and y=8.<br />When t=-21, x=9 and y=3.<br /><br /><p align="center"><a href="http://www.myspacejunks.com"><br /><img src="http://i64.photobucket.com/albums/h191/curion123/funnyanimations/myspace-codes-animations-75.gif" border="0"><br><font size="1">Myspace Graphics</font></a></p>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com1tag:blogger.com,1999:blog-3682890904091571482.post-60501067224820340882009-03-29T22:53:00.000-07:002009-04-09T05:00:09.391-07:00Euclidean AlgorithmTheorem 3<br />Given integers a and b,not both of which are zero,there exist integers x and y such that gcd(a,b) = ax+by.<br /><br />Theorem 4<br />If gcd (a,b) = d,then <a href="http://www.codecogs.com/eqnedit.php?latex=gcd(\frac{a}{d},\frac{b}{d})=1" target="_blank"><img title="gcd(\frac{a}{d},\frac{b}{d})=" src="http://latex.codecogs.com/gif.latex?gcd(\frac{a}{d},\frac{b}{d})=1" /></a>.<br /><br />Proof – Since gcd(a,b) is equal to d,then from theorem 3,exist x and y such that d = ax + by. Divide both sides by d to obtain 1=(a/d)x+(b/d)y. Because d is the common divisor of a and b,then it is sure that a/d and b/d is integers. Thus,<a href="http://www.codecogs.com/eqnedit.php?latex=gcd(\frac{a}{d},\frac{b}{d})=1" target="_blank"><img title="gcd(\frac{a}{d},\frac{b}{d})=" src="http://latex.codecogs.com/gif.latex?gcd(\frac{a}{d},\frac{b}{d})=1" /></a><br /><br />Euclidean Algorithm<br /><br />The Euclidean algorithm, also called Euclid's algorithm, is an <a class="Hyperlink" href="http://mathworld.wolfram.com/Algorithm.html">algorithm</a> for finding the <a class="Hyperlink" href="http://mathworld.wolfram.com/GreatestCommonDivisor.html">greatest common divisor</a> of two numbers a and b . The algorithm can also be defined for more general <a class="Hyperlink" href="http://mathworld.wolfram.com/Ring.html">rings</a> than just the integers. There are even <a class="Hyperlink" href="http://mathworld.wolfram.com/PrincipalRing.html">principal rings</a> which are not <a class="Hyperlink" href="http://mathworld.wolfram.com/EuclideanRing.html">Euclidean</a> but where the equivalent of the Euclidean algorithm can be defined. The algorithm for rational numbers was given in Book VII of Euclid's <a class="Hyperlink" href="http://mathworld.wolfram.com/Elements.html">Elements</a>. The algorithm for reals appeared in Book X, making it the earliest example of an <a class="Hyperlink" href="http://mathworld.wolfram.com/IntegerRelation.html">integer relation</a> algorithm (Ferguson et al. 1999).<br /><br />Example 1 - Find the gcd of 754 and 56<br /><br />765 - (13) * (56) = 37<br />56 - (1) * (37) = 19<br />37 - (1) * (19) = 18<br />19 - (1) * (18) = 1<br />18 - (18) * (1) = 0<br /><br /> So gcd of 765 and 56 is 1<br /><br />Example 2 - Find the gcd of 342 and 76<br /><br /> 342 - (4) * (76) = 38<br />76 - (2) * (38) = 0<br /><br /> So gcd of 342 and 76 is 38<br /><br /><p align="center"><a href="http://www.myspacejunks.com"><br /><img src="http://i64.photobucket.com/albums/h191/curion123/funnyanimations/myspace-codes-animations-134.gif" border="0"><br><font size="1">Myspace Graphics</font></a></p>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com3tag:blogger.com,1999:blog-3682890904091571482.post-51823293915018682242009-03-29T22:27:00.000-07:002009-04-07T06:37:35.683-07:00Diophantine EquationIntroduction<br /><br />In <a title="Mathematics" href="http://en.wikipedia.org/wiki/wiki/Mathematics">mathematics</a>, a Diophantine equation is an <a title="Indeterminate equation" href="http://en.wikipedia.org/wiki/wiki/Indeterminate_equation">indeterminate</a> <a title="Polynomial" href="http://en.wikipedia.org/wiki/wiki/Polynomial">polynomial</a> <a title="Equation" href="http://en.wikipedia.org/wiki/wiki/Equation">equation</a> that allows the variables to be <a title="Integer" href="http://en.wikipedia.org/wiki/wiki/Integer">integers</a> only. Diophantine problems have fewer equations than unknown variables and involve finding integers that work correctly for all equations. In more technical language, they define an <a title="Algebraic curve" href="http://en.wikipedia.org/wiki/wiki/Algebraic_curve">algebraic curve</a>, <a title="Algebraic surface" href="http://en.wikipedia.org/wiki/wiki/Algebraic_surface">algebraic surface</a> or more general object, and ask about the <a title="Lattice point" href="http://en.wikipedia.org/wiki/wiki/Lattice_point">lattice points</a> on it.