0=2, math inside.

[Hiphophippo]

I don't know if you know this, but those are two different numbers. As a side note, I would suspect every programmer ever would disagree with you.

 

[Auron225]

You can get numbers to say anything if you do it right. For example.

X=Y+1
Y=X+1

Those equations cant work together.

Substitute Y=X+1 into that first equation and you get X = X + 1 + 1, which is X = X + 2, which doesnt work.

 

[Nivag the Owl]

Mathematics as an expression isn't always 100% perfect and you can occasionally stumble on the "proof" for ridiculous claims like this. I remember reading some one 1 = 2. I remember it said a good way to disprove it is to actually apply it to a physical experiment.

 

[Xvito]

Lets not start this again.

Almost as bad as the infamous 0.9 repeating = 1 thread.

I hate whoever decided infinite 0.9 = 1. By that logic you could say that every number is the same.

No, you can't say that.

0.999... = 1

Without this we wouldn't, as Xenon (ancient Greece) pointed out, be able to run past people who are slower than us.

 

[Glademaster_v1legacy]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

For starters there is no value for X that is the problem right there you just seem to be pulling the value out of your arse without it being given as cos of a variable is not always equal to 1 neither is sin of a variable always equal to 0.

 

[dont_blink]

Okay, here is your problem. Cos^2(x) =/= (Cosx)^2. What you were doing was saying EXACTLTY that. You pretty much made a fallacy like saying cos(x) + cos (x) = cos (2x).

i think what you mean is cos[sup]2[/sup]x =/= cosx[sup]2[/sup]
because i think cos[sup]2[/sup](x) = (cosx)[sup]2[/sup]

sorry, correct me if i'm wrong; this whole thread has bent my brain a little...

 

[MurderousToaster]

Try telling that to binary.

 

[dont_blink]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

For starters there is no value for X that is the problem right there you just seem to be pulling the value out of your arse without it being given as cos of a variable is not always equal to 1 neither is sin of a variable always equal to 0.

x = pi

Try telling that to binary.

0=10

>>

 

[Doug]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

You can't root cos[sup]2[/sup]x by just removing the square of cos, since you're treating sin[sup]2[/sup]x as one element.

Agreed. You can't square root both sides of an equation. You could divide both sins of the equation by cos x, and get cos x = (1 - sin[sup]2[/sup] x) / cos x

At x = pi,
-1 = (1 - 0) / -1
-1 = -1

 

[Glademaster_v1legacy]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

For starters there is no value for X that is the problem right there you just seem to be pulling the value out of your arse without it being given as cos of a variable is not always equal to 1 neither is sin of a variable always equal to 0.

x = pi

Try telling that to binary.

0=10

>>

Well there is the problem cos pi is not equal to 1 and sin pi is not equal to 0.

 

[Glademaster_v1legacy]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

Cos pi is not equal to 1 and sin pi is not equal to zero problem sovled.

 

[Asturiel]

How do you justify these two?

X=Y+1
Y=X+1

That just makes no sense.

Those equations cant work together.

Substitute Y=X+1 into that first equation and you get X = X + 1 + 1, which is X = X + 2, which doesnt work.

Holy crap the math police questioned my 5=7 equation RUN AWAY!*Runs while flailing arms*

 

[dont_blink]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

For starters there is no value for X that is the problem right there you just seem to be pulling the value out of your arse without it being given as cos of a variable is not always equal to 1 neither is sin of a variable always equal to 0.

x = pi

Try telling that to binary.

0=10

>>

Well there is the problem cos pi is not equal to 1 and sin pi is not equal to 0.

YOU FOUND IT!! =D
cos[pi]=-1
i looked it up =D

well done. you win this thread.

 

[benylor]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

For starters there is no value for X that is the problem right there you just seem to be pulling the value out of your arse without it being given as cos of a variable is not always equal to 1 neither is sin of a variable always equal to 0.

Sorry, but sin^2(x) + cos^2(x) = 1 for all x.

Make a circle with radius 1 centred on the origin. Any point on the circle will be 1 unit away from the origin. Now, if you make a right-angled triangle there with the hypotenuse going from the origin to any point on the circle you can see that the co-ordinate of that point will be (cos x, sin x). According to Pythagoras' Theory, then cos^2(x) + sin^2(x) = 1^2 = 1, for all x.

So there's no problems there.

Mathematics as an expression isn't always 100% perfect and you can occasionally stumble on the "proof" for ridiculous claims like this. I remember reading some one 1 = 2. I remember it said a good way to disprove it is to actually apply it to a physical experiment.

No. If you stumble on that sort of proof, then you've made an error in your calculations. Check for divisions by zero, not being careful with square roots, etc. The only time when maths breaks down is if you're trying to apply a model to describe a physical experiment (because there will always be factors the model will ignore). If you're trying to do something strictly with numbers, without using experimental data at all, and you get a result telling you 1 = 2, then you've made a mistake.

Maths is perfect. We haven't discovered everything we need to know, but never be so arrogant as to blame mathematics for being inconsistant. It's not maths that's wrong, it's YOUR maths that are wrong. It's when you apply the maths to real things when you get errors. Look at trying to predict the weather, for example.

