Given : a=b
then
ab=b^2
ab-a^2=b^2-a^2
a(b-a)=(b-a)(b+a)
a=b+a
a=a+a
a=2a
1=2
prove it wrong and best answer gets 10 points
2006-10-28 12:27:33 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Good job all of u
u caught the mistake
2006-10-28 12:38:56 · update #1
When you went from a(b-a) = (b-a)(b+a) to
a = b+a
You divided both sides by (b-a). The problem is if a = b then b-a = 0, so you really divided by 0. As you know division by zero is not a legal math operation.
2006-10-28 12:32:02 · answer #1 · answered by teacher2006 3 · 2⤊ 0⤋
in the third line you say 0=0 , stop there, dividing both sides by 0 makes the whole thing indeterminate and the next line effectively says infinity is infinity.
Best of Luck - Mike
2006-10-28 12:33:20 · answer #2 · answered by Anonymous · 1⤊ 0⤋
You divided by zero (b-a = 0 because a=b). That's a no-no.
2006-10-28 12:31:08 · answer #3 · answered by just♪wondering 7 · 1⤊ 0⤋
a=b
then:
ab = b^2
ab-a^2= b^2 - a^2 = 0
a(b-a)= ab - a^2 = a^2 - a^2 = 0
Therefore:
aâ b+a
2006-10-28 12:45:39 · answer #4 · answered by smarties 6 · 1⤊ 0⤋
That is very typical: it is a problem of indetermination, you cannot divide by zero.
The problem is in step a(b-a)=(b-a)(b+a) --> a=b+a
You divide in both sides by (b-a) and b-a=0.
2006-10-28 12:33:33 · answer #5 · answered by Anonymous · 1⤊ 0⤋
You divided by zero! (b-a)=0, if b=a.
2006-10-28 13:02:24 · answer #6 · answered by smagu 2 · 0⤊ 0⤋
you divided by 0=b-a
2006-10-28 12:39:44 · answer #7 · answered by locuaz 7 · 1⤊ 1⤋
simple algebra a(a-b) does not equal (b-a)(b+a)
2006-10-28 12:33:19 · answer #8 · answered by RichUnclePennybags 4 · 1⤊ 2⤋