<?xml version='1.0' encoding='UTF-8'?><?xml-stylesheet href="http://www.blogger.com/styles/atom.css" type="text/css"?><feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearchrss/1.0/' xmlns:georss='http://www.georss.org/georss' xmlns:gd='http://schemas.google.com/g/2005' xmlns:thr='http://purl.org/syndication/thread/1.0'><id>tag:blogger.com,1999:blog-8723800153584323203</id><updated>2012-02-16T17:03:16.130-08:00</updated><title type='text'>Extreme mathamatical proofs</title><subtitle type='html'></subtitle><link rel='http://schemas.google.com/g/2005#feed' type='application/atom+xml' href='http://killmaths.blogspot.com/feeds/posts/default'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default?max-results=100'/><link rel='alternate' type='text/html' href='http://killmaths.blogspot.com/'/><link rel='hub' href='http://pubsubhubbub.appspot.com/'/><author><name>Raghuram M V</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><generator version='7.00' uri='http://www.blogger.com'>Blogger</generator><openSearch:totalResults>6</openSearch:totalResults><openSearch:startIndex>1</openSearch:startIndex><openSearch:itemsPerPage>100</openSearch:itemsPerPage><entry><id>tag:blogger.com,1999:blog-8723800153584323203.post-8010623711809491277</id><published>2010-09-28T09:14:00.000-07:00</published><updated>2010-09-28T09:35:39.888-07:00</updated><title type='text'>To prove 1+1 = 0</title><content type='html'>we all know that 1+1 =2 but using compelex numbers like i which is equal to root of -1 we can prove 1+1 = 0&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;let&lt;br /&gt;&lt;br /&gt;1  +   1   =  1    +  √1&lt;br /&gt;&lt;br /&gt;as we can take the value of √1 for 1&lt;br /&gt;&lt;br /&gt;              =   1    +  √-1 *  -1&lt;br /&gt;&lt;br /&gt;because  -1 * -1 is + 1&lt;br /&gt;&lt;br /&gt;              =   1    +  √-1   *   √ -1&lt;br /&gt;&lt;br /&gt;   now we can separate the multipliers and finally  as we can denote √-1 with complex charecter i&lt;br /&gt;&lt;br /&gt;then&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;              =   1    +   i   *   i&lt;br /&gt;&lt;br /&gt;which can be written with squares like this&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;             =    1    +   i²&lt;br /&gt;&lt;br /&gt;as we know  that   i =  √-1&lt;br /&gt;&lt;br /&gt;and   (√a)² is a with it we can write it like this now&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;            =     1    +  (√-1)²&lt;br /&gt;&lt;br /&gt;            =     1    +  (-1)&lt;br /&gt;&lt;br /&gt;as   + *  - is -&lt;br /&gt;&lt;br /&gt;            =     1    -   1&lt;br /&gt;&lt;br /&gt;            =      0&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;this is how we can prove 1 + 1 = 0 and usign the prvious method of proofs given in previous posts also we can prove it but this is a different style of proof&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/8723800153584323203-8010623711809491277?l=killmaths.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://killmaths.blogspot.com/feeds/8010623711809491277/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://killmaths.blogspot.com/2010/09/to-prove-11-0.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/8010623711809491277'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/8010623711809491277'/><link rel='alternate' type='text/html' href='http://killmaths.blogspot.com/2010/09/to-prove-11-0.html' title='To prove 1+1 = 0'/><author><name>Raghuram M V</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-8723800153584323203.post-8567146055580957936</id><published>2008-11-04T08:59:00.000-08:00</published><updated>2008-11-04T09:22:14.455-08:00</updated><title type='text'>to prove i equals infinity/zero</title><content type='html'>&lt;span style="color: rgb(51, 102, 255);"&gt;let&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;u=5,v=4,w= -1,x=zero,y=infinity,z=1 and i = square root of (-1)&lt;/span&gt;&lt;br /&gt; &lt;br /&gt; &lt;span style="color: rgb(51, 102, 255);"&gt;firstly take&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; - 20 = - 20&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;which can be written as&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt; 16-36 = 25-45&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;by adding (81/4) on both sides we get&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt; 16-36+(81/4) = 25-45+(81/4)&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;by writing the above eqn like this&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;                                   ((4)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2  &lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;-(2*4*(9/2)) +(9/2)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-family:verdana;font-size:78%;"  &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;)      =                          ((5)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; -(2*5*(9/2)) +(9/2)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;)  &lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;which is of the form&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;       &lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:100%;" &gt; (a-b)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; =  a&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; + b&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; -2ab&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;here&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt; a = 4 a = 5&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; b = 9/2 b = 9/2&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;from this we can write the above equation as&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;                                   (4-(9/2))&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt; 2&lt;span style="font-size:100%;"&gt;                                   =                         (5-(9/2))&lt;span style="font-size:78%;"&gt;2&lt;/span&gt;     &lt;br /&gt;by taking square root on both sides we get&lt;br /&gt;=&gt; +- (4-(9/2)) = +- (5-(9/2))&lt;br /&gt;according to axioms when both sides are equal and having the same signs on both sides then both sides are positively equated to each other&lt;br /&gt;=&gt; 4 - (9/2) = 5 - (9/2)&lt;br /&gt;by shifting -(9/2) from left side to right side we get&lt;br /&gt;=&gt; 4 = 5 - (9/2) +(9/2)&lt;br /&gt;then as 9/2 gets cancelled we get&lt;br /&gt;=&gt; &lt;span style="font-weight: bold; font-style: italic;"&gt;4 = 5 &lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;&lt;span style="font-size:100%;"&gt;then substituting corresponding variables for result we get&lt;br /&gt;=&gt; v = u &lt;/span&gt;&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;&lt;span style="font-size:100%;"&gt; ........................................eqn1    &lt;/span&gt;&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;&lt;span style="font-size:100%;"&gt; then by shifting v we get&lt;br /&gt;=&gt; 0 = u-v                          &lt;br /&gt;here by entering values of u and v we prove that 0=1&lt;br /&gt;i.e  0 = 5-4 =&gt; &lt;span style="font-weight: bold; font-style: italic;"&gt;0 = 1 &lt;/span&gt;  ..............................eqn2&lt;br /&gt;again here by substituting the corresponding variables to result we get&lt;br /&gt;=&gt; x = z&lt;br /&gt;here by cross multiplying x to z quadrant we get&lt;br /&gt;=&gt; 1 = (z/x)&lt;br /&gt;then again by substituting corresponding values of z and x we get&lt;br /&gt;=&gt; 1 = (1/0)&lt;br /&gt;as we know that (1/0) is infinity&lt;br /&gt;=&gt; &lt;span style="font-weight: bold; font-style: italic;"&gt;1 = infinity&lt;/span&gt; ................................eqn3         &lt;br /&gt;again by representing result with corresponding variables we get&lt;br /&gt;=&gt; z = y&lt;br /&gt;then by mathematical laws&lt;br /&gt;if a=b and b=c then a=c we get&lt;br /&gt;=&gt;x = y&lt;br /&gt;here by substituting there values we get&lt;br /&gt;=&gt; zero = infinity&lt;br /&gt;from eqn 1 by moving u to the other side we get&lt;br /&gt;=&gt; v-u = 0&lt;br /&gt;which is equal to&lt;br /&gt;=&gt; 4-5 = 0 =&gt; &lt;span style="font-weight: bold; font-style: italic;"&gt;-1 = 0 &lt;/span&gt;....................eqn 4&lt;br /&gt;again by using laws here we can write&lt;br /&gt;-1 = 0 and 0 = infinity therfore&lt;br /&gt;&lt;span style="font-weight: bold; font-style: italic;"&gt;-1 = infinity &lt;/span&gt;..........................eqn 5&lt;br /&gt;again in eqn 4&lt;br /&gt;=&gt; -1 = 0  by taking square root on both sides we get&lt;br /&gt;=&gt; &lt;span style="font-weight: bold; font-style: italic;"&gt;i = 0&lt;/span&gt; .............................eqn6&lt;br /&gt;therefore i = 0&lt;br /&gt;similarly&lt;br /&gt;by eqn 5&lt;br /&gt;=&gt; -1 = 0 and by taking square root of both sides we  get &lt;br /&gt;=&gt; i = root of infinity (which is undefined)&lt;br /&gt;again back with eqn 6 i.e&lt;br /&gt;=&gt; i = 0&lt;br /&gt;by mathematical laws we can write it as&lt;br /&gt;=&gt; &lt;span style="font-weight: bold; font-style: italic;"&gt;i = infinity&lt;/span&gt; (because i = 0 and 0 = infinity so i = infinity)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold; font-style: italic;"&gt;&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Extra tips---&gt;&lt;br /&gt;1)by using eqn 2 any number can be proved to zero (i.e by multiplying both sides by any number)&lt;br /&gt;2)by using eqn 3 any number can be proved to infinity &lt;/span&gt;&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;&lt;span style="font-size:100%;"&gt;(i.e by multiplying both sides by any number)&lt;/span&gt;&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;&lt;span style="font-size:100%;"&gt; &lt;br /&gt;3)by using eqns above u can  prove which ever  number  present to which  ever number u wan't&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;isn't this amazing&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/8723800153584323203-8567146055580957936?l=killmaths.