+1694 Given x = 1 and y = 1, we have
x = y
Multiplying each side by x
x2 = xy
Subtracting y2 from each side
x2 - y2 = xy - y2
Factoring each side
(x + y)(x - y) = y (x - y)
Dividing out the common term ((x - y) we have
x + y = y
Substituting the given values
1 + 1 = 1
Or
2 = 1
What is wrong with this proof?
+583 after you factor you equation, you got (x+y)(x-y)=y(x-y),then you divided the both sides of the equation by x-y. note that given x=1 and y=1 so x-y=0.
From both sides of the equation by dividing or multiplying the same number (divisor can not be zero), the resulting equation is still valid.$$\Rightarrow$$ if a=b(c not equal to 0),then $${\frac{{\mathtt{a}}}{{\mathtt{c}}}} = {\frac{{\mathtt{b}}}{{\mathtt{c}}}}$$
So it not vaild to divide 0 from both sides of an equation.
addition explantion from wiki -In ordinary arithmetic, the expression has no meaning, as there is no number which, multiplied by 0, gives a (assuming a≠0), and so division by zero is undefined.
+1694 @fiora:/ I didn't even touch it; I took this one off some website, and just pasted it here.
+583 after you factor you equation, you got (x+y)(x-y)=y(x-y),then you divided the both sides of the equation by x-y. note that given x=1 and y=1 so x-y=0.
From both sides of the equation by dividing or multiplying the same number (divisor can not be zero), the resulting equation is still valid.$$\Rightarrow$$ if a=b(c not equal to 0),then $${\frac{{\mathtt{a}}}{{\mathtt{c}}}} = {\frac{{\mathtt{b}}}{{\mathtt{c}}}}$$
So it not vaild to divide 0 from both sides of an equation.
addition explantion from wiki -In ordinary arithmetic, the expression has no meaning, as there is no number which, multiplied by 0, gives a (assuming a≠0), and so division by zero is undefined.
