second try 2=3.41

saundo · 2009-10-21 00:46:50

i have proven not that 1=2 but that 2=3.41 i am only 16 but im pretty sure there are no faults in this. what i did and expalnations on how are written below
x=2 y=2
x=y
x^2=y+x
√(x^2 )=√(y)+(x)
x=1.41+x
2=1.41+2
2=3.14
line1 x=2 y=2
line 2 therefore x=y
line 3 add x to both sides( since x=2 and 2+2 Is the same as 2^2 it become x^2)
line 4 square root both sides of the equation
line 5 we are left with x=1.41+x because √2 is 1.41 to 2 decimal places
line 6 substitute the pro-numeral x with the its value (2) leaves us with 2=1.41+2
line 7 the answer is 2=3.41
find any faults now?

bobbym · 2009-10-21 00:54:36

Hi saundo;

line 4 to 5 is incorrect. When you take the square root of both sides in line 4 you should get:

√(x^2 )=√(y+x)

saundo · 2009-10-21 01:02:45

bobbym wrote:

Hi saundo;
line 4 to 5 is incorrect. When you take the square root of both sides in line 4 you should get:
√(x^2 )=√(y+x)

that is why they were seperate √(y)+(x)

bobbym · 2009-10-21 01:08:43

Hi saundo;

But you cannot separate them. The rules of algebra have to be adhered to.

saundo · 2009-10-21 01:13:51

bobbym wrote:

Hi saundo;
But you cannot separate them. The rules of algebra have to be adhered to.

im pretty sure there is another rule that what you do too 1 side you do to all parts of the other which means that √x^2=√y+√x thus proving my first equation 2=2.82 correct

Last edited by saundo (2009-10-21 01:15:28)

bobbym · 2009-10-21 01:21:37

Hi saundo;

Sorry, I wished that were true. Would make a lot of my work more meaningful. Unfortunately it is not, this is correct:

√(x^2 )=√(y+x)

this is not:

√x^2=√y+√x

Last edited by bobbym (2009-10-21 01:21:58)

saundo · 2009-10-21 01:25:55

bobbym wrote:

Hi saundo;
Sorry, I wished that were true. Would make a lot of my work more meaningful. Unfortunately it is not, this is correct:
√(x^2 )=√(y+x)
this is not:
√x^2=√y+√x

ah i didnt put those brakets in but try these
x=2 y=2
x=y
x^2=y+x
√x^2 =√y+x
x=1.41+1.41
x=2.82
2=2.82

or

x=2 y=2
x=y
x^2=y+x
√x^2 =√y+x
x=1.41+1.41
x=2.82
1.41=2.82

bobbym · 2009-10-21 01:30:26

Hi saundo;

√x^2 =√y+x this is also incorrect. What you do to one side you must do to the other side of an equation.

√(x^2) =√(y+x) is correct. Surely you know that 2 does equal 2.82. If I gave you 2 dollars would you give me 2.82 back? The fact that you are getting this contradiction proves that √x^2 =√y+x is incorrect.

Last edited by bobbym (2009-10-21 01:31:24)

saundo · 2009-10-21 01:37:54

hi bobbym
It's ok i understand now but im happy i got as as close as i did since i havent finished high school yet
but thank you for pointing out the flaw now that i know it i can see if my teacher believes it and if my teacher can point out the flaw.:P
thanks anyway bobbym
p.s if you gave me two dollars i would prove in a practical way that 2 = 0

Last edited by saundo (2009-10-21 01:38:55)

bobbym · 2009-10-21 01:41:42

Hi saundo;

Yes, I know how to prove that 2 = 0 in that case. Glad to help!

Math Is Fun Forum

#1 2009-10-21 00:46:50

second try 2=3.41

#2 2009-10-21 00:54:36

Re: second try 2=3.41

#3 2009-10-21 01:02:45

Re: second try 2=3.41

#4 2009-10-21 01:08:43

Re: second try 2=3.41

#5 2009-10-21 01:13:51

Re: second try 2=3.41

#6 2009-10-21 01:21:37

Re: second try 2=3.41

#7 2009-10-21 01:25:55

Re: second try 2=3.41

#8 2009-10-21 01:30:26

Re: second try 2=3.41

#9 2009-10-21 01:37:54

Re: second try 2=3.41

#10 2009-10-21 01:41:42

Re: second try 2=3.41

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