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Alex Shepherd
20-11-09, 21:57
I was trying to post something hard question for the math thead, and now I found this.

x=0
(find x)

x=0
--> x (x-1) = 0 (Because when you multiply, you will get the same result is equal to zero, as for x multiply by x and then by -1 will get x-x which is equal to zero)
--> x-1 = 0 (Is when you define the two x with two different numbers)
--> x = 1 (When it comes to the different side, it sign will change)
So as a result

x=0
x(x-1) = 0
x-1 = 0
x= 1
1 = 0 (Is that possible?)


Another example

http://upload.wikimedia.org/math/b/e/c/bece7a4ff69a1ca64899e8693b697dc3.png

1. Let a and b be equal non-zero quantities

http://upload.wikimedia.org/math/9/4/7/947fc392e1843e655a04e4ccedbf8146.png

2. Multiply through by a

http://upload.wikimedia.org/math/0/4/a/04ae6e1f708633c55b0f2910229c8218.png

3. Subtract

http://upload.wikimedia.org/math/0/d/a/0da776c891aae2ddcea10de4168e80d7.png

4. Factor both sides

http://upload.wikimedia.org/math/f/3/a/f3ab9270d9c7c080e69391947e4d9f0a.png

5. Divide out

http://upload.wikimedia.org/math/8/e/6/8e629bc66a8fd064dadff7e31fb9ab0d.png

6. Observing that

http://upload.wikimedia.org/math/d/a/c/dac22e850802fe51e4e3c55b41a1cbc8.png

7. Combine like terms on the left

http://upload.wikimedia.org/math/3/7/d/37d0811a86406b970245d48f50015ec0.png

8. Divide by the non-zero b

http://upload.wikimedia.org/math/8/f/7/8f747d1e8bf6521547fe9fe7f8a5d45a.png

Hows that possible?

Raider Man

toxicraider
20-11-09, 22:08
I was trying to post something hard question for the math thead, and now I found this.

x=0
(find x)

x=0
--> x (x-1) = 0 (Because when you multiply, you will get the same result is equal to zero, as for x multiply by x and then by -1 will get x-x which is equal to zero)
--> x-1 = 0 (Is when you define the two x with two different numbers)
--> x = 1 (When it comes to the different side, it sign will change)
So as a result

x=0
x(x-1) = 0
x-1 = 0
x= 1
1 = 0 (Is that possible?)
This is just quadratics;
Try drawing a graph
y=x(x-1)

when y = 0
x= 0 or 1
This is shown as the graph will cross the x axis (y=0) at two points (0,0) and (1,0)

Saying 0=1 is wrong, because x is not a constant value; it can be any number, which coincides with the y value (in this case is 0)

the correct way to say it would be

f(0) = f(1)
f(x) is the 'function' of x, i.e, what you have to do to x, to get y. f(x) = x(x-1) = y
when x = 0, you write f(0), and etc.
so f(0) = 0(0-1) = 0
and f(1) = 1(1-1) = 0

0=0 so f(0) = f(1)


Another example

http://upload.wikimedia.org/math/b/e/c/bece7a4ff69a1ca64899e8693b697dc3.png

You cannot square root negative numbers, so 1 and -1 are not equal.


1. Let a and b be equal non-zero quantities

http://upload.wikimedia.org/math/9/4/7/947fc392e1843e655a04e4ccedbf8146.png

2. Multiply through by a

http://upload.wikimedia.org/math/0/4/a/04ae6e1f708633c55b0f2910229c8218.png

3. Subtract

http://upload.wikimedia.org/math/0/d/a/0da776c891aae2ddcea10de4168e80d7.png

4. Factor both sides

http://upload.wikimedia.org/math/f/3/a/f3ab9270d9c7c080e69391947e4d9f0a.png

5. Divide out

http://upload.wikimedia.org/math/8/e/6/8e629bc66a8fd064dadff7e31fb9ab0d.png

6. Observing that

http://upload.wikimedia.org/math/d/a/c/dac22e850802fe51e4e3c55b41a1cbc8.png

7. Combine like terms on the left

http://upload.wikimedia.org/math/3/7/d/37d0811a86406b970245d48f50015ec0.png

8. Divide by the non-zero b

http://upload.wikimedia.org/math/8/f/7/8f747d1e8bf6521547fe9fe7f8a5d45a.png

Hows that possible?

Raider Man

if a=b
then (a-b) = 0
so you divided by 0 in step 5, (= infinity) which caused it to go wrong.

