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Say that "A" and "B" are Rational, and C is irrational, would the solution to "A(B+C)" be Rational or Irrational? An example for clarification would be wonderful.

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    $\begingroup$ As long as $A$ is not zero, then $A(B+C)$ is irrational. $\endgroup$ – Thomas Andrews Oct 29 '15 at 18:03
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Remember how to add, subtract, multiply, and divide fractions: $$ \frac p q \pm \frac r s = \frac{ps \pm qr}{qs}, \qquad \frac p q \cdot \frac r s = \frac{pr}{qs}, \qquad \frac{p/q}{r/s} = \frac{ps}{qr}. $$

You have $B+C$ and $B$ is rational and $C$ is irrational. Say $B = \dfrac p q$ and $p,q$ are integers. If $B+C$ is rational, then $B+C = \dfrac r s$ for some integers $r,s$, and then $$ \frac pq + C = \frac r s, \text{ and therefore } C = \frac r s - \frac p q =\cdots \quad\text{(So $C$ would be rational.)} $$

Hence $B+C$ must be irrational.

Next we ask whether $A(B+C)$ might be rational. Since $A$ is rational, there are integers $k,\ell$ such that $A=k/\ell$. If $A(B+C)$ is rational then there are integers $m,n$ such that $$ \frac k \ell (B+C) = A(B+C) = \frac m n, \text{ and so } B+C = \frac{m/n}{k/\ell} = \frac{m\ell}{kn}, \text{ so } B+C \text{ would be rational}. $$

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HINT:

Sum of a rational and an irrational number is always irrational $$\text{AND}$$ Product of a rational and an irrational number is a non-zero rational.

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  • $\begingroup$ Non-zero rational for the second part. :) $\endgroup$ – Thomas Andrews Oct 29 '15 at 18:03
  • $\begingroup$ @ThomasAndrews Sorry but I couldn't get you :) $\endgroup$ – SchrodingersCat Oct 29 '15 at 18:05
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    $\begingroup$ The product of a rational and an irrational is not always irrational. $0\cdot \sqrt{2}$ is not irrational. $\endgroup$ – Thomas Andrews Oct 29 '15 at 18:05

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