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://killmaths.blogspot.com/feeds/8567146055580957936/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://killmaths.blogspot.com/2008/11/to-prove-i-equals-infinityzero.html#comment-form' title='4 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/8567146055580957936'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/8567146055580957936'/><link rel='alternate' type='text/html' href='http://killmaths.blogspot.com/2008/11/to-prove-i-equals-infinityzero.html' title='to prove i equals infinity/zero'/><author><name>Raghuram M V</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>4</thr:total></entry><entry><id>tag:blogger.com,1999:blog-8723800153584323203.post-7446427360862285157</id><published>2008-11-04T08:31:00.000-08:00</published><updated>2008-11-04T08:58:02.756-08:00</updated><title type='text'>To prove zero equals infinity</title><content type='html'>&lt;span style="color: rgb(51, 102, 255);"&gt;let&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;u=5,v=4,x=zero,y=infinity,z=1&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;firstly take&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; - 20 = - 20&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;which can be written as&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt; 16-36 = 25-45&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;by adding (81/4) on both sides we get&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt; 16-36+(81/4) = 25-45+(81/4)&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;by writing the above eqn like this&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;                                   ((4)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2  &lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;-(2*4*(9/2)) +(9/2)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-family:verdana;font-size:78%;"  &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;)      =                          ((5)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; -(2*5*(9/2)) +(9/2)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;)  &lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;which is of the form&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;       &lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:100%;" &gt; (a-b)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; =  a&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; + b&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; -2ab&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;here&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt; a = 4 a = 5&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; b = 9/2 b = 9/2&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;from this we can write the above equation as&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;                                   (4-(9/2))&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt; 2&lt;span style="font-size:100%;"&gt;                                   =                         (5-(9/2))&lt;span style="font-size:78%;"&gt;2&lt;/span&gt;     &lt;br /&gt;by taking square root on both sides we get&lt;br /&gt;=&gt; +- (4-(9/2)) = +- (5-(9/2))&lt;br /&gt;according to axioms when both sides are equal and having the same signs on both sides then both sides are positively equated to each other&lt;br /&gt;=&gt; 4 - (9/2) = 5 - (9/2)&lt;br /&gt;by shifting -(9/2) from left side to right side we get&lt;br /&gt;=&gt; 4 = 5 - (9/2) +(9/2)&lt;br /&gt;then as 9/2 gets cancelled we get&lt;br /&gt;=&gt; 4 = 5&lt;br /&gt;then substituting corresponding variables for result we get&lt;br /&gt;=&gt; v = u&lt;br /&gt;then by shifting v we get&lt;br /&gt;=&gt; 0 = u-v   ........................................eqn1                            &lt;br /&gt;here by entering values of u and v we prove that 0=1&lt;br /&gt;i.e  0 = 5-4 =&gt; 0 = 1   ..............................eqn2&lt;br /&gt;again here by substituting the corresponding variables to result we get&lt;br /&gt;=&gt; x = z&lt;br /&gt;here by cross multiplying x to z quadrant we get&lt;br /&gt;=&gt; 1 = (z/x)&lt;br /&gt;then again by substituting corresponding values of z and x we get&lt;br /&gt;=&gt; 1 = (1/0)&lt;br /&gt;as we know that (1/0) is infinity&lt;br /&gt;=&gt; 1 = infinity ................................eqn3         &lt;br /&gt;again by representing result with corresponding variables we get&lt;br /&gt;=&gt; z = y&lt;br /&gt;then by mathematical laws&lt;br /&gt;if a=b and b=c then a=c we get&lt;br /&gt;=&gt;x = y&lt;br /&gt;here by substituting there values we get&lt;br /&gt;=&gt; zero = infinity&lt;br /&gt;hence proved.