If the rational number is zero, climate the an outcome will be rational. So have the right to we conclude the in general, us can"t decide, and it relies on the reasonable number?

Any nonzero rational number times an irrational number is irrational. Permit $r$ be nonzero and also rational and also $x$ be irrational. If $rx=q$ and also $q$ is rational, climate $x=q/r$, i m sorry is rational. This is a contradiction.

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

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

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

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

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Irrational times non-zero reasonable is irrational number.If not, mean a is a irrational number and also b is non-zero rational number such that ab=c, where c is a rational number.As repertoire of every rational number develops field.so any non-zero rational is invertible.So the would imply a is reasonable number--which is no true.

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Proof verification: permit $a$ it is in an irrational number and also $r$ it is in 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 linear function?

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

deserve to the sum of irrational square roots of two different rational number be another irrational square root of a reasonable number?

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