Maths problem. take a look.

[Kingofjam]

Well I go to an after school maths thing (basically its just a way for use to get an as level in statistics early ( i am in year 10 at the mo )) and we did something different this time. basically our teacher gave us this equation which proves that 1 is equal to 2. there's a problem with it and we didn't notice because we were too amazed by the fact that 1 could possibly equal 2. Well I will post the equation and a pat on the back goes to anyone who can tell me what is wrong with it.

x=1
y=1


x=y
we times by x
x(squared) = xy
we times by -y(squared)
x(squared) - y(squared) = xy - y(squared)
we factorise (most people should know what this means)
(x - y) (x + y) = y(x - y)
We divide by x-y
x+y=y
and now we simplify
y+y=y
2y+y
2+1

as I said first one to figure it out gets a pat on the back. Someone will probably get this straight away but I was just bored.

also sorry for the (squared) but I don't know how to use superscript or whatever it is on this.

[ZOIDBERG]

If x = y, dividing by x - y is dividing by zero. You can't do that.

Problem solved. (again)

[feeshy]

Pick mee!

The problem is 'We divide by x-y'. If x and y=1 then x-y is 0, and dividing something by zero equals infinity. Or it just doesn't work.

Edit- Pfft, damn my split-second slower typing... But at least I had the courtesy to hide the answer from the unenlightened ones :p

[Sebas]

If x = y, dividing by x - y is dividing by zero. You can't do that.

Problem solved. (again)
Toooo easy :D

In fact, just plain old. :p

[Kingofjam]

yeah I thought something like this would happen. Hmm okay for something new, prove that 0.9999999999 recurring is equal to 1. Most people should be able to do this easily.

[tom93]

x = 0.999999
10x = 9.99999
10x - x = 9
9x = 9
x = 9/9 = 1

[Mash89]

Right, this is an age old one, and I'm not sure if I can remember it totally, but this is what I think it is:

You're on a game show. You have three doors in front of you. One has a posh car behind it and the other two have nothing behind and you have to pick one. So you do. The presenter says you can change your mind if you'd like. Now, mathematically, should you change your mind or stick with the one you've chosen in the first place?

Correct me if that's not right.

[Taekwon-joe]

x = 0.999999
10x = 9.99999
10x - x = 9
9x = 9
x = 9/9 = 1
10x-x=8.999991

edit:
Right, this is an age old one, and I'm not sure if I can remember it totally, but this is what I think it is:

You're on a game show. You have three doors in front of you. One has a posh car behind it and the other two have nothing behind and you have to pick one. So you do. The presenter says you can change your mind if you'd like. Now, mathematically, should you change your mind or stick with the one you've chosen in the first place?

Correct me if that's not right.
If you have to pick 1 then you pick number 1.

[feeshy]

Nope, normally it goes mroe like you pick a door and the host opens one of the other doors to reveal nothing. Should you stay or switch to have the highest probability of winning?

[Mash89]

Er, no. Try using a calculator...

1.1 - 0.1 = 1
1.23 - 0.23 = 1
1.11111 - 0.11111 - 1
9.99 - 0.99 = 9
9.99999999 - 0.99999999 = 9

See?

[esquilax]

Also, the last line should read 2=1. :eng101:

[Taekwon-joe]

Er, no. Try using a calculator...

1.1 - 0.1 = 1
1.23 - 0.23 = 1
1.11111 - 0.11111 - 1
9.99 - 0.99 = 9
9.99999999 - 0.99999999 = 9

See?
Count the amount of 9s after the decimal for x and 10x

[Mash89]

Count the amount of 9s after the decimal for x and 10x

Talk about pedantic...

[GregHeikel]

Nope, normally it goes mroe like you pick a door and the host opens one of the other doors to reveal nothing. Should you stay or switch to have the highest probability of winning?

You should always change. I can't remember why exactly but it's in Curious Incident of the Dog in the Night Time. I need to be able to draw to explain it properly.

[LeoZ]

this get posted every couple of months or so.

[Smartie]

Is the answer 482?

Yes?

I thought so.

[Mash89]

42

[stalefish]

Right, this is an age old one, and I'm not sure if I can remember it totally, but this is what I think it is:

You're on a game show. You have three doors in front of you. One has a posh car behind it and the other two have nothing behind and you have to pick one. So you do. The presenter says you can change your mind if you'd like. Now, mathematically, should you change your mind or stick with the one you've chosen in the first place?

Correct me if that's not right.You missed a bit: After you choose a door, the presenter opens another door to deliberately reveal nothing. Now the solution:

For simplicity, let us assume the prize is behind door 1.

If you choose the door with the car behind it, then switching will make you lose.
If you choose a door with nothing behind it, then switching will make you win.

Though it seems like there is a 50:50 chance of getting a car or nothing, it is actually in your favour to switch:

1. You choose door 1. Either door 2 or 3 is opened. If you switch, you go to door 2/3 and lose.
2. You choose door 2. Door 3 is opened. If you switch, you go to door 1 and win.
3. You choose door 3. Door 2 is opened. If you switch, you go to door 1 and win.

Thus, there is a 2/3 chance of winning if you switch, compared to a 1/3 chance of winning if you stick.

[Mash89]

That was it. What was it called? The Monty Hall puzzle?

[EyeBall]

A real gambler willtell yo the odds are 50/50 no matter what.

[duckmanTTF]

You missed a bit: After you choose a door, the presenter opens another door to deliberately reveal nothing. Now the solution:

For simplicity, let us assume the prize is behind door 1.

If you choose the door with the car behind it, then switching will make you lose.
If you choose a door with nothing behind it, then switching will make you win.

Though it seems like there is a 50:50 chance of getting a car or nothing, it is actually in your favour to switch:

1. You choose door 1. Either door 2 or 3 is opened. If you switch, you go to door 2/3 and lose.
2. You choose door 2. Door 3 is opened. If you switch, you go to door 1 and win.
3. You choose door 3. Door 2 is opened. If you switch, you go to door 1 and win.

Thus, there is a 2/3 chance of winning if you switch, compared to a 1/3 chance of winning if you stick.

Thats pretty much it. As an aside, some people have tried to apply this to deal or no deal - but in this case it doesn't work, when you get to the last 2 boxes (or cases) its 50/50 as to which contains the higher value. The difference is that in that in DoND you choose at random throughout the game. In the monty hall problem the host knows which door to open.