Math and logic geniuses, show us your stuff!

[Bat-mite]

No fair if you already know the answer.

I can prove that 2 =1. Can you show me how I'm wrong?

Assume a = b != 0. IOW, a and b are two equal quantities which are not zero.

If a = b

then a^2 = ab

and a^2 - b^2 = ab -b^2

factorizing

(a - b)(a + b) = b(a-b)

divide both sides by (a - b)

a + b = b

Since a = b

then 2b = b

and dividing both sides by b leaves

2 = 1.

This is obviously wrong, but where does the proof veer off course?

 

[KennB]

Dividing by zero -- (a-b)=0 -- is one spot.
Are there others?

 

[mbeels]

divide both sides by (a - b)

Well... if a=b, then dividing by zero is undefined.

 

[Bat-mite]

Well... if a=b, then dividing by zero is undefined.

Gee, that didn't take long. Congrats!

 

[Bat-mite]

Dividing by zero -- (a-b)=0 -- is one spot.
Are there others?

Nope, that's the one. Congrats.

 

[jsdemar]

If you take each side of the equality:
(a - b)(a + b) = b(a-b)
and divide by (a-b), let x=(a-b) then take the limit of x->0:

To solve these, you can try applying the product rule: "the limits of a products is the product of the limits". Unfortunately, the product rule only applies when none of the terms approach 0. So, the full solution has to be done applying calculus, solving both from the left and the right.

Both of these limit problems have a different solution as x approaches zero from the negative side versus the positive side, therefore both general limits do not exist at x=0 (there's a discontinuity). Since the two equations do not have limits that exist as (a-b)->0, they can not be used in an equality to simplify further.

Another interesting thing is that (a^2 - b^2)/x is a Laurent Series. At x=0, the solutions to the series expansion only exists in the complex plain.

 

[tomsteve]

theres not enough bacon involved for me to answer.

 

[dhbarr]

One isn't prime. I just found that out the other day, still trying to come to terms with it.

Maybe I just won't do anything fancy with numbers smaller than three, seems safer.

 

[blackjack2564]

According to professor Giddy-Yup at the Jethro Bodine school of chiphering 1+1=11
Well....it looks right.

 

[KennB]

Thanks to Jim Flis for the image.

 

[adrian]

If the answer is "FF", what is the Shakespearean question?

Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and let t = a + b. Then
a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b

 

[Bat-mite]

Forgive my poor math skills, but how are you getting from

a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4

to

(a - t/2)^2 = (b - t/2)^2

 

[SecondRow]

He just factored. Look at the left side and work backwards:
(a - t/2)^2
(a - t/2)*(a - t/2)
a^2 - at/2 - at/2 + t^2/4
a^2 - at + t^2/4

Works for the right side as well. By adding the t^2/4 term in the previous line, he used a algebraical tool called “completing the square.”

 

[jderimig]

According to professor Giddy-Yup at the Jethro Bodine school of chiphering 1+1=11
Well....it looks right.

1+1=10

 

[Bat-mite]

He just factored. Look at the left side and work backwards:
(a - t/2)^2
(a - t/2)*(a - t/2)
a^2 - at/2 - at/2 + t^2/4
a^2 - at + t^2/4

Works for the right side as well. By adding the t^2/4 term in the previous line, he used a algebraical tool called “completing the square.”

When using real integers, however, this is where the proof breaks down. I chose a = 5 and b = 6. I end up (5 - 11/2)^2 = (6 - 11/2)^2

 

[SecondRow]

You asked for how he got from line 5 to line 6. I was explaining how. There’s no error there. The error in the proof is from from line 6 to line 7. He’s taking the square root of a negative number.

 

[adrian]

Not so. Going from line 5 to line 6 converts the equations into squares. There being no complex numbers at line 5 (unless you chose to use them for a or b), both the squares on line 6 must be positive.

Bat-Mite has almost cracked it...

