Can you spot the flaw (maths trick)
#1
Scooby Regular
Thread Starter
Join Date: Apr 2002
Location: North Ayrshire (sometimes)
Posts: 512
Likes: 0
Received 0 Likes
on
0 Posts
x - y = 0
2x - 2y = 0
x - y = 2x - 2y
1(x - y) = 2(x - y)
1 = 2
if x minus y equals zero
that means 2 times x minus 2 times y also equals zero right?
so that would mean x minus y equals 2 times x minus 2 times y
if you cross divide both sides of the equation by x minus y
your left with 1 equals two.
worked out the flaw yet.
both x and y are the number 0. Any number multiplied by zero will always equal zero.
2x - 2y = 0
x - y = 2x - 2y
1(x - y) = 2(x - y)
1 = 2
if x minus y equals zero
that means 2 times x minus 2 times y also equals zero right?
so that would mean x minus y equals 2 times x minus 2 times y
if you cross divide both sides of the equation by x minus y
your left with 1 equals two.
worked out the flaw yet.
both x and y are the number 0. Any number multiplied by zero will always equal zero.
#7
Scooby Regular
Join Date: Apr 2001
Location: in a town with bad roads
Posts: 514
Likes: 0
Received 0 Likes
on
0 Posts
That's wrong of course
x - y = 0
2x - 2y = 0
x - y = 2x - 2y
1(x - y) = 2(x - y)
1 = 2
if
x - y = 0 / +y >
x = y
then of course is
2x - 2y =0 / +2y >
2x = 2y / :2 >
x = y
but now
x - y = 2x - 2y / -x; +2y >
x - x -y + 2y = 2x - x - 2y + 2y
y = x
here comes something undefined:
1(x - y) = 2(x - y) which is 1 * 0 = 2 * 0 which is undefined IIRC
so the next step of
1 = 2
is not correct
shid
Math day today?
Ralf
x - y = 0
2x - 2y = 0
x - y = 2x - 2y
1(x - y) = 2(x - y)
1 = 2
if
x - y = 0 / +y >
x = y
then of course is
2x - 2y =0 / +2y >
2x = 2y / :2 >
x = y
but now
x - y = 2x - 2y / -x; +2y >
x - x -y + 2y = 2x - x - 2y + 2y
y = x
here comes something undefined:
1(x - y) = 2(x - y) which is 1 * 0 = 2 * 0 which is undefined IIRC
so the next step of
1 = 2
is not correct
shid
Math day today?
Ralf
Trending Topics
#9
Scooby Regular
Join Date: Feb 2001
Location: Hemel Hempstead
Posts: 7,953
Likes: 0
Received 0 Likes
on
0 Posts
no, it works like this, tis quite straight forward actualy
x - y = 2x - 2y / -x; +2y >
x - x -y + 2y = 2x - x - 2y + 2y
y + x - 2x = x
then if
x - y = 0
x - 2y = 0
x - y = 2x - 2y
1(x - y) = 2(x - y)
then surley
2x - 2y =0 / +2y >
2x = 2y / +2 >
x + 2y - x= y
so you get
x - y = 2x - 2y / -x; +2y
x - y = 0 / +y
x - y = 2x - 2y
x - x -y + 2y = 2x - x - 2y + 2y
y = x
Simple when you really think about it.
P
x - y = 2x - 2y / -x; +2y >
x - x -y + 2y = 2x - x - 2y + 2y
y + x - 2x = x
then if
x - y = 0
x - 2y = 0
x - y = 2x - 2y
1(x - y) = 2(x - y)
then surley
2x - 2y =0 / +2y >
2x = 2y / +2 >
x + 2y - x= y
so you get
x - y = 2x - 2y / -x; +2y
x - y = 0 / +y
x - y = 2x - 2y
x - x -y + 2y = 2x - x - 2y + 2y
y = x
Simple when you really think about it.
P
#10
Scooby Regular
Join Date: Nov 2001
Location: Leeds - It was 562.4bhp@28psi on Optimax, How much closer to 600 with race fuel and a bigger turbo?
Posts: 15,239
Likes: 0
Received 1 Like
on
1 Post
you'd expect paul to be good at maths (has to count a lot of cash when buying and selling his cars.)
David
David
#12
1*0 and 2*0 are not undefined, they're 0
The problem arises that you divide both sides by (x-y) which is zero. It's the dividing by zero that's undefined.
Of course if you got to the point where you had 2(x-y)=(x-y) you should pretty rapidly conclude that x=y anyway
The problem arises that you divide both sides by (x-y) which is zero. It's the dividing by zero that's undefined.
Of course if you got to the point where you had 2(x-y)=(x-y) you should pretty rapidly conclude that x=y anyway
Thread
Thread Starter
Forum
Replies
Last Post
ATWRX
Full Cars Breaking For Spares
88
01 February 2016 07:28 PM