Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (goformit100):

Prove 2 = 3 . ( using algebraic equations )

Parth (parthkohli):

There are a lot 2 * 0 = 3 * 0 0 = 0 Therefore 2 = 3

Parth (parthkohli):

But none of these are true.

Parth (parthkohli):

The number '0' shouldn't really be used in isolation.

Parth (parthkohli):

Dividing 0 ===> Undefined Multiplying 0 ===> Gives us a = b where a \(\ne\) b Subtracting and adding 0 ====> no need to subtract or add zero as that serves nothing

OpenStudy (goformit100):

Anyone any other method please

Parth (parthkohli):

We can prove 2 \(\ne\) 3 btw

OpenStudy (experimentx):

2^0 = 3^0

Parth (parthkohli):

^ lol another example of not using 0 in isolation

Parth (parthkohli):

TWO = 3 3 letters = 3 One of the joke methods

OpenStudy (unklerhaukus):

2 = a number 3= a number 2=3

Parth (parthkohli):

a + a = a a + a + a = a + a 3a = 2a 3 = 2

OpenStudy (goformit100):

is there any method of solving it by this method |dw:1341240307957:dw|

Parth (parthkohli):

There are a lot of the joke proofs related to assumption, but they all involve operations of the number 0.

OpenStudy (goformit100):

????

Parth (parthkohli):

http://www.astahost.com/info/tiiiss-ramanujams-proof-flaws.html This is a proof from a very famous Indian Mathematician that has flaws.

Parth (parthkohli):

The flaw here is that when a^2 = b^2, then a is not necessarily = b

Parth (parthkohli):

Why so? Because we have negative square roots :D Take -8 and 8 as a and b. \((8)^2 = 8^2\) But \(-8 \ne 8\)

OpenStudy (anonymous):

If you're looking through false algrebraeic proofs, they almost all involve one of the two mistakes: Division by 0 \[\sqrt{1}=1\]

OpenStudy (zarkon):

\[i=\sqrt{-1}=\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{i}=-i\] so \[i=-i\Rightarrow 1=-1\] so \[2=0\] divide by 2 \[1=0\] add 2 to both sides \[3=2\]

Parth (parthkohli):

lol there are a lot of proofs like these. :P

OpenStudy (lgbasallote):

let a = b ab = b^2 ab - a^2 = b^2 - a^2 a(b-a) = (b+a)(b-a) a = b + a a + a = b + a + a a + a = a + a + a 2a = 3a 2 = 3

OpenStudy (anonymous):

Good one, Zarkon =) The error of course being: \[\sqrt{1}=1\]

OpenStudy (lgbasallote):

no "good one, lgba =)"? :(

OpenStudy (goformit100):

Thanks Os users.

OpenStudy (goformit100):

thank you

OpenStudy (anonymous):

Good one, lgba. The mistake of course being at this step a(b-a) = (b+a)(b-a) a = b + a You divided by (b-a), but since a=b, b-a =0 Division by 0 makes me a sad panda.

OpenStudy (goformit100):

thanks every one & @lgbasallote

OpenStudy (lgbasallote):

<tips hat>

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!