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1. Show that the following numbers are irrational.(iii)

    \[\mathbf{6}\text{ }+\text{ }\surd \mathbf{2}\]

CBSE | Class 10 | Maths | RD Sharma | Real Numbers
1. Show that the following numbers are irrational.(iii)

    \[\mathbf{6}\text{ }+\text{ }\surd \mathbf{2}\]

Solution:

Let’s assume on the contrary that

    \[6+\surd 2\]

is a rational number. Then, there exist co prime positive integers a and b such that

    \[6\text{ }+\text{ }\surd 2\text{ }=\text{ }a/b\]

    \[\Rightarrow \surd 2\text{ }=\text{ }a/b\text{ }\text{ }6\]

    \[\Rightarrow \surd 2\text{ }=\text{ }\left( a\text{ }\text{ }6b \right)/b\]

⇒

    \[\surd 2\]

is rational [∵ a and b are integers ∴

    \[\left( a-6b \right)/b\]

is a rational number]

This contradicts the fact that

    \[\surd 2\]

 is irrational. So, our assumption is incorrect.

Hence,

    \[6\text{ }+\text{ }\surd 2\]

 is an irrational number.

Solution:

Let’s assume on the contrary that

    \[3-\surd 5\]

is a rational number. Then, there exist co prime positive integers a and b such that

    \[3-\surd 5\text{ }=\text{ }a/b\]

    \[\Rightarrow \surd 5\text{ }=\text{ }a/b\text{ }+\text{ }3\]

⇒

    \[\surd 5\text{ }=\text{ }\left( a\text{ }+\text{ }3b \right)/b\]

             

⇒

    \[\surd 5\]

is rational [∵ a and b are integers ∴

    \[\left( a+3b \right)/b\]

is a rational number]

This contradicts the fact that

    \[\surd 5\]

is irrational. So, our assumption is incorrect.

Hence,

    \[3-\surd 5\]

is an irrational number.

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