<br /><br />While individual equations present a kind of puzzle and have been considered throughout history, the formulation of general theories of Diophantine equations (beyond the theory of <a title="Quadratic form" href="http://en.wikipedia.org/wiki/wiki/Quadratic_form">quadratic forms</a>) was an achievement of the twentieth century.<br /><br />Linear Diophantine equations take the form ax + by = c. If c is the <a title="Greatest common divisor" href="file://wiki/Greatest_common_divisor">greatest common divisor</a> of a and b then this is Bézout's identity, and the equation has infinitely many solutions. These can be found by applying the <a title="Extended Euclidean algorithm" href="file://wiki/Extended_Euclidean_algorithm">extended Euclidean algorithm</a>. It follows that there are also infinitely many solutions if c is a multiple of the greatest common divisor of a and b. If c is not a multiple of the greatest common divisor of a and b, then the Diophantine equation ax + by = c has no solutions.<br /><br />Conditions for Diophantine equation to have solution.<br />i. The greatest common divisor of a and b divide c.<br />ii. No three of a,b,c,d have a common factor 1.<br />iii. Intergers (x,y) is a solution of the linear Diophantine equation ax+by=c if and only if the (x,y) is a lattice point in the plane that lies in line ax+by=c.<br /><br />Method to find the solution of ax+by=c .<br /><br />i. Given an equation of ax+by=c. First,find the greatest common divisor of a and b using Euclidean algorithm.<br /><br />Example : Determine all solution in intergers of<br />24x+138y=18<br />The computation is as follows :<br />138 = 5•24 + 18<br />24 = 1•18 + 6<br />18 = 3•6 + 0<br />So,the gcd(24,138)=d=6.<br /><br />ii. Check whether gdc(a,b) divide c or not. With regards to example above,since 618,hence the equation 24x+138y=18 has a solution.<br /><br /><br />iii. Obtain c in linear combination of a and b.<br /><br />To obtain 6 in linear combination of 24 and 138,we work backward through the previous computation as follows :<br />138 = 5•24 + 18 <a href="http://www.codecogs.com/eqnedit.php?latex=\to" target="_blank"><img title="\to" src="http://latex.codecogs.com/gif.latex?\to" /></a> 18 = 138 - 5•4<br />24 = 1•18 + 6 <a href="http://www.codecogs.com/eqnedit.php?latex=\to" target="_blank"><img title="\to" src="http://latex.codecogs.com/gif.latex?\to" /></a> 6 = 24 - 1•18<br />18 = 3•6 + 0<br /><br />6 = 24 - 1•18<br />= 24 – 1 (138 - 5•4)<br />= 24 - 1•138 + 5•4<br />= 6•24 - 1•138<br />= 6•24 + (-1)•138<br />Hence,6 in linear combination of 24 and 138 is 6= 6•24 + (-1)•138.<br /><br />iv. Continuing from example above,upon multiplying the 6= 6•24 + (-1)•138 by 3,we obtain<br />18 = 18•24 + (-3)•138<br />Thus,<a href="http://www.codecogs.com/eqnedit.php?latex=x_{0}=18" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x_{0}=18" title="x_{0}=18" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=y_{0}=-3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y_{0}=-3" title="y_{0}=-3" /></a>.<br /><br />Other solution are obtained by using formulae <a href="http://www.codecogs.com/eqnedit.php?latex=x=x_{0}@plus;\frac{b}{d}t" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x=x_{0}+\frac{b}{d}t" title="x=x_{0}+\frac{b}{d}t" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=y=y_{0}-\frac{a}{d}t" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=y_{0}-\frac{a}{d}t" title="y=y_{0}-\frac{a}{d}t" /></a><br /><br />The other solution are given by x = 18 + 23t and y = -3 – 4t for <a href="http://www.codecogs.com/eqnedit.php?latex=t=\pm 1,\pm 2,\pm 3.." target="_blank"><img src="http://latex.codecogs.com/gif.latex?t=\pm 1,\pm 2,\pm 3.." title="t=\pm 1,\pm 2,\pm 3.." /></a>.....and so on.Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-80949598066192018612009-03-29T22:16:00.000-07:002009-04-07T06:44:30.758-07:00Division AlgorithmTheorem 1 (Division Algorithm)<br /><br />Given integers a and b,with b>0,there exist unique integers q and r satisfying a = qb + r. The integers q and r are called,respectively,the quotient and remainder in the division of a by b.<br />Example,<br />a = 8 , b = 3, q = 2, r = 2 <br />8 = 2 x 3 + 2 ( a = qb + r)<br /><br />Definition.<br /><br />An integer b is said to be divisible by an integer <a href="http://www.codecogs.com/eqnedit.php?latex=a\neq 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a\neq 0" title="a\neq 0" /></a>, in symbols a|b, if there exist some integer c such that b = ac.<br />3|6 indicate 3 divide 6 , which exist c=2.<br /><br />Theorem 2<br />For integers a , b , c , the following hold :-<br />1. a|0 , 1|a , a|a<br /><br />2. a|1 iff <a href="http://www.codecogs.com/eqnedit.php?latex=a=\pm 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a=\pm 1" title="a=\pm 1" /></a><br /><br />3. if a|b and c|d , then ac|bd<br /><br />4. if a|b and b|c , then a|c<br /><br />5. a|b and b|a iff <a href="http://www.codecogs.com/eqnedit.php?latex=a=\pm b" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a=\pm b" title="a=\pm b" /></a><br /><br />6. if a|b and b|c , then a|(bx+cy) for some integers x and y.<br /><br />Proof<br />3. since a|b and c|d,then exist e and f such that b = ac and d = cf. Because bd = ae•cf = ac•ef , thus ac|bd.<br /><br />4. Since a|b and b|c,then exist e and f such that b = ae and c = bf. Because c = bf =aef,thus a|c.<br /><br />5. Since a|b and a|c,then exist e and f such that b = ae and c = af. Hence,<br />bx + cy = aex + afy = a(ex+fy). Let (ex+fy) = w. Hence, bx + cy = aw. Thus,a|(bx+cy) for some x and y.<br /><br />Definition<br /><br />Let a anb b be given integers,with at least one of them different from zero. The greatest common divisor (gcd) of a and b,denoted by gcd(a,b) is positive integer d satisfying the following;<br />a) d|a and d|b (d is a common divisor)<br />b) if c|a and c|b,then c|d.Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com0tag:blogger.com,1999:blog-3682890904091571482.post-22379665928015094592009-03-29T09:09:00.000-07:002009-04-09T04:58:01.331-07:00Introduction To Number TheoryDivisibility Theory in the Integers<br /><br />Number Theory is a study of numbers,which involve integers only such as the numbers -2,-1,0,1,2,3,4 and so on. This numbers is called real numbers.<br />Another kind of numbers are as follows:<br />Numbers such as <a href="http://www.codecogs.com/eqnedit.php?latex=%5Cfrac%7B4%7D%7B5%7D,%5Cfrac%7B45%7D%7B76%7D,%5Cfrac%7B546%7D%7B978%7D" target="_blank"><img src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B4%7D%7B5%7D,%5Cfrac%7B45%7D%7B76%7D,%5Cfrac%7B546%7D%7B978%7D" title="\frac{4}{5},\frac{45}{76},\frac{546}{978}" /></a> and so on are known as fraction.<br />Numbers such as <a href="http://www.codecogs.com/eqnedit.php?latex=%5Csqrt%7B-1%7D" target="_blank"><img src="http://latex.codecogs.com/gif.latex?%5Csqrt%7B-1%7D" title="\sqrt{-1}" /></a> are known as complex number.<br /><br />Well-ordering principle.<br /><br />Every non-empty set, S of non-negative integers contains a least element ; that is,there is some integer a in S such that <a href="http://www.codecogs.com/eqnedit.php?latex=a%5Cleq%20b" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a%5Cleq%20b" title="a\leq b" /></a> for all b’s belonging to S.<br />An example to illustrate this principle,<br />S = { 1 , 2 , 3 , 5 , 7 , 9 , 12 }<br /><br /><p align="center"><a href="http://www.myspacejunks.com"><br /><img src="http://i87.photobucket.com/albums/k156/jnds/funnygif/3.gif" border="0"><br><font size="1">Myspace Graphics</font></a></p>Number Theory Brothershttp://www.blogger.com/profile/12997719989483364640noreply@blogger.com1