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Prove that between every rational number and every irrational number, there is an irrational number. Provide which type of a proof you are using.

User Metaxal
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Final answer:

To prove that between every rational number and every irrational number, there is an irrational number, we can use a proof by contradiction.

Step-by-step explanation:

To prove that between every rational number and every irrational number, there is an irrational number, we can use a proof by contradiction. Let's assume that there is a rational number and an irrational number, such that there is no irrational number between them.

Let's consider two numbers: a rational number 'r' and an irrational number 'i'. Since 'r' and 'i' are different, we can assume that 'r' is less than 'i'.

Now, we can find a rational number 'q' such that 'r < q < i'. Since 'r' and 'q' are both rational, their average will also be rational. Let's call the average 'm'. This means that 'r < m < q'. But 'm' is also between 'r' and 'i', and 'm' is rational, which contradicts our assumption that there is no rational number between 'r' and 'i'.

Therefore, we have proved that there is always an irrational number between every rational number and every irrational number.

User Stefan Zhelyazkov
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