1 = 2 revisited

[bonanova]

It is a beginner-level bit of mathematical magic to prove that 1=2.

This demonstration probably occurs multiple times in the history of this forum.

Its fallacy depends on embedding, and cleverly hiding somewhere in the "analysis," a division by zero.

Which of course is not permitted. So regarding this matter our world is still a safe abode.

Nonetheless, this equality still has a life for puzzle solvers.

By the application of a familiar proof, the addition of a single well-known mathematical symbol,

and without resorting to prohibited mathematical operations, 1=2 can still be shown to be true.

Have fun, and please use spoilers.

[vistaptb]

Put a slash through the equal sign?

[phil1882]

the most familiar proof i can think of is a^2 +b^2 = c^2.

if i take the logarithm of both sides; i get

log(a^2 +b^2) = log(c^2)

log(a^2 +b^2) = 2*log©

log(a^2 +b^2)/log© = 2

how to show this is also equal to 1 however has me stymied.

[superprismatic]

We will prove a more general result using a simple induction argument.

We prove: Any set of numbers are all equal.

Proof:
(Base Case) -- If we have a set consisting of one number, it is clear
that all the numbers in this set are equal.
(Inductive Step) -- We assume that any set of N numbers are equal.
Now, we will show that any set of N+1 numbers are all equal. Let us
take one number out of the set of N+1. We are now left with a set
of N numbers and, by our induction hypothesis, all N of these are
equal. So, we put the number we took out (call this number X)
back in the set to again make it a set of N+1 numbers. We now
take a different number out of this set (call this number Y).
Now, we once again have a set of N numbers which, by the induction
hypothesis, have all equal numbers. The only way this can happen
is if X=Y because each of X and Y are equal to the other numbers in
the set. So, any set of N+1 numbers must all be equal.
(Therefore) By induction, any set of numbers must contain numbers
which are all equal.

In particular, the set {17.3,1,-6,2} must contain numbers which are all
equal. Thus 1=2.

[bushindo]

We will prove a more general result using a simple induction argument.

We prove: Any set of numbers are all equal.

Proof:

(Base Case) -- If we have a set consisting of one number, it is clear

that all the numbers in this set are equal.

(Inductive Step) -- We assume that any set of N numbers are equal.

Now, we will show that any set of N+1 numbers are all equal. Let us

take one number out of the set of N+1. We are now left with a set

of N numbers and, by our induction hypothesis, all N of these are

equal. So, we put the number we took out (call this number X)

back in the set to again make it a set of N+1 numbers. We now

take a different number out of this set (call this number Y).

Now, we once again have a set of N numbers which, by the induction

hypothesis, have all equal numbers. The only way this can happen

is if X=Y because each of X and Y are equal to the other numbers in

the set. So, any set of N+1 numbers must all be equal.

(Therefore) By induction, any set of numbers must contain numbers

which are all equal.

In particular, the set {1,2} must contain numbers which are all

equal. Thus 1=2.

Counter example to this proof =)

We use the same set- {1, 2}. Following the reasoning above, then

X = 1

Y = 2

but it doesn't follow that X = Y. I guess N=2 is the weakest link of this induction chain.

[superprismatic]

We will prove a more general result using a simple induction argument.

We prove: Any set of numbers are all equal.

Proof:

(Base Case) -- If we have a set consisting of one number, it is clear

that all the numbers in this set are equal.

(Inductive Step) -- We assume that any set of N numbers are equal.

Now, we will show that any set of N+1 numbers are all equal. Let us

take one number out of the set of N+1. We are now left with a set

of N numbers and, by our induction hypothesis, all N of these are

equal. So, we put the number we took out (call this number X)

back in the set to again make it a set of N+1 numbers. We now

take a different number out of this set (call this number Y).

Now, we once again have a set of N numbers which, by the induction

hypothesis, have all equal numbers. The only way this can happen

is if X=Y because each of X and Y are equal to the other numbers in

the set. So, any set of N+1 numbers must all be equal.

(Therefore) By induction, any set of numbers must contain numbers

which are all equal.

In particular, the set {1,2} must contain numbers which are all

equal. Thus 1=2.

Counter example to this proof =)

We use the same set- {1, 2}. Following the reasoning above, then

X = 1

Y = 2

but it doesn't follow that X = Y. I guess N=2 is the weakest link of this induction chain.

it's the ONLY weak link....Give a guy a little break here!

[bonanova]

Good thinking so far, but it's not as complicated as it looks.

You'll just have to go a bit more outside the box.

[bonanova]

Put a slash through the equal sign?

Actually not bad.

But not the solution I had in mind.

[superprismatic]

Let T be the square root of 2. Consider the infinite continued

exponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearly
have X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4
because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets
1=2. QED

[superprismatic]

Start with the trig identity

cos(2x)=cos²(x)-sin²(x)
re-arrange the equation to
cos(2x)+sin²(x)=cos²(x)
take square roots
sqrt(cos(2x)+sin²(x))=cos(x)
divide by sqrt(2) to get
sqrt(cos(2x)+sin²(x))/sqrt(2)=cos(x)/sqrt(2)
add (3/2) to both sides
(3/2)+sqrt(cos(2x)+sin²(x))/sqrt(2)=(3/2)+cos(x)/sqrt(2)
evaluate at x=(3*pi/4) to get
2=1.
Voila!

[bonanova]

Let T be the square root of 2. Consider the infinite continuedexponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearlyhave X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets1=2. QED

This is great - beautiful, even.

Except, the power tower unfortunately does not converge for both cases (for 2, but alas not for 4 as I remember.)

I made a graph of the divergence point sometime back, I will look for it.

When the records are written, however, this def gets Honorable Mention.

