√3 is an irrational number. Prove it
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Answer:
Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N. ... Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational number.
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Answer:
Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N. We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational number.
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