 

[benylor]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

For starters there is no value for X that is the problem right there you just seem to be pulling the value out of your arse without it being given as cos of a variable is not always equal to 1 neither is sin of a variable always equal to 0.

x = pi

Try telling that to binary.

0=10

>>

Well there is the problem cos pi is not equal to 1 and sin pi is not equal to 0.

Sin pi is equal to 0. Real maths doesn't use degrees, switch to radians and it works.

Cos pi doesn't equal 1, I'll give you that, but the OP does not say it equals 1 at any point. He says -1. That is correct. As remarked before, the problem is that he's forgotten that sqrt[cos^2(x)] = EITHER cos(x) OR -cos(x).

 

[Glademaster_v1legacy]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

For starters there is no value for X that is the problem right there you just seem to be pulling the value out of your arse without it being given as cos of a variable is not always equal to 1 neither is sin of a variable always equal to 0.

Sorry, but sin^2(x) + cos^2(x) = 1 for all x.

Make a circle with radius 1 centred on the origin. Any point on the circle will be 1 unit away from the origin. Now, if you make a right-angled triangle there with the hypotenuse going from the origin to any point on the circle you can see that the co-ordinate of that point will be (cos x, sin x). According to Pythagoras' Theory, then cos^2(x) + sin^2(x) = 1^2 = 1, for all x.

So there's no problems there.

Mathematics as an expression isn't always 100% perfect and you can occasionally stumble on the "proof" for ridiculous claims like this. I remember reading some one 1 = 2. I remember it said a good way to disprove it is to actually apply it to a physical experiment.

No. If you stumble on that sort of proof, then you've made an error in your calculations. Check for divisions by zero, not being careful with square roots, etc. The only time when maths breaks down is if you're trying to apply a model to describe a physical experiment (because there will always be factors the model will ignore). If you're trying to do something strictly with numbers, without using experimental data at all, and you get a result telling you 1 = 2, then you've made a mistake.

Maths is perfect. We haven't discovered everything we need to know, but never be so arrogant as to blame mathematics for being inconsistant. It's not maths that's wrong, it's YOUR maths that are wrong. It's when you apply the maths to real things when you get errors. Look at trying to predict the weather, for example.

But it isn't sin^2 or cos^2 as cos^2 has been canceled while sin^2 remains so we are left with a sin^2 and a cos so cos pi is not 1.

 

[shroomz]

Agreed. You can't square root both sides of an equation. You could divide both sins of the equation by cos x, and get cos x = (1 - sin[sup]2[/sup] x) / cos x

At x = pi,
-1 = (1 - 0) / -1
-1 = -1

Yes you can, otherwise a lot of maths we base technology on would be instantly wrong

 

[Glademaster_v1legacy]

First off, this particular proof is search bar approved, and does not contain division by zero.

cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2


So where is the error?

EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.

For starters there is no value for X that is the problem right there you just seem to be pulling the value out of your arse without it being given as cos of a variable is not always equal to 1 neither is sin of a variable always equal to 0.

x = pi

Try telling that to binary.

0=10

>>

Well there is the problem cos pi is not equal to 1 and sin pi is not equal to 0.

Sin pi is equal to 0. Real maths doesn't use degrees, switch to radians and it works.

Cos pi doesn't equal 1, I'll give you that, but the OP does not say it equals 1 at any point. He says -1. That is correct. As remarked before, the problem is that he's forgotten that sqrt[cos^2(x)] = EITHER cos(x) OR -cos(x).

I'm sorry what do you define as rel maths degrees are perfectly fine. Also if he is using degrees or radians this should have already been stated or it is assumed degrees are used.

 

[benylor]

SNIP
But it isn't sin^2 or cos^2 as cos^2 has been canceled while sin^2 remains so we are left with a sin^2 and a cos so cos pi is not 1.

You misunderstand, sir.

The first equation in the problem is cos^2(x) = 1-sin^2(x). This is true for ANY x. You can pick any value of x you like for that equation, and it will still be true - it's more than an equation, it's an identity.

So, the flaw in the maths can not be that he has pulled the value of x out of his arse, as you have said. The first line is true for any x.

 

[Glademaster_v1legacy]

SNIP
But it isn't sin^2 or cos^2 as cos^2 has been canceled while sin^2 remains so we are left with a sin^2 and a cos so cos pi is not 1.

You misunderstand, sir.

The first equation in the problem is cos^2(x) = 1-sin^2(x). This is true for ANY x. You can pick any value of x you like for that equation, and it will still be true - it's more than an equation, it's an identity.

So, the flaw in the maths can not be that he has pulled the value of x out of his arse, as you have said. The first line is true for any x.

If you really want to go down that root then you can write square root of 2 as a fraction and get 2=1. While it may be true what he was done when he used X as number(pi) is not true as he has not stated that he is using radians so the general world will assume he is using degree even taking that into account. As you said the flaw is also a square root could be the number+ or -.