Rational and Irrational Numbers

Let $x$ be rational and $y$ irrational.

Suppose $x+y$ were rational.

Then

would be the difference of two rational numbers and therefore rational, a contradiction.

Hence

Similarly, if $x-y$ were rational, then

would be rational, again a contradiction.

Thus

Let $x$ be rational and $y$ irrational.

If

then

which is rational.

But if $x\ne0$ and $xy$ were rational, then

would be rational, a contradiction.

Thus

Therefore, the table correctly says

Let $x$ be rational and $y$ irrational. Since an irrational number cannot be $0$,

If $x=0$,

which is rational.

If $x\ne0$ and $\frac xy$ were rational, say

then $r\ne0$ and

which would be rational, a contradiction.

Therefore,

So again,

For the reverse quotient, if $x\ne0$, then $\frac yx$ must be irrational. Otherwise, multiplying a rational value of $\frac yx$ by the nonzero rational number $x$ would make $y$ rational, a contradiction.

Rewrite:

Since $x$ is irrational and nonzero, $\frac3x$ is irrational.

Therefore,

is irrational.

If $x\ne0$, then $x\sqrt2$ is irrational, so

is irrational.

Therefore, we must have

Indeed, $x=0$ gives $3$.

Hence

Assume, for the sake of contradiction, that

for some rational $r$.

Then

Squaring,

Thus

Since $r>0$,

so $\sqrt n$ is rational. Hence $n$ must be a perfect square.

Similarly, $\sqrt{n+1}=r-\sqrt n$ is rational, so $n+1$ must also be a perfect square.

But two positive consecutive integers cannot both be perfect squares.

Contradiction.

Therefore,

for every positive integer $n$.

Suppose such rational numbers $a,b$ exist.

Squaring,

If $ab\ne0$, then

is irrational, while

is rational, which is impossible.

Thus

If $b=0$, then $a=\sqrt3$, contradicting $a\in\mathbb{Q}$.

If $a=0$, then

so

which is irrational.

Therefore, no such rational numbers $a,b$ exist.