Once you start playing with infinity, by dividing by zero, then all numbers are irrelevant.

These aren't really illusions, the second 2 are simply breaking mathmatical rules :p.

Lizard of Oz
20-11-09, 22:14
This is just quadratics;
Try drawing a graph
y=x(x-1)

when y = 0
x= 0 or 1
This is shown as the graph will cross the x axis (y=0) at two points (0,0) and (1,0)
the 1=0 is simply bad notation
It would actually be
f(1) = f(0)
Which takes the equation into account.



You cannot square root negative numbers, so 1 and -1 are not equal.


if a=b
then (a-b) = 0
so you divided by 0 in step 5, (= infinity) which caused it to go wrong.

Once you start playing with infinity, by dividing by zero, then all numbers are irrelevant.

:D
lol i dont get it but ok

Lost_Soul
20-11-09, 22:16
toxicraider...wow...just wow. I didn't even understand the questions. Then again I did pretty much fail math at school.

Your_Envy*
20-11-09, 22:25
Well, I am always surprised but anyway.. :)


1 111 111 1 111 111 = 1 234 567 654 32

When I first saw this, I was like WUT? :eek: :vlol:

Cochrane
20-11-09, 22:28
I was trying to post something hard question for the math thead, and now I found this.

x=0
(find x)

x=0
--> x (x-1) = 0 (Because when you multiply, you will get the same result is equal to zero, as for x multiply by x and then by -1 will get x-x which is equal to zero)

At this point, you divide by zero (as x is zero), so all that follows is wrong.
--> x-1 = 0 (Is when you define the two x with two different numbers)
--> x = 1 (When it comes to the different side, it sign will change)
So as a result

x=0
x(x-1) = 0
x-1 = 0
x= 1
1 = 0 (Is that possible?)


Another example

http://upload.wikimedia.org/math/b/e/c/bece7a4ff69a1ca64899e8693b697dc3.png
Not possible, as you are trying to take the square root of -1, which is not defined. You can do that only in imaginary numbers, where i is defined as sqrt(-1), which means i*i = -1. But even then, saying sqrt((-1)*(-1)) = sqrt(-1)*sqrt(-1) is just wrong.

1. Let a and b be equal non-zero quantities

http://upload.wikimedia.org/math/9/4/7/947fc392e1843e655a04e4ccedbf8146.png

2. Multiply through by a

http://upload.wikimedia.org/math/0/4/a/04ae6e1f708633c55b0f2910229c8218.png

3. Subtract

http://upload.wikimedia.org/math/0/d/a/0da776c891aae2ddcea10de4168e80d7.png

4. Factor both sides

http://upload.wikimedia.org/math/f/3/a/f3ab9270d9c7c080e69391947e4d9f0a.png

5. Divide out

http://upload.wikimedia.org/math/8/e/6/8e629bc66a8fd064dadff7e31fb9ab0d.png

6. Observing that

http://upload.wikimedia.org/math/d/a/c/dac22e850802fe51e4e3c55b41a1cbc8.png

7. Combine like terms on the left

http://upload.wikimedia.org/math/3/7/d/37d0811a86406b970245d48f50015ec0.png

8. Divide by the non-zero b

http://upload.wikimedia.org/math/8/f/7/8f747d1e8bf6521547fe9fe7f8a5d45a.png

Hows that possible?

Raider Man

toxicraider already pointed out the division by zero.

Alex Shepherd
20-11-09, 23:02
This is just quadratics;
Try drawing a graph
y=x(x-1)

when y = 0
x= 0 or 1
This is shown as the graph will cross the x axis (y=0) at two points (0,0) and (1,0)

Saying 0=1 is wrong, because x is not a constant value; it can be any number, which coincides with the y value (in this case is 0)

the correct way to say it would be

f(0) = f(1)
f(x) is the 'function' of x, i.e, what you have to do to x, to get y. f(x) = x(x-1) = y
when x = 0, you write f(0), and etc.
so f(0) = 0(0-1) = 0
and f(1) = 1(1-1) = 0

0=0 so f(0) = f(1)


You cannot square root negative numbers, so 1 and -1 are not equal.


if a=b
then (a-b) = 0
so you divided by 0 in step 5, (= infinity) which caused it to go wrong.

Once you start playing with infinity, by dividing by zero, then all numbers are irrelevant.

These aren't really illusions, the second 2 are simply breaking mathmatical rules :p.

:vlol: Discovery hero :vlol:
I didn't break the mathmatical rules, just half of it :whi:
But good one :vlol:

Raider Man