Answer
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Hint: In this particular question use the concept that for infinite series, for example $\sqrt {a + \sqrt {a + \sqrt {a + ....} } } = b$ this will be written as $\sqrt {a + b} = b$ so use this concept to reach the solution of the question.
Complete step-by-step answer:
Given equation
$\sqrt {x + \sqrt {x + \sqrt {x + .......} } } = \sqrt {x\sqrt {x\sqrt {x......} } } $
Now we have to prove that
x + y = xy, for all $y \in N.$
Proof –
So, let $\sqrt {x + \sqrt {x + \sqrt {x + .......} } } = \sqrt {x\sqrt {x\sqrt {x......} } } $ = y........................ (1), for all $y \in N.$
Therefore, $\sqrt {x + \sqrt {x + \sqrt {x + .......} } } $ = y
Now the LHS of above expression is an infinite series so the above series is also written as
$ \Rightarrow \sqrt {x + y} = y$
Now squaring on both sides we have,
$ \Rightarrow {\left( {\sqrt {x + y} } \right)^2} = {y^2}$
$ \Rightarrow x + y = {y^2}$................. (2)
Now from equation (1) we also have,
$\sqrt {x\sqrt {x\sqrt {x......} } } $ = y
Now again the LHS of above expression is an infinite series so the above series is also written as
$ \Rightarrow \sqrt {xy} = y$
Now squaring on both sides we have,
$ \Rightarrow {\left( {\sqrt {xy} } \right)^2} = {y^2}$
$ \Rightarrow xy = {y^2}$................. (3)
Now from equation (2) and (3) we can say that,
$ \Rightarrow $ x + y = xy = ${y^2}$
Therefore, x + y = xy, for all $y \in N.$
Hence Proved.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall that if y belongs to natural number than it can be any number starting from negative infinity to positive infinity, that why we assume the given expression equal to y, then simplify the expression as above we will get the required answer.
Complete step-by-step answer:
Given equation
$\sqrt {x + \sqrt {x + \sqrt {x + .......} } } = \sqrt {x\sqrt {x\sqrt {x......} } } $
Now we have to prove that
x + y = xy, for all $y \in N.$
Proof –
So, let $\sqrt {x + \sqrt {x + \sqrt {x + .......} } } = \sqrt {x\sqrt {x\sqrt {x......} } } $ = y........................ (1), for all $y \in N.$
Therefore, $\sqrt {x + \sqrt {x + \sqrt {x + .......} } } $ = y
Now the LHS of above expression is an infinite series so the above series is also written as
$ \Rightarrow \sqrt {x + y} = y$
Now squaring on both sides we have,
$ \Rightarrow {\left( {\sqrt {x + y} } \right)^2} = {y^2}$
$ \Rightarrow x + y = {y^2}$................. (2)
Now from equation (1) we also have,
$\sqrt {x\sqrt {x\sqrt {x......} } } $ = y
Now again the LHS of above expression is an infinite series so the above series is also written as
$ \Rightarrow \sqrt {xy} = y$
Now squaring on both sides we have,
$ \Rightarrow {\left( {\sqrt {xy} } \right)^2} = {y^2}$
$ \Rightarrow xy = {y^2}$................. (3)
Now from equation (2) and (3) we can say that,
$ \Rightarrow $ x + y = xy = ${y^2}$
Therefore, x + y = xy, for all $y \in N.$
Hence Proved.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall that if y belongs to natural number than it can be any number starting from negative infinity to positive infinity, that why we assume the given expression equal to y, then simplify the expression as above we will get the required answer.
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