&lt;br /&gt;&lt;br /&gt;Extra tips---&gt;&lt;br /&gt;1)by using eqn 2 any number can be proved to zero (i.e by multiplying both sides by any number)&lt;br /&gt;2)by using eqn 3 any number can be proved to infinity &lt;/span&gt;&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;&lt;span style="font-size:100%;"&gt;(i.e by multiplying both sides by any number)&lt;/span&gt;&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;&lt;span style="font-size:100%;"&gt; &lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;isn't this amazing&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/8723800153584323203-7446427360862285157?l=killmaths.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://killmaths.blogspot.com/feeds/7446427360862285157/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://killmaths.blogspot.com/2008/11/to-prove-zero-equals-infinity.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/7446427360862285157'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/7446427360862285157'/><link rel='alternate' type='text/html' href='http://killmaths.blogspot.com/2008/11/to-prove-zero-equals-infinity.html' title='To prove zero equals infinity'/><author><name>Raghuram M V</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-8723800153584323203.post-4536755531854674581</id><published>2008-11-03T21:13:00.000-08:00</published><updated>2008-11-03T21:53:44.283-08:00</updated><title type='text'>To prove 4 = 5</title><content type='html'>&lt;span style="color: rgb(51, 102, 255);"&gt;let&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;                                      - 20                                                                                             =                                           - 20&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;which can be written as&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;                                   16-36                                                                                  =                                                25-45&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;by adding (81/4) on both sides we get&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;                                   16-36+(81/4)                                                      =                                                25-45+(81/4)&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;by writing the above eqn like this&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;                                   ((4)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2  &lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;-(2*4*(9/2)) +(9/2)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-family:verdana;font-size:78%;"  &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;)      =                          ((5)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; -(2*5*(9/2)) +(9/2)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;)  &lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;which is of the form&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;       &lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:100%;" &gt; (a-b)&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; =  a&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; + b&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt;2&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);"&gt; -2ab&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;here&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;        a =  4                                                                                                                                                                                                   a = 5&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;             b =  9/2                                                                                                            b = 9/2&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;from this we can write the above equation as&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;=&gt;                                   (4-(9/2))&lt;/span&gt;&lt;span style="color: rgb(51, 102, 255);font-size:78%;" &gt; 2&lt;span style="font-size:100%;"&gt;                                   =                         (5-(9/2))&lt;span style="font-size:78%;"&gt;2&lt;/span&gt;      &lt;br /&gt;by taking square root on both sides we get&lt;br /&gt;=&gt;                                                          +- (4-(9/2))                                     =                    +- (5-(9/2))&lt;br /&gt;according to axioms when both sides are equal and having the same signs on both sides then both sides are positively