${y \in \mathbb{N}}$ but, ${y}$ comes out to be zero, Why?

Question

${x \in \mathbb{N}, y \in \mathbb{N}, \mathbb{K} \gt 0, \mathbb{A} \gt 0}$, Where, ${\mathbb{N}}$ represents 'Natural Numbers', ${\mathbb{A}}$ and ${\mathbb{K}} \in {\mathbb{R}^{+}}$,
Two Equations are given, $${Equation 1 : \mathbb{K}(x+y) = xy}$$ $${Equation 2 : \mathbb{A}(x-y) = xy}$$ By Using Equation (1) and Equation (2), $${Equation 3 : \mathbb{K} = \frac{xy}{(x+y)}}$$ $${Equation 4 : \mathbb{A} = \frac{xy}{(x-y)}}$$ Dividing Equation (3) by Equation (4), $${Equation 5 : \frac{\mathbb{K}}{\mathbb{A}} = \frac{(x-y)}{(x+y)}}$$ Multiplying RHS with (-1) and dividing the same by (-1) must not effect it, $${\therefore \frac{\mathbb{K}}{\mathbb{A}} = \frac{(x-y)\times(-1)}{(x+y)\times(-1)}}$$ $${\therefore \frac{\mathbb{K}}{\mathbb{A}} = \frac{(x+y)}{(x-y)}}$$ But, by Equation (5), $${\frac{(x+y)}{(x-y)} = \frac{\mathbb{A}}{\mathbb{K}}}$$ $${\therefore \frac{\mathbb{K}}{\mathbb{A}} = \frac{\mathbb{A}}{\mathbb{K}}}$$ $${Equation 6 : \mathbb{K}^{2} = \mathbb{A}^{2}}$$ Now, By Squaring Equation (1), $${\mathbb{K}^{2}(x+y)^{2} = x^{2}y^{2}}$$ By Substituting value of ${\mathbb{K}}$ from Equation (6), $${\therefore \mathbb{A}^{2}(x+y)^{2} = x^{2}y^{2}}$$ As all the variables represent positive numbers, the square root will remain positive, $${\therefore \mathbb{A}(x+y) = xy}$$ By using Equation (2), $${\mathbb{A}(x+y) = \mathbb{A}(x-y)}$$ $${\therefore x+y = x-y}$$ $${\therefore y = -y}$$ $${\therefore 2y = 0}$$ $${\therefore y = 0}$$ But, according to the statement, ${y \in \mathbb{N}}$ which is contradictory to the above solution. Why ?

Answer 1

I don't really understand what $\mathbb{N}, \mathbb{A}, \mathbb{K}$ are since usually they denote sets rather than values.

Also, at the step when you are multiplying by $(-1)$ you have a mistake - it should be $\frac{y-x}{-x-y}$ rather than $\frac{x+y}{x-y}$