[bushindo]

Let T be the square root of 2. Consider the infinite continuedexponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearlyhave X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets1=2. QED

This is great - beautiful, even.

Except, the power tower unfortunately does not converge for both cases (for 2, but alas not for 4 as I remember.)

I made a graph of the divergence point sometime back, I will look for it.

When the records are written, however, this def gets Honorable Mention.

Great proof as well. I have no beef with the power tower, but isn't the argument similar to this one

Let's say that a variable x satisfies the following relationship,

x^2 - 6x + 8 = 0

Since x=2 and x=4 both satisfy the equation, then 2 = 4, or 1=2. It should be easy to find the weak point now.

[bonanova]

Start with the trig identitycos(2x)=cos

²(x)-sin²(x)re-arrange the equation tocos(2x)+sin²(x)=cos²(x)take square rootssqrt(cos(2x)+sin²(x))=cos(x)divide by sqrt(2) to getsqrt(cos(2x)+sin²(x))/sqrt(2)=cos(x)/sqrt(2)add (3/2) to both sides(3/2)+sqrt(cos(2x)+sin²(x))/sqrt(2)=(3/2)+cos(x)/sqrt(2)evaluate at x=(3*pi/4) to get2=1.Voila!

Ah, the old dazzle 'em with double angles and parentheses ploy!

Remember it well from my college days.

Did you think I would actually evaluate all those functions?

Unfortunately that is unnecessary. It came out wrong by a factor of 4.

While the problem asked for a "proof" that 1=2; you ended up with 2=1; I.e., that 4=2.

In a dyslexic world, where a skeptic is unsure there is such a thing as a dog, that might suffice.

But I'm afraid the quest must go on.

Efforts should be rewarded, nonetheless. Thus, a clue.

If you get far enough out of the box, it might be the start of a relaxing hour after work.

[bonanova]

Let T be the square root of 2. Consider the infinite continuedexponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearlyhave X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets1=2. QED

This is great - beautiful, even.Except, the power tower unfortunately does not converge for both cases (for 2, but alas not for 4 as I remember.)I made a graph of the divergence point sometime back, I will look for it.When the records are written, however, this def gets Honorable Mention.

Great proof as well. I have no beef with the power tower, but isn't the argument similar to this one

Let's say that a variable x satisfies the following relationship,x^2 - 6x + 8 = 0Since x=2 and x=4 both satisfy the equation, then 2 = 4, or 1=2. It should be easy to find the weak point now.

Nicely debunked Bushindo.

However, as I stand now the the company of two great logicians/mathematicians, I tread lightly regarding weak points, knowing the desired solution might likewise be so described. Its redeeming quality being only that it fits the given conditions so nicely.

[bonanova]

Write: For x > 0,

x + x + x + ... + x (x terms) = x²

Differentiate with respect to x:

1 + 1 + 1 + ... + 1 (x terms) = 2x

x = 2x.

Divide by x > 0.

1 = 2.

No. Sorry that's not it, either. Keep trying.

[BobbyGo]

I remember reading somewhere (it might have even been this site, for all I know) that 1 = 2 was used in some kind of physics conversion. Ohms to amps maybe? Or something close-about?

[CaptainEd]

Yes, but that's physics.

My father was a nuclear physicist in the 50's, in the slide rule era. When I was in high school, he explained to me a couple of important principles of applied mathematics:

* For large enough N, anything = 1

* For really large enough N, anything = 0

As a corollary, he pointed out (for a different problem than this one), that "2 is one of the larger numbers that is equal to 1".

But that's physics...

[bonanova]

Uhhh ... There! I just cut a king-size hole in the box so we can all leave..

Since we're looking for a proof and a symbol, I considered saying what the symbol is.

Suffice it that the symbol would probably bring the search to an end, and specifying the proof most certainly would.

So I'll just let that information itself be the clue.

Relax, and let the solution come to you.

[Prime]

I haven't read the entire thread, not sure if someone has not suggested that already.

(1 - 1.5)

² = 0.25
sqrt((1 - 1.5)²)= sqrt(0.25)
1 - 1.5 = 0.5
1 = 2

[TimeSpaceLightForce]

1 unit = 2 units vice versa

Time dilation: Special Relativity

A and B are reference frames moving at v m/s.
So that:
A observed (compares) B clock at 1 second
mark while his clock is at 2 seconds mark.
Same is true to:
B observed (compares) A clock at 1 second
mark while his clock is at 2 seconds mark.

Physics are describe by mathematical means
Like Lorentz transformations.

[plasmid]

If we're looking for a proof and a mathematical symbol

the proof is a pitcher of 10 proof lager

the symbol is ∞ (infinity if your font doesn't match mine)

and with beer goggles, 1 can definitely be 2

[bonanova]

If we're looking for a proof and a mathematical symbol

the proof is a pitcher of 10 proof lager

the symbol is ∞ (infinity if your font doesn't match mine)

and with beer goggles, 1 can definitely be 2

Thanks plasmid for taking us out of the box.

With the pitcher and goggles, and infinity tho, I lost track of where the equation fit in.

You might be on the right track, but we're not quite there yet.

[James33]

How about considering it all modulo 1?

[bonanova]

my OPs give all the clues necessary to solve.

Let me point out two for this case.

Note that:

1. The OP contrasts clever math - like which
   I think is very clever - with the answer needed to solve this puzzle.
   So the answer is not of that type.
   
2. It asserts the equation can stand, with the addition of two things.
   Could it be as simple as saying 1 ten-dollar bill = 2 five-dollar bills?
   That would be adding two things, but those two things are so similar
   it really would not make a puzzle.
   

The "aha" moment should be forthcoming shortly.

[bonanova]

Add a percent sign.

[bonanova]

Add a percent sign.

1% = 2 proof