If the rational number is zero, then the result will it is in rational. So have the right to we conclude that in general, us can"t decide, and also it relies on the reasonable number?

Any nonzero reasonable number time an irrational number is irrational. Permit $r$ be nonzero and rational and also $x$ be irrational. If $rx=q$ and $q$ is rational, climate $x=q/r$, i beg your pardon is rational. This is a contradiction.

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If $a$ is irrational and also $b e0$ is rational, climate $a,b$ is irrational. Proof: if $a,b$ were equal to a rational $r$, then us would have actually $a=r/b$ rational.

**Claim:**If $x$ is irrational and $r e 0$ is rational, then $xr$ is irrational.

**Proof:** suppose that $xr$ were rational. Then, $x = fracxrr$ would certainly be rational (as the quotient of 2 rationals). This plainly contradicts the assumption that $x$ is irrational. Therefore, $xr$ is irrational.

The $r = 0$ instance is special, and also the over argument doesn"t work.

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Irrational times non-zero rational is irrational number.If not, expect a is a irrational number and also b is non-zero rational number such the ab=c, whereby c is a reasonable number.As repertoire of every rational number forms field.so any kind of non-zero rational is invertible.So the would indicate a is reasonable number--which is no true.

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Proof verification: permit $a$ it is in an irrational number and $r$ be a nonzero rational number. If $s$ is a rational number climate $ar$ + $s$ is irrational

$f$ differentiable, $f(x)$ reasonable if $x$ rational; $f(x)$ irrational if $x$ irrational. Is $f$ a direct function?

given that $f(x) = -1$ if $f$ is irrational and $f(x)=1$ if $f$ is rational, present that $f$ is not constant anywhere.

deserve to the sum of irrational square root of two various rational number be an additional irrational square source of a reasonable number?

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