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Sia ? 6 years, 4 months ago
Suppose that {tex} \sqrt { p } + \sqrt { q }{/tex} is a rational number equal to {tex} \frac { a } { b }{/tex}, where a and b are integers having no common factor.
Now, {tex} \sqrt { p } + \sqrt { q } = \frac { a } { b }{/tex}
{tex} \Rightarrow \sqrt { p } = \frac { a } { b } - \sqrt { q }{/tex} (squaring both side)
{tex} \Rightarrow \quad ( \sqrt { p } ) ^ { 2 } = \left( \frac { a } { b } - \sqrt { q } \right) ^ { 2 }{/tex}
{tex} \Rightarrow \quad p = \frac { a ^ { 2 } } { b ^ { 2 } } - 2 \left( \frac { a } { b } \right) \sqrt { q } + q{/tex}
{tex} \Rightarrow \quad 2 \left( \frac { a } { b } \right) \sqrt { q } = \frac { a ^ { 2 } } { b ^ { 2 } } + q - p{/tex}
{tex} \Rightarrow \quad 2 \frac { a } { b } \sqrt { q } = \frac { a ^ { 2 } + b ^ { 2 } ( q - p ) } { b ^ { 2 } }{/tex}
{tex} \Rightarrow \quad \sqrt { q } = \frac { a ^ { 2 } + b ^ { 2 } ( q - p ) } { 2 a b }{/tex}
{tex} \Rightarrow \sqrt { q }{/tex} is a rational number. (because sum of two rational numbers is always rational)
This is a contradiction as {tex} \sqrt { q }{/tex} is an irrational number.
Hence, {tex} \sqrt { p } + \sqrt { q }{/tex} is an irrational number.
Posted by Bhavik Girdharwal 7 years, 7 months ago
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Sia ? 6 years, 4 months ago
Let the coordinates of the required point be (x, y). Then,
{tex}x = \frac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}}{/tex}
{tex}= \frac{{(2)(4) + (3)( - 1)}}{{2 + 3}}{/tex}
{tex}= \frac{{8 - 3}}{5} = \frac{5}{5} = 1{/tex}
{tex}y = \frac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}{/tex}
{tex} = \frac{{(2)( - 3) + (3)(7)}}{{2 + 3}}{/tex}
{tex} = \frac{{ - 6 + 21}}{5} = \frac{{15}}{5} = 3{/tex}
Hence, the required point is (1, 3).
Posted by Farhan Alam 7 years, 7 months ago
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Dhruvrajsinh Vala 7 years, 7 months ago
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Sanjana Sharma 7 years, 7 months ago
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Anshika Mittal 7 years, 7 months ago
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