A very amusing Paradox

[eastsidebagel]

This doesn't make any sense at all. I want to commit suicide!

 

[Narz]

Shoot 'em all and let God sort it out.

 

[Flying Pig]

Another good one which proves that x=0:

x=y

x2=xy

Multiply by y​

x2 - y2 = xy - y2

Subtract y2​

(x-y)(x+y) = y(x-y)

As x2 - y2 = (x-y)(x+y​

(x+y)=y

Cancel the (x-y)s​

x=0

Remove the brackets and simplify​

 

[Flying Pig]

That didn't take long... OK, another one:

3>2

3 log(1/2) > 2 log(1/2)

log (1/2)^3 > log (1/2)^2

(1/2)^3 > (1/2)^2

1/8>1/4

 

[Souron]

If we're doing fake proofs, my favorite is:

4-10=9-15
4-10+25/4=9-15+25/4            --add 25/4 to both sides
(2-5/2)(2-5/2)=(3-5/2)(3-5/2)  --factor both sides
2-5/2=3-5/2                    --square root both sides
2=3                            --add 5/2 to both sides

 

[Flying Pig]

thats not a parradox, just a series of incorrect statements. You can only keep an equality after having done an opperation to both sides of it, if that opperation is a continuously increasing functio.
ln(1/2) is negative, so the second line is wrong, and the one's below that are too.

Correct: that's the big flaw. By the way, there is no method of getting from a logical premise via a series of logical steps to an illogical conclusion, so every paradox is just a series of incorrect statements

If we're doing fake proofs, my favorite is:

Code:
4-10=9-15
4-10+25/4=9-15+25/4 --add 25/4 to both sides
(2-5/2)(2-5/2)=(3-5/2)(3-5/2) --factor both sides
2-5/2=3-5/2 --square root both sides
2=3 --add 5/2 to both sides

Square Root both sides is an invalid operation as you are actually multiplying/dividing (depending on how you look at it) by different numbers on either side

 

[Souron]

Square Root both sides is an invalid operation as you are actually multiplying/dividing (depending on how you look at it) by different numbers on either side

That's not quite it. There is no multiplying or dividing, and even if there were, it would be by equivalent expressions. Taking the square of both sides would certainly be valid, because you are just multiplying both sides by equivalent expressions.

You are right that that is the trouble line, and that taking the square root is not done this way, but your reasons for why not is wrong.

Spoiler hint :

What set of two numbers can you multiply to get (3-5/2)(3-5/2)?

 

[Flying Pig]

If we're doing fake proofs, my favorite is:

4-10=9-15
4-10+25/4=9-15+25/4            --add 25/4 to both sides
(2-5/2)(2-5/2)=(3-5/2)(3-5/2)  --factor both sides
2-5/2=3-5/2                    --square root both sides
2=3                            --add 5/2 to both sides

Square root of x = x^1/2 which is a multiplication or a division depending on how you look at it. Supposing it was x and y ^3, and x was 2 and y 3, then it would be 2*2*2 = 8 against 3*3*3 = 27

 

[Souron]

Square root of x = x^1/2 which is a multiplication or a division depending on how you look at it. Supposing it was x and y ^3, and x was 2 and y 3, then it would be 2*2*2 = 8 against 3*3*3 = 27

You need to start with an equality For example this is valid for all x, y:
x=y^3 -> x^2=(y^3)^2=y^6

Specifically try (x=1, y=1), (x=8 y=2), and (x=3, x=2).

Multiplying, dividing, adding, and subtracting can both be done by applying the operation to both sides by with representations of equal expressions. This is essentially the same as multiplying by the same expression, then substituting it on one side for the equivalent alternative.

But taking the square root is not multiplication or division.

 

[Flying Pig]

x^1/2 means "x multiplied by itself half a time", does it not? Anyway, it seems that one has been done to death a bit - any more?

 

[Souron]

That's just the thing, it doesn't. If it did, then it would be valid, just like taking the square is.

Spoiler answer :

The problem is that there are two answers to taking the square root, the positive, and the negative root. In our problem 3-5/2 is the positive root (1/2), and 2-5/2 is the negative root. (-1/2) Therefore if you want to take the square root of both sides, you have to also take the absolute value of the result. This absolute value would render the last step invalid.

 

[Flying Pig]

That's just the thing, it doesn't. If it did, then it would be valid, just like taking the square is.

Spoiler answer :

The problem is that there are two answers to taking the square root, the positive, and the negative root. In our problem 3-5/2 is the positive root (1/2), and 2-5/2 is the negative root. (-1/2) Therefore if you want to take the square root of both sides, you have to also take the absolute value of the result. This absolute value would render the last step invalid.

Oh, OK. I only did O-level maths...