How can we prove that 1=2 when x=1 and y=1?

Question

please help me to solve my homework.

Answer 1

This is a (false) proof that 1 = 2:

Let x = 1 and y = 1. Then,

x = y

If we square both sides,

x^2 = y^2

Therefore

x^2 - y^2 = 0

Back to x = y, if we multiply both sides of this equation by y, we obtain

xy = y^2

Therefore

xy - y^2 = 0

We have two equations that are equal to 0 now, so we can equate them to each other.

x^2 - y^2 = xy - y^2

Let's factor both sides.

(x - y)(x + y) = y(x - y)

Now, let's divide both sides of the equation by (x - y).

x + y = y

Plugging in the values x = 1 and y = 1 back in the equation

1 + 1 = 1, OR
2 = 1

*****
This is, of course, a false proof. The invalid step was when we divide both sides by (x - y), because (x - y) is equal to 0 (since
x = 1 and y = 1). All it takes is an invalid step in order to invalidate a claim.

Many people don't recognize this idea of an "invalid step" when presented with the proof that 0.9(repeating infinitely) = 1.

Answer 2

I think you can only "prove" that the assumption 1=2 is wrong by using a counterexample like an answer above mine.

Answer 3

I believe mmcall made an error in factoring the eqation.

a2 - b2 = ab-b2
(a+b)(a-b)=b(a-2)
assuming a=b ...
2a=0
a=0 b=0
...doesn't work

Answer 4

1=1
x^2=y^1
when the base no.s are we can compare the powers so,
2=1
proved.

Answer 5

Assume a = b then:
a2 = ab
a2 - b2 = ab-b2
(a-b)(a+b) = b(a-b)
a+b = b
b+b = b
2b = b
2 = 1

Answer 6

x=y=1

x=y
xx=yx or xsquare=x product y
subtract yy from both sides we get xx-yy=xy-yy
factorise and we get y(x-x)=(x+y)(x-y)
divide by(x-y){actyually, it means divide by zero} we get y=x+y
hence, 1=2
this is because you are actually dividing by 0 which gives this problem

Answer 7

Question was not clear

Answer 8

are you completely insane????? that is a false equation!!!!!!