equated to each other&lt;br /&gt;=&gt;                                                                4 - (9/2)                                                                        =                                                5 - (9/2)&lt;br /&gt;by shifting -(9/2) from left side to right side we get&lt;br /&gt;=&gt;                                    4                                                  =                          5 - (9/2) +(9/2)&lt;br /&gt;then as 9/2 gets cancelled we get&lt;br /&gt;=&gt;                                    4                                                                                                  =                          5&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;isn't this amazing&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/8723800153584323203-4536755531854674581?l=killmaths.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://killmaths.blogspot.com/feeds/4536755531854674581/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://killmaths.blogspot.com/2008/11/to-prove-4-5.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/4536755531854674581'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/4536755531854674581'/><link rel='alternate' type='text/html' href='http://killmaths.blogspot.com/2008/11/to-prove-4-5.html' title='To prove 4 = 5'/><author><name>Raghuram M V</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-8723800153584323203.post-5263116583058452671</id><published>2008-11-03T09:29:00.000-08:00</published><updated>2010-09-28T09:13:17.442-07:00</updated><title type='text'>tips to prove any number to any number directly</title><content type='html'>&lt;ul&gt;&lt;li&gt;Select the numbers which are to be proved equal to each other ex: 2 and 7&lt;/li&gt;&lt;li&gt;Then write down the selected number multiples with negative sign&lt;br /&gt;&lt;/li&gt;&lt;/ul&gt;           i.e   -14    =     -14              &lt;br /&gt;&lt;ul&gt;&lt;li&gt;Then write the numbers on both lhs and rhs in the form of square of selected number substracted by some number so that their results produce the numbers which are written above them i.e             4-18          =          49-63&lt;/li&gt;&lt;li&gt;Then add square of a number on both sides which must be derived from dividing the substracting number by selected number and again by 2    i.e                 (-18)/(2)=-9 again divided by 2 gives -9/2 and similarly  -63/7 = -9 again divided by 2 gives -9/2 note that the number obtained on both sides will be the same number.Then square the number and add it to both sides   i.e       4-18+(81/4)      =      49-63+(81/4)&lt;/li&gt;&lt;li&gt;Then the present numbers will be of the form a&lt;span style="font-size:78%;"&gt;2&lt;/span&gt; + b&lt;span style="font-size:78%;"&gt;2&lt;/span&gt; -2ab which is equal to (a-b)&lt;span style="font-size:78%;"&gt;2&lt;/span&gt; next write down the numbers present in the above form i.e  2&lt;span style="font-size:78%;"&gt;2&lt;/span&gt; + (9/2)&lt;span style="font-size:78%;"&gt;2&lt;/span&gt; - 2*2*(9/2) =                7&lt;span style="font-size:78%;"&gt;2&lt;/span&gt; + (9/2)&lt;span style="font-size:78%;"&gt;2&lt;/span&gt; -2*7*(9/2)&lt;br /&gt;&lt;/li&gt;&lt;li&gt;Then write them in their consized form like (2-(9/2))&lt;span style="font-size:78%;"&gt;2&lt;/span&gt;   =    (7-(9/2))&lt;span style="font-size:78%;"&gt;2&lt;/span&gt;&lt;br /&gt;&lt;/li&gt;&lt;li&gt;Take square root on both sides so we get     (2-(9/2))     =    (7-(9/2))&lt;/li&gt;&lt;li&gt;Then move the calculated number from lhs to rhs so that the number gets cancled.Which produces the desired result i.e  2  =   7  -(9/2)+(9/2)  =&gt; 2  = 7  .Hence proved                                                                                                                   &lt;br /&gt;&lt;/li&gt;&lt;/ul&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/8723800153584323203-5263116583058452671?l=killmaths.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://killmaths.blogspot.com/feeds/5263116583058452671/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://killmaths.blogspot.com/2008/11/tips-to-prove-any-nuber-to-any-nuber.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/5263116583058452671'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/5263116583058452671'/><link rel='alternate' type='text/html' href='http://killmaths.blogspot.com/2008/11/tips-to-prove-any-nuber-to-any-nuber.html' title='tips to prove any number to any number directly'/><author><name>Raghuram M V</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-8723800153584323203.post-496864447002041461</id><published>2008-11-02T01:10:00.000-07:00</published><updated>2008-11-02T01:18:53.295-07:00</updated><title type='text'>Proof of     -*-  =   +</title><content type='html'>&lt;p style="color: rgb(51, 102, 255);"&gt;Product of two negative numbers always gives a positive number...