Rameez said:

I think it's Impossible

its possible.but challenging

cute4cute said:

eveything is possible now

then prove this question cute....

The point which I highlighted in red is where you would have to divide each side of
(a+b)(a-b)=b(a-b)
by (a-b) to get
(a+b)=b

But you will notice that if the first part of the proof (a=b) is true, then (a-b) would equal 0.
And, as we know, we can't divide by 0, so from that point on this "proof" that 1=2 is relegated to the non-valid, but interesting, heap of other things which people try to use to prove things which are also false.

Ayazgaad said:

a = b
multiply both sides by a
a^2 = a*b
subtract b^2 from both sides
a^2-b^2 = a*b-b^2
apply the distributive law to both sides
(a+b)(a-b) = b(a-b)
divide both sides by (a-b)
(a+b) = b
substitute all a's for b's (remember, if a = b you can do this)/
a+a = a
regroup the two a's in the left side, and rename it 2a
2a = a
divide both sides by a
2 = 1

The point which I highlighted in red is where you would have to divide each side of
(a+b)(a-b)=b(a-b)
by (a-b) to get
(a+b)=b

But you will notice that if the first part of the proof (a=b) is true, then (a-b) would equal 0.
And, as we know, we can't divide by 0, so from that point on this "proof" that 1=2 is relegated to the non-valid, but interesting, heap of other things which people try to use to prove things which are also false.

wah kya bat ha...

Ayazgaad said:

a = b
multiply both sides by a
a^2 = a*b
subtract b^2 from both sides
a^2-b^2 = a*b-b^2
apply the distributive law to both sides
(a+b)(a-b) = b(a-b)
divide both sides by (a-b)
(a+b) = b
substitute all a's for b's (remember, if a = b you can do this)/
a+a = a
regroup the two a's in the left side, and rename it 2a
2a = a
divide both sides by a
2 = 1

The point which I highlighted in red is where you would have to divide each side of
(a+b)(a-b)=b(a-b)
by (a-b) to get
(a+b)=b

But you will notice that if the first part of the proof (a=b) is true, then (a-b) would equal 0.
And, as we know, we can't divide by 0, so from that point on this "proof" that 1=2 is relegated to the non-valid, but interesting, heap of other things which people try to use to prove things which are also false.

Baijan constant cannot be proof by variables, and you want to proof 2=1, You must firstly take one side either L.H.S or R.H.S
think its Maths not joke

Rameez said:

Ayazgaad said:

a = b
multiply both sides by a
a^2 = a*b
subtract b^2 from both sides
a^2-b^2 = a*b-b^2
apply the distributive law to both sides
(a+b)(a-b) = b(a-b)
divide both sides by (a-b)
(a+b) = b
substitute all a's for b's (remember, if a = b you can do this)/
a+a = a
regroup the two a's in the left side, and rename it 2a
2a = a
divide both sides by a
2 = 1

The point which I highlighted in red is where you would have to divide each side of
(a+b)(a-b)=b(a-b)
by (a-b) to get
(a+b)=b

But you will notice that if the first part of the proof (a=b) is true, then (a-b) would equal 0.
And, as we know, we can't divide by 0, so from that point on this "proof" that 1=2 is relegated to the non-valid, but interesting, heap of other things which people try to use to prove things which are also false.

Baijan constant cannot be proof by variables, and you want to proof 2=1, You must firstly take one side either L.H.S or R.H.S
think its Maths not joke

acha rameez muja tu pata he nahi ta ka ye proof nahi hu sakta.......

Assumptions

i = Sqrt (-1)
Sqrt (a.b) = Sqrt a x Sqrt b

Now...
1+1 = 1 + Sqrt 1
1 + Sqrt 1 = 1 + Sqrt [(-1)(-1)]
1 + Sqrt [(-1)(-1)] = 1 + Sqrt (-1) x Sqrt (-1)
1 + Sqrt (-1) x Sqrt (-1) = 1 + i x i
1+ i^2 = 1 + (-1)
1 + (-1) = 1-1
So..
1 - 1 = 0

ShahidRhm said:

1+1 = 0

Assumptions

i = Sqrt (-1)
Sqrt (a.b) = Sqrt a x Sqrt b

Now...
1+1 = 1 + Sqrt 1
1 + Sqrt 1 = 1 + Sqrt [(-1)(-1)]
1 + Sqrt [(-1)(-1)] = 1 + Sqrt (-1) x Sqrt (-1)
1 + Sqrt (-1) x Sqrt (-1) = 1 + i x i
1+ i^2 = 1 + (-1)
1 + (-1) = 1-1
So..
1 - 1 = 0

wah engineer bhai

agha11 said:

no not possible

kyu its possible