 

[Bat-mite]

Not so. Going from line 5 to line 6 converts the equations into squares. There being no complex numbers at line 5 (unless you chose to use them for a or b), both the squares on line 6 must be positive.

Bat-Mite has almost cracked it...

Got it, I think, but I was inspired by SecondRow, so give him the credit.

From line 6 to line 7, you taking the square root of both sides. But a square root can be positive or negative; i.e., 144 = 144, so 12 = -12. Since there are possible logical paths from a square root, the proof is insufficient and thus breaks down.

 

[BDB]

The first issue I noticed was the word "factorizing."

 

[SecondRow]

Yikes. You’re right. I got caught up in realizing
a-t/2 = -(b - t/2)
that I thought he was taking the square root of a negative number, forgetting that -1^2 = 1.

 

[adrian]

Well done, Bat-Mite and SecondRow.

Now, how about the other question, about the Shakespearean FF?

And how high can you count on your fingers? (This isn't a trick word question, though the Shakespearean one might be. You're allowed to use your thumbs if you like, in whatever way you see fit.)

 

[SecondRow]

My thought on the Shakespeare question is that it’s a play on
To be OR NOT to be
In the Boolean sense of those terms, but I can’t make it work right.

And if the finger counting is in binary, you can get pretty high. 2^5 - 1 = 31 with one hand. 2^10 - 1 with both hands.

 

[Bat-mite]

And how high can you count on your fingers?

I guess if I were in a jetliner cruising at 35,000', I could still count on my fingers.

This isn't a trick word question

D'oh!

 

[adrian]

My thought on the Shakespeare question is that it’s a play on
To be OR NOT to be
In the Boolean sense of those terms, but I can’t make it work right.

It is right. "2B OR NOT 2B, that is the question."

And if the finger counting is in binary, you can get pretty high. 2^5 - 1 = 31 with one hand. 2^10 - 1 with both hands.

Winner!

Though it could be a bit tricky keeping all the fingers in their correct states. A more robust system was in use in old times. Each finger has three parts, so you use your thumb to point at each part in turn and you can easily count up to 12 on one hand.

Someone else's turn to set a question...

 

[SecondRow]

It is right. "2B OR NOT 2B, that is the question.

But isn't "A OR NOT A" a tautology? I can't figure out why the answer would be "FF".

 

[SecondRow]

Ok. Here's one I've always liked...

My son's school has 1000 students. There are 1000 lockers in the school, numbered 1 to 1000. The principal decides to run an experiment (it's the end of the year and all the kids have already passed their state mandated tests, amazing I know). Before he starts, the principal makes sure all the lockers are closed and has all the students line up outside the school. The principal has the first student in line go through and open up every locker. Then, the second student closes every second locker (2, 4, 6,...). Then, the third student goes through and changes the state of every third locker. That is, the student goes to locker numbers 3, 6, 9, 12 ..... If the locker is closed, the student opens it. If it's opened, the student closes it. This continues on throughout the day until the last student finally gets to locker #1000 and changes its state.

At the end of the experiment, how many lockers are open?

Edit: Feel free to ask clarifying questions. I may have worded this poorly.

 

[Mushtang]

31

 

[SecondRow]

Yep. And could you solve it quickly for a larger number of lockers and students, say 10 million? What special property do all the open lockers have?

 

[jsdemar]

In Matlab:

L = zeros(1000,1); % all closed to begin
for S=1:1000 % for each student...
for N=1:1000 % check each locker number...
if mod(N,S) == 0 % if modulus locker/student is 0
L(N) = not(L(N)); % toggle the locker door
end
end
end
sum(L,'all') % sum of open lockers
​

ans = 31

 

[SecondRow]

Heh. This is my fault for wording the question the way I did. I was hoping a computer wouldn’t be used. There’s a way to quickly get to the solution with just pencil and paper. I’d still like to see someone do it that way. I’ll leave it open for a little while longer before I post that method.