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Prove by contradiction that if "a" is a nonzero rational number and

"r" is an irrational number, then "ar" is an irrational
number.

User Pinky
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Final answer:

To prove that the product of a nonzero rational number and an irrational number is irrational, one can use proof by contradiction. Assuming the product is rational leads to a contradiction that an irrational number can be expressed as a fraction, which is false. Thus, the product must be irrational.

Step-by-step explanation:

To prove by contradiction that if a is a nonzero rational number and r is an irrational number, then ar is an irrational number, we can begin with the following steps:


  1. Assume that a is a nonzero rational number, which means a can be expressed as a fraction of two integers, p/q where p and q are integers and q ≠ 0.

  2. Assume that r is an irrational number, meaning r cannot be expressed as a fraction of two integers.

  3. Now, suppose for the sake of contradiction that ar is rational. This would mean that ar can be expressed as a fraction of two integers, let's say m/n where m and n are integers and n ≠ 0.

  4. From step 3, if ar = m/n, then r = m/(an). But we initially assumed that r is irrational and cannot be expressed as a fraction. We have a contradiction because we found that r can be expressed as a fraction, which goes against our initial assumption that r is irrational.

  5. Therefore, our initial assumption that ar is rational is false, which means ar must be irrational.

By using proof by contradiction, we have shown that the product of a nonzero rational number and an irrational number is always irrational.

User Hardik Kothari
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