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## Question

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?

## 11 answers to this question

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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

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...perhaps...

Edited by sandstorm

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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?

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

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?

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

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

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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.

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

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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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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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