the Wookie

10-16-2003, 03:52 PM

a=b

a^2 = ab

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

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

(a+b)=b

(a+a)=a

2a = a

2 = 1

valid?

a^2 = ab

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

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

(a+b)=b

(a+a)=a

2a = a

2 = 1

valid?

View Full Version : 2=1, is this valid?

the Wookie

10-16-2003, 03:52 PM

a=b

a^2 = ab

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

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

(a+b)=b

(a+a)=a

2a = a

2 = 1

valid?

a^2 = ab

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

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

(a+b)=b

(a+a)=a

2a = a

2 = 1

valid?

PJYelton

10-16-2003, 03:56 PM

No, because a-b equals zero, therefore you are dividing by zero which is a no-no. Its like saying zero times three equals zero times four, cancel the zeroes and you get three equals four.

JaWiB

10-16-2003, 03:57 PM

Where did you get 2=1? As far as I can tell, a and b both equal 0

the Wookie

10-16-2003, 03:58 PM

well if you said a=b=2

PJ: i see it now. cant divide by zero

PJ: i see it now. cant divide by zero

joshdick

10-16-2003, 05:51 PM

This thread gets two thumbs down.

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