UPSKILL MATH PLUS
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Learn moreProve that \sqrt{3} + \sqrt{5} is irrational.
Ans:
Let's prove \sqrt{3} + \sqrt{5} is an irrational number.
Now prove by contradiction method.
| 1. | Assume \sqrt{3} + \sqrt{5} is | |
| 2. | Therefore, it can be written as | |
| 3. | And p and q are | |
| 4 |
Squaring on both sides, we get
|
|
| 5. | Simplifying the term, | |
| 6. | This implies that, | |
| 7. | This | contradicts our assumption |
| 8. | Thus, \sqrt{3} + \sqrt{5} is |
Answer variants:
co-primes
an irrational number
a rational number
\sqrt{3} + \sqrt{5}=\frac{p}{q} where p, q are integers and q = 0
\sqrt{3} + \sqrt{5}=\frac{p}{q} where p, q are integers and q \neq 0
(\sqrt{3} - \sqrt{5})^2 = \left(\frac{p}{q} \right)^2
\sqrt{3} + \sqrt{5}=\frac{p}{q} where p, q are integers and p \neq 0
(\sqrt{3} + \sqrt{5})^2 = \left(\frac{p}{q} \right)^2
\sqrt{15} =\frac{p^2 - 8q^2}{2q^2}
\sqrt{3} + \sqrt{5}
contradicts
\sqrt{15} is a rational number
(\sqrt{3} + \sqrt{5})^2 = \left(\frac{q}{p} \right)^2
\sqrt{15} is an rational number
Satisfies
\sqrt{5} =\frac{p^2 - 8q^2}{2q^2}
composites
\sqrt{15} =\frac{8p^2 - q^2}{2q^2}
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