- #1

- 24

- 0

Let a = b where a, b are any 2 real numbers

a^2 = ab

a^2-b^2=ab-b^2

(a-b)(a+b)=b(a-b)

a+b=b

But a = b therefore b = 2b for any real number b

enjoy! dont curse me for wasting ur time when u find the mistae!

- Thread starter toocool_sashi
- Start date

- #1

- 24

- 0

Let a = b where a, b are any 2 real numbers

a^2 = ab

a^2-b^2=ab-b^2

(a-b)(a+b)=b(a-b)

a+b=b

But a = b therefore b = 2b for any real number b

enjoy! dont curse me for wasting ur time when u find the mistae!

- #2

- 258

- 1

...or a=0...

- #3

- 486

- 0

a+b=b;

a=b;

so b+b=b

so b=a=0?

a=b;

so b+b=b

so b=a=0?

- #4

rachmaninoff

Answer in white:

"(a-b)(a+b)=b(a-b)"

a-b=0, so it is meaningless to divide both sides by (a-b)

"(a-b)(a+b)=b(a-b)"

a-b=0, so it is meaningless to divide both sides by (a-b)

- #5

- 356

- 0

a

Subtract both sides with b

Simplify ==> (a-b)(a+b)=b(a-b)

divide with (a-b) ==> a+b = b

a=b gives ==> 2b = b

And the error is as said very simple.

- #6

- 54

- 0

to get eq. (2) from eq. (1)toocool_sashi said:(1) (a-b)(a+b)=b(a-b)

(2) a+b=b

you have to divide both sides by (a-b)

since a=b

therefore, (a-b)=0...

in addition, you have divided two sides by zero...

a number divided by zero is undefined in math...

got ya!

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Basic Math Problem of the Week 1/02/2017

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