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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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Posted · Report post

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 :P

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Posted (edited) · Report post

...perhaps...

Edited by sandstorm
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Posted · Report post

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?

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

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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 ...

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Posted · Report post

From (5) O*E is E not O ...

Thank you, I was trying to do this too quickly.

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Posted · Report post

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 :P

You didn't expand the binomial correctly

(odd+odd)*(odd*even) = (odd*odd) + (odd*odd) + (odd*even) + (odd*even)

That is not true.

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Posted · Report post

Wow, I'm glad I posted this here. I would have never gotten the answer otherwise!

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Posted · Report post

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.

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.

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Posted · Report post

v. O*E = E ==> E/E=O

The inference is not always true:

18/6 = 3

24/6 = 4

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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 ...

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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.

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