I was trying to solve questions from "Mathematical Analysis" by "Apostol" and came across this question which states

""If $a, b, c$ and $d$ are rational and if $x$ is irrational, prove that $\dfrac{\left( ax + b \right)}{\left( cx + d \right)}$ is usually irrational.""

Now, the problem with this is that how can we prove something is usually True? All we can do is to show some cases where this is not true. Like in this example, we can give the solution as:-

The only time it will be rational is when the numerator is some integer times the denominator or vice - versa. The rest of the times, when this is not true, it will be irrational.

Is this the correct way of proof or there is some other war to prove something usually true?

Note: My question is not about finding the possibilities that the given expression is rational. The question is about how to prove something usually True? Because, as far as I have learnt about logic, we may prove something to be True (completely) or False (by just giving one counter example). So, here, I am asking that if such a statement comes, what does it mean to prove it? In fact, I have solved the question and its solution is written in the explaination. All I want to know is that if my solution is correct or is there a proper way to prove something usually True?


marked as duplicate by zipirovich, dxiv, Rohan, Paramanand Singh, Did real-analysis Jan 27 '18 at 9:03

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  • $\begingroup$ Can you check for what $x$ it is rational. $\endgroup$ – user312648 Jan 27 '18 at 4:23

$$\frac{ax+b}{cx+d}=\frac{(a+\frac{b}{x})x}{(c+\frac{d}{x})x}=\frac{a+\frac{b}{x}}{c+\frac{d}{x}}$$ Here, we see that it is only rational if $\frac{a}{c}=\frac{b}{d}$.

  • 5
    $\begingroup$ we see that ... How do you "see" that in any way that is more direct than the original fraction, before dividing by $\,x\,$? $\endgroup$ – dxiv Jan 27 '18 at 5:02
  • $\begingroup$ How could you see that? $\endgroup$ – Aniruddha Deshmukh Jan 27 '18 at 8:25

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