frosch45 Posted March 23, 2008 Share Posted March 23, 2008 does anyone know how to do this? my teacher said that she can prove it, she said that in ceartain instances, numbers may be forced to do things that they are not supposed to do.... alternatively, does anyone have any other simple problems like this that you could prove like 1+5=5 or something seriously, any ideas at all, no matter how absurd OR complex, would be appreciated Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted March 23, 2008 Share Posted March 23, 2008 I can prove that 2 = 1. Let's assume that we have two numbers, a and b, and a = b. We can see that [math]a^2 = a \times b[/math] because a = b. Subtracting b2 from both sides... [math]a^2 - b^2 = ab - b^2[/math] and then factoring: [math](a + b)(a - b) = b(a - b)[/math] There's an (a - b) on both sides, so it can be canceled out, leaving [math]a + b = b[/math] Because a = b: [math]b + b = b[/math] [math]2b = b[/math] Divide by b: [math]2 = 1[/math] There is, of course, an error, but it's tough to find. [hide]You can't cancel out (a - b). To cancel out, you have to divide by (a - b) -- and (a - b) = 0. Dividing by zero isn't allowed.[/hide] Link to comment Share on other sites More sharing options...
5614 Posted March 23, 2008 Share Posted March 23, 2008 Hopefully you know that √(-1)=i Therefore: 1 + 1 = 1 + √1 = 1 + √(-1*-1) [as you know √(ab) = √a√b] = 1 + √(-1)√(-1) = 1 + i*i [as √(-1)=i, i²=-1] = 1 - 1 = 0 Thus 1+1=0. QED. Of course like always there's a flaw in it. Now I've shown off the one I know(!), here's a good link: http://en.wikipedia.org/wiki/Invalid_proof also Google searches will give you more, if you want. Link to comment Share on other sites More sharing options...
Daecon Posted March 24, 2008 Share Posted March 24, 2008 I can prove that 2 = 1. Let's assume that we have two numbers, a and b, and a = b. We can see that [math]a^2 = a \times b[/math] because a = b. Subtracting b2 from both sides... [math]a^2 - b^2 = ab - b^2[/math] and then factoring: [math](a + b)(a - b) = b(a - b)[/math] There's an (a - b) on both sides, so it can be canceled out, leaving [math]a + b = b[/math] Because a = b: [math]b + b = b[/math] [math]2b = b[/math] Divide by b: [math]2 = 1[/math] There is, of course, an error, but it's tough to find. [hide]You can't cancel out (a - b). To cancel out, you have to divide by (a - b) -- and (a - b) = 0. Dividing by zero isn't allowed.[/hide] If a and b have exactly the same value, then why bother using two different letters for it? Does that sequence of equations still work if you only use one letter? Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted March 24, 2008 Share Posted March 24, 2008 It's more obvious what the problem is if you use one letter, and hence people notice more quickly. Link to comment Share on other sites More sharing options...
frosch45 Posted March 24, 2008 Author Share Posted March 24, 2008 cool thanks any others?? Link to comment Share on other sites More sharing options...
YT2095 Posted March 24, 2008 Share Posted March 24, 2008 `+` is frequently use to represent OR. if you look at the Truth Table for an OR gate you`ll see that 1+1 is True or 1 (unless it`s an XOR). Link to comment Share on other sites More sharing options...
Daecon Posted March 24, 2008 Share Posted March 24, 2008 I thought "^" was OR? Link to comment Share on other sites More sharing options...
YT2095 Posted March 24, 2008 Share Posted March 24, 2008 `V` Is using Formal logic symbology. but `+` and `.` are also used to represent OR and AND respectively. here: http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/diglog.html and here: http://www.asic-world.com/digital/boolean1.html#Function_Definitions Link to comment Share on other sites More sharing options...
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