*(a-b)=b(a-b)
step 5; (a+b)=b (cancelling (a-b) from both sides)
step 6; 2*b=b (as a=b from step 1)
therefore, 2=1 (cancelling b from both the side)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestthnx for answering

Log in to reply

step 5 is wrong due to dividing by 0

Log in to reply

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.

Log in to reply

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

Log in to reply