What is the mistake in this
let a and b be any two numbers and the mean of them is c then,
a+b/2=c
a+b=2c
multiply both sides with a-b
a^2-b^2=2ac- 2bc
adding b^2-2ac+c^2 on both sides
a^2-b^2+b^2-2ac+c^2=2ac- 2bc+b^2-2ac+c^2
a^2-2ac+c^2=b^2-2bc+c^2
(a-c)^2=(b-c)^2
(a-c)=(b-c)
a=b
ie;all numbers are equal
Anonymous:
(a-c) is not equal to (b-c)
Answers
Answered by
1
according to your proof,
i substitute the value of a as 6,b as 4 and c as 5
(6+4)/2=5
6+4=2*5
6^2-4^2=10(6-4)=60-40
6^2-4^2+4^2-60+5^2=60-40+4^2-60+5^2
hence, (6-5)^2=(4-5)^2
i.e,1^2=(-1)^2
but,if we take 6-5=4-5,itgoes wrong eventhough their squares are same
because,in this case,(a-c)=(b-c) applies=> 1= -1,which is wrong
hence, (a-c)≠(b-c)
i substitute the value of a as 6,b as 4 and c as 5
(6+4)/2=5
6+4=2*5
6^2-4^2=10(6-4)=60-40
6^2-4^2+4^2-60+5^2=60-40+4^2-60+5^2
hence, (6-5)^2=(4-5)^2
i.e,1^2=(-1)^2
but,if we take 6-5=4-5,itgoes wrong eventhough their squares are same
because,in this case,(a-c)=(b-c) applies=> 1= -1,which is wrong
hence, (a-c)≠(b-c)
Answered by
0
if you multiply a-b on both sides the theleft hand side will be equal to a square - b square
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