Real numbers
Grade Class 10
Question:
<p>If a is a non-zero rational and <span class="math-tex">\(\sqrt b \)</span> is irrational, then <span class="math-tex">\(a\sqrt b \)</span> is:</p>
<p style="display:inline">a natural number</p>
<p style="display:inline">a rational number</p>
<p style="display:inline">an integer</p>
<p style="display:inline">an irrational number</p>
Step-by-Step Solution
Key Concept: Prove the irrationality of the product by assuming it is rational and demonstrating that this forces the irrational factor to be rational.
<p>If possible let <span class="math-tex">\(a\sqrt b \)</span> be rational.<br />
Then <span class="math-tex">\(a\sqrt b = \frac{p}{q},\)</span> where p and q are non-zero integers, having no common factor other than 1.<br />
Now, <span class="math-tex">\(a\sqrt b = \frac{p}{q}\)</span><br />
<span class="math-tex">\( \Rightarrow \)</span> <span class="math-tex">\(\sqrt b = \frac{p}{{aq}}\)</span> ... (i)<br />
But, <span class="math-tex">\(p\)</span> and <span class="math-tex">\(aq\)</span> are both rational and aq <span class="math-tex">\(\neq\)</span> 0<br />
<span class="math-tex">\(\because \)</span> <span class="math-tex">\(\frac{p}{{aq}}\)</span> is rational.<br />
Therefore, from eq. (i), it follows that <span class="math-tex">\(\sqrt b \)</span> is rational.<br />
The contradiction arises by assuming that <span class="math-tex">\(a\sqrt b \)</span> is rational.<br />
Hence, <span class="math-tex">\(a\sqrt b \)</span> is irrational.</p>
Correct Answer: D