NBVegita Posted November 5, 2006 Report Posted November 5, 2006 So I was reading some posts and saw rootbears sig and looked at it...and it does not work out. a=1 b=1 a=b a*a=a*b Multiply 'a' to both sides. a^2=ab Simplify. a^2-b^2=ab-b^2 Subtract 'b^2' from both sides.(a+ (a- =b(a- Factor. a+b=b Divide both sides by 'a-b' 1+1=1 2=1 When theoretically proving a math equation, you may always multiply both sides of the equation, now it must be both sides, with the exact same value, as done in line 3. But you may never add a value to the equation, because as you can see from the resultant it (can) return a value that is false, and cannot be. It is a basic rule of theoretical Mathematics. Sorry just bugged me lol
Deathboy-evil Posted November 5, 2006 Report Posted November 5, 2006 Word. NB, you got too much time on your hands.
»SOS Posted November 5, 2006 Report Posted November 5, 2006 However, you are absolutely wrong Adding/subtracting is perfectly fine. It is a basic rule of 3rd grade mathematics The issue here is this:Divide both sides by 'a-b' This is valid only if a != b, which is not true. Therefore, it is invalid (division by zero). Another way to look at it would be to view the operation as a function - you will see that as (a - b ) approaches 0 (as they approach equality), the value of the function approaches infinity (which happens in case of divide by zero). The later part is basic calculus
Samapico Posted November 5, 2006 Report Posted November 5, 2006 So I was reading some posts and saw rootbears sig and looked at it...and it does not work out. a=1 b=1 a=b a*a=a*b Multiply 'a' to both sides. a^2=ab Simplify. a^2-b^2=ab-b^2 Subtract 'b^2' from both sides.(a+B )(a-B )=b(a-B ) Factor. a+b=b Divide both sides by 'a-b' 1+1=1 2=1 When theoretically proving a math equation, you may always multiply both sides of the equation, now it must be both sides, with the exact same value, as done in line 3. But you may never add a value to an equation, because as you can see from the resultant it (can) return a value that is false, and cannot be. It is a basic rule of theoretical Mathematics. Sorry just bugged me lol actually, that's not the point. You CAN add/substract the same value on both sides of the equation.if a = a, then a + b = a + b The reason why this thing above is wrong is because you divide by 0 at some point.When you divide both sides by ( a - b ) , you are actually dividing by 0 because a = b. This proves that dividing by 0 makes no sense, mathematically speaking. edit: SOS beat me to it...
NBVegita Posted November 5, 2006 Author Report Posted November 5, 2006 Sorry for it being early, what I was trying to state was that adding something to both sides of an equation can not help you solve that equation. And as for the division, the function, with predifined variables would be invalid, but the theory behind the equation would still be valid. I was thinking two things and they got melded into one sentence.
Vile Requiem Posted November 6, 2006 Report Posted November 6, 2006 The limit of nerds in a math topic as the posts approach infinity is: NBV(SOS + Samapico) + Vile. (Believe it or not, that equation actually fits. Think about it.)
X`terrania Posted November 6, 2006 Report Posted November 6, 2006 *walken impression* okay --licks lips- the premise for this question, amazing, it is, really, okay. What the !@#$%^&*? I had over 666 posts, and now I'm down to 600?
»SOS Posted November 6, 2006 Report Posted November 6, 2006 It ("the theory" as you call it) would be invalid even without predefined parameters. It would be OK only for pre-highschool level problems where nobody cares about issues like this. Why am I multiplied with NBV, Vile?
Vile Requiem Posted November 6, 2006 Report Posted November 6, 2006 As you and Sama generally would reply just to NBV, his postings are a a scaling factor of how much you will post. Or something like that.
»SOS Posted November 6, 2006 Report Posted November 6, 2006 But then it should be the inverse, since we post less if only replying to him! Of course, another option is that NBV is just less than one
rootbear75 Posted November 8, 2006 Report Posted November 8, 2006 wow, i leave the forums for 4 days and look what happens! lol im so famous!
rootbear75 Posted November 9, 2006 Report Posted November 9, 2006 heres something to think about,take away the predefined variables. you still end up with a+b=balso, if you take away the predefined variables, you must make the rule, a!=b ; a!=0 ; b!=0
Samapico Posted November 9, 2006 Report Posted November 9, 2006 you must make the rule, a!=bwhich is in complete contradiction with the first statement of the "proof":a=b actually, a=1 and b=1 isn't needed to "prove" that 2=1at then end you have:a + b = bb + b = b (because a = b )2 = 1
rootbear75 Posted November 9, 2006 Report Posted November 9, 2006 you must make the rule, a!=bwhich is in complete contradiction with the first statement of the "proof":a=boops xD
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