Logic, writing proof

Question

i)Suppose that $x$ and $y$ are real numbers. Prove that if $x\neq 0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$

ii)Suppose that $x$ and $y$ are real numbers. Prove that if $x^2y=2x+y$, then if $y\neq 0$ then $x\neq 0$

What I did is

i)Suppose that $y\neq3\Rightarrow \frac{3x^2+2y}{x^2+2}\neq 3\Leftrightarrow 3x^2+2y\neq 3x^2+6\Rightarrow y=3$ which is a contradition then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$

Is this right? I need to make some consideration about $x$?

ii)Suppose that $y=0$ and $x=0$ thus $x^2+y\neq2x+y\Leftrightarrow 0\neq0$, which is a contradition then $x^2y=2x+y$, then if $y\neq 0$ then $x\neq 0$

Answer 1

i) Given $$y=\frac{3x^2+2y}{x^2+2}$$ cross multiplying we have, $$y(x^2+2)=3x^2+2y$$ $$\implies yx^2+2y=3x^2+2y$$ canceling $2y$ in both side we have, $$yx^2=3x^2$$ Since $x\neq 0 \implies x^2\neq0$ So dividing both side by x^2, we have $$y=3$$

ii) On contrary suppose that $x=0$ then we have $x^2=0$ so $$0\cdot y=2\cdot 0 +y$$ $$\implies y=0$$ Which is a contradiction since given that $y\neq 0$

Answer 2

Sketch for $(i)$: Suppose $x\not=0$ and $y=\frac{3x^2+2y}{x^2+2}$.

1) Observe $x^2+2$ is never zero (must be proven).

2) Multiply both sides by $x^2+2$ to get $yx^2+2y=3x^2+2y$.

3) Subtract $2y$ from both sides to get $yx^2=3x^2$.

4) Divide both sides by $x^2$ (which you can do because $x\not=0$, here is where we use that assumption - you might need to prove that $x^2\not=0$ if you want to be pedantic).

5) You are left with $y=3$.

Note: You could, alternately at step $4$, subtract $3x^2$ from both sides and factor to get $x^2(y-3)=0$. Since a product equals zero, one of the factors is zero, but since $x\not=0$, $y-3=0$ or $y=3$.

Answer 3

For ii), a better starting approach is to assume that $y=0$ and $x \neq 0$. You are trying to find a contradiction.

Then for the left hand side you would have:

$$ x^2 y = x^2 (0) = 0$$

On the right hand side:

$$ 2x + y = 2x + 0 = 2x \neq 0 \text{$\qquad$ by assumption that $x \neq 0$}$$

Answer 4

Since the OP has stated that contradiction is needed, here's another answer.

Consider the statement: If $x\not=0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$.

The hypothesis is $x\not=0$ and the conclusion is if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$

To prove this by contradiction, you need to assume the hypothesis and the negation of the conclusion. The negation of the conclusion is the negation of an if-then statement. Since $A\rightarrow B\equiv \neg A\vee B$, its negation is $A\wedge \neg B$.

Therefore, you should assume $x\not=0$, $y=\frac{3x^2+2y}{x^2+2}$, and $y\not=3$ and try to find a contradiction.