from which step does the error begins,and what is the error

step 1; a=b step 2; a^2=ab (multiplying both sides by a) step 3; a^2-b^2=ab-b^2 (subtracting both sides by b^2) step 4; (a+b)(a-b)=b(a-b) step 5; (a+b)=b (cancelling (a-b) from both sides) step 6; 2b=b (as a=b from step 1) therefore, 2=1 (cancelling b from both the side)

#Math

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

step 5 is wrong because you are cancelling (a-b) from both sides which according to step 1 is equal to zero. Dividing by zero is incorrect.

Surya Prakash - 8 years, 1 month ago

yes the step 5 is incorrect and the procedure includes diving by 0 i.e.(a-b) is absolutely wrong...leading to wrong result...:)

Sayan Chaudhuri - 8 years, 1 month ago

step 5 is wrong due to dividing by 0

bhuvnesh goyal - 8 years, 1 month ago

thnx for answering

Haque Farazul - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$