Guest Posted March 20, 2008 Report Share Posted March 20, 2008 I got this one out of a book by the name of Math Magic (I think). It did not provide an answer, so I don't even know what it is. Give it a try! “An odd number plus an odd number is an even number, and an even number plus an odd number is an odd number. OK?” “OK.” “An even number plus an even number is an even number. OK?” “Of course.” “An odd number times an odd number is an odd number, and an odd number times an even number is an even number. OK?” “Yes.” “Then an even number times an even number is an odd number. OK?” “No! It is an even number.” “No! It is an odd number! I can prove it!” How? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 20, 2008 Report Share Posted March 20, 2008 Reveal hidden contents i. odd+odd=even ii. even+even=even iii. odd*odd=odd vi. odd*even=even We want to show that even*even=odd. replacing (i) and (vi) for even: (odd+odd)*(odd*even) = (odd*odd) + (odd*odd) + (odd*even) + (odd*even) = (odd + odd) + (odd + odd) = odd + odd = odd QED Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 20, 2008 Report Share Posted March 20, 2008 (edited) ...perhaps... Edited March 20, 2008 by sandstorm Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 20, 2008 Report Share Posted March 20, 2008 sandstorm said: I got this one out of a book by the name of Math Magic (I think). It did not provide an answer, so I don't even know what it is. Give it a try! “An odd number plus an odd number is an even number, and an even number plus an odd number is an odd number. OK?” “OK.” “An even number plus an even number is an even number. OK?” “Of course.” “An odd number times an odd number is an odd number, and an odd number times an even number is an even number. OK?” “Yes.” “Then an even number times an even number is an odd number. OK?” “No! It is an even number.” “No! It is an odd number! I can prove it!” How? Reveal hidden contents 1. O+O=E 2. E+O=O OK 3. E+E=E OK 4. O*O=O 5. O*E=E OK Q: E*E=O? by 1- E-O=O by 2 O-E=O, E-O=O by 3 E-E=E by 4 O/O=O by 5 E/O=E E/E=O so how about E*E=(O or E): (E/O)*(E/O)=(O or E): E*E=O*(O or E) E*E=O*O or O*E, so E*E=O or O so E*E=O Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 20, 2008 Report Share Posted March 20, 2008 statman said: 5. O*E=E OK Q: E*E=O? so how about E*E=(O or E): (E/O)*(E/O)=(O or E): E*E=O*(O or E) E*E=O*O or O*E, so E*E=O or O so E*E=O From (5) O*E is E not O ... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 20, 2008 Report Share Posted March 20, 2008 brhan said: From (5) O*E is E not O ... Thank you, I was trying to do this too quickly. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 20, 2008 Report Share Posted March 20, 2008 brhan said: Reveal hidden contents i. odd+odd=even ii. even+even=even iii. odd*odd=odd vi. odd*even=even We want to show that even*even=odd. replacing (i) and (vi) for even: (odd+odd)*(odd*even) = (odd*odd) + (odd*odd) + (odd*even) + (odd*even) = (odd + odd) + (odd + odd) = odd + odd = odd QED You didn't expand the binomial correctly (odd+odd)*(odd*even) = (odd*odd) + (odd*odd) + (odd*even) + (odd*even) That is not true. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 21, 2008 Report Share Posted March 21, 2008 Wow, I'm glad I posted this here. I would have never gotten the answer otherwise! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 21, 2008 Report Share Posted March 21, 2008 Noct said: You didn't expand the binomial correctly (odd+odd)*(odd*even) = (odd*odd) + (odd*odd) + (odd*even) + (odd*even) That is not true. Thanks, I agree it is not true. This one should work. Reveal hidden contents i. O+O = E ii. O+E = E iii. E+E = E iv. O*O = O v. O*E = E ==> E/E=O We want to show E*E=O. Combining (i) and (iii), E*E*E = (O+O)*[(O+O)*(E+E)] = (O+O)*(O*E+O*E+O*E+O*E) = (O+O)*E ... this is b/c O*E=E and E+E+E+E=E. ==> E*E = [(O+O)*E]/E ... dividing both sides by E. ==> E*E = [(O*E)+(O*E)]/E ... distributive property ==> E*E = (E+E)/E ... from (v), O*E=E. ==> E*E = E/E ... from (iii), E+E=E ==> E*E = O ... from (v) above. QED. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 21, 2008 Report Share Posted March 21, 2008 Quote v. O*E = E ==> E/E=O The inference is not always true: 18/6 = 3 24/6 = 4 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 21, 2008 Report Share Posted March 21, 2008 bonanova said: The inference is not always true: 18/6 = 3 24/6 = 4 Correct..... But, the conclusion E*E=O is also not true. I think that is where you play the game ... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 21, 2008 Report Share Posted March 21, 2008 bonanova said: The inference is not always true: 18/6 = 3 24/6 = 4 I think the moral of the proof is if you follow a reasoning which is not always true, you end up with a conclusion which will never be true. O*E=E and E*E=E ... but as you said, it is not always true E/E=E or E/E=O. The conclusion was E*E=O ... which will never be true. Quote Link to comment Share on other sites More sharing options...
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Guest
I got this one out of a book by the name of Math Magic (I think). It did not provide an answer, so I don't even know what it is. Give it a try!
“An odd number plus an odd number is an even number, and an even number
plus an odd number is an odd number. OK?”
“OK.”
“An even number plus an even number is an even number. OK?”
“Of course.”
“An odd number times an odd number is an odd number, and an odd number
times an even number is an even number. OK?”
“Yes.”
“Then an even number times an even number is an odd number. OK?”
“No! It is an even number.”
“No! It is an odd number! I can prove it!”
How?
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