&lt;br /&gt;&lt;br /&gt;How ?&lt;br /&gt;&lt;br /&gt;Here how it is.... "-" always means the opposite direction..&lt;br /&gt;&lt;br /&gt;Consider the following number line&lt;br /&gt;&lt;br /&gt;----|-----|-----|-----|-----|-----|---...&lt;br /&gt;-3   -2   - 1     0    1      2    3     4     5     6&lt;br /&gt;&lt;br /&gt;Now we know that what multiplication means...&lt;br /&gt;&lt;br /&gt;2X3 means 2 added 3 times 2+2+2=6&lt;br /&gt;&lt;br /&gt;Incase of a negative number&lt;br /&gt;-1X2 means -1 added 2 times -1-1=-2&lt;br /&gt;&lt;br /&gt;Incase of both the negatives...&lt;br /&gt;-1X-3 is going in the opposite direction on the number line&lt;br /&gt;&lt;br /&gt;----|-----|-----|-----|-----|-----|---...      (-3 times from 0)&lt;br /&gt;-3   -2   - 1     0    1      2    3     4     5     6&lt;br /&gt;                |____________|&lt;br /&gt;              &lt;br /&gt;So, 0-(-3)=3&lt;br /&gt;&lt;br /&gt;graphical explanation&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&lt;h2 style="color: rgb(51, 102, 255);"&gt;Direction &lt;/h2&gt;       &lt;p style="color: rgb(51, 102, 255);"&gt;It is all about direction. Remember the &lt;a href="http://www.mathsisfun.com/number-line.html"&gt;Number Line&lt;/a&gt;?&lt;/p&gt;              &lt;p style="color: rgb(51, 102, 255);" align="center"&gt;&lt;img src="http://www.mathsisfun.com/images/number-line.gif" height="52" width="591" /&gt;   &lt;/p&gt;       &lt;p style="color: rgb(51, 102, 255);"&gt;Well here we have Baby krishna taking his first steps. He takes 2 paces at a time, and does this three times, so he moves 3 x 2 = 6 steps forward:&lt;/p&gt;       &lt;p style="color: rgb(51, 102, 255);" align="center"&gt;&lt;img src="http://www.mathsisfun.com/images/multiply-p2p3.gif" height="150" width="588" /&gt; &lt;/p&gt;    &lt;p style="color: rgb(51, 102, 255);"&gt;Now, Baby krishna can also step backwards (he is a clever little guy). His Dad puts him back at the start and then krishna steps backwards 2 steps, and does this three times:&lt;/p&gt;        &lt;p style="color: rgb(51, 102, 255);" align="center"&gt;&lt;img src="http://www.mathsisfun.com/images/multiply-n2p3.gif" height="152" width="589" /&gt;&lt;/p&gt;    &lt;p style="color: rgb(51, 102, 255);"&gt;Once again krishna's Dad puts him back at the start, but facing the other way.krishna takes 2 steps forward (for him!) but he is heading in the negative direction. He does this 3 times: &lt;/p&gt;    &lt;p style="color: rgb(51, 102, 255);" align="center"&gt;&lt;img src="http://www.mathsisfun.com/images/multiply-p2n3.gif" height="151" width="589" /&gt;      &lt;/p&gt;    &lt;p style="color: rgb(51, 102, 255);"&gt;Back at the start again (thanks Dad!), still facing in the negative direction, he tries his backwards walking, once again taking two steps at a time, and he does this three times: &lt;/p&gt;    &lt;p style="color: rgb(51, 102, 255);" align="center"&gt;&lt;img src="http://www.mathsisfun.com/images/multiply-n2n3.gif" height="153" width="588" /&gt;&lt;/p&gt;    &lt;p style="color: rgb(51, 102, 255);"&gt;So, by walking backwards, while facing in the negative                direction, he moves in the positive direction. &lt;/p&gt;       &lt;p style="color: rgb(51, 102, 255);"&gt;And That's All !&lt;/p&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;and finally&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;Axiom based proofs&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;1) a*0 = a*0 + 0 = a*(b + -b) + ab + -(ab) = a(b + -b + b) + -(ab) = ab + -(ab) = 0&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;2) 0*a = 0*a + 0 = (b + -b)*a + ba + -(ba) = (b+ -b + b)a + -(ba) = ba + -(ba) = 0&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;3) a*-a = a*-a + 0 = a*-a + (a*a + -(aa)) = a(-a + a) + -(aa) = a*0 + -(aa) = -(aa)&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(51, 102, 255);"&gt;4) -a*-a = -a*-a + 0 = -a*-a + (a*-a + aa) = (-a + a)*-a + aa = 0*-a + aa = aa&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;p style="color: rgb(51, 102, 255);"&gt;&lt;br /&gt;&lt;/p&gt;&lt;p style="color: rgb(51, 102, 255);"&gt;&lt;br /&gt;&lt;/p&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/8723800153584323203-496864447002041461?l=killmaths.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://killmaths.blogspot.com/feeds/496864447002041461/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://killmaths.blogspot.com/2008/11/proof-of.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/496864447002041461'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/8723800153584323203/posts/default/496864447002041461'/><link rel='alternate' type='text/html' href='http://killmaths.blogspot.com/2008/11/proof-of.html' title='Proof of     -*-  =   +'/><author><name>Raghuram M V</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry></feed>
