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Ask QuestionPosted by Yash Rana 7 years, 3 months ago
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Kunal Waldia 7 years, 3 months ago
Posted by Yash Rana 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Consider {tex}\sqrt { p } + \sqrt { q }{/tex} is rational and can be represented as {tex}\sqrt { p } + \sqrt { q }{/tex} = a
{tex}\Rightarrow ( \sqrt { p } ) = a - \sqrt { q }{/tex}
{tex}\Rightarrow ( \sqrt { p } ) ^ { 2 } = ( a - \sqrt { q } ) ^ { 2 }{/tex} (squaring both sides)
⇒ p = a2 + {tex}\left(\sqrt q\right)^2{/tex} - 2 a {tex}\sqrt { q }{/tex}
⇒ p = a2 + q - 2 a {tex}\sqrt { q }{/tex}
⇒ 2a {tex} \sqrt { q }{/tex} = a2 + q - p
{tex}\Rightarrow \sqrt { q } = \frac { a ^ { 2 } + q - p } { 2 a }{/tex}
As q is prime so {tex}\sqrt { q }{/tex} is not rational but {tex}\frac { a ^ { 2 } + q - p } { 2 a }{/tex} is rational because a, p, q are non-zero integers which contradicts our consideration.
Hence, {tex}\sqrt { p } + \sqrt { q }{/tex} is irrational where p and q are primes.
Posted by Bableen Singh 7 years, 3 months ago
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Posted by Samir Raza 7 years, 3 months ago
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Posted by P Mahato 7 years, 3 months ago
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Posted by Atharva Kothawade 7 years, 3 months ago
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Posted by Vivek Bhandari 7 years, 3 months ago
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Shulabh Rawat 7 years, 3 months ago
Posted by Kushal C 7 years, 3 months ago
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Posted by Abhi Shek 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
If the points A(- 2, 1), B(a, b) and C(4, 1) are collinear and a - b = 1, we have to find a and b.
If three points are collinear, then area covered by given points = 0.
{tex}\therefore{/tex} Area = {tex}\frac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( y _ { 1 } - y _ { 2 } \right) \right]{/tex}=0
Here, (x1, x2, x3) = (- 2, a, 4)
and (y1, y2, y3) = (1, b, 1)
Area ={tex}\frac { 1 } { 2 } [ - 2 ( b - 1 ) + a ( 1 - 1 ) + 4 ( 1 - b ) ]{/tex}
Area = {tex}\frac { 1 } { 2 } [ - 2 b + 2 + 0 + 4 ( 1 - b ) ]{/tex}
0 = -3b + 3
or, 3b = 3 or, b = 1
Given, a - b = 1
i.e, a - 1 = 1
a = 2
(a, b) = (2, 1)
Posted by Aditya Hatwar 7 years, 3 months ago
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Posted by Aditya Hatwar 7 years, 3 months ago
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Posted by Ajay K 5 years, 8 months ago
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Posted by Akshit Rajoria 7 years, 3 months ago
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Ankit Raj 7 years, 3 months ago
Posted by Asmi Lamba 7 years, 3 months ago
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Asmi Lamba 7 years, 3 months ago
Posted by Mahadevan.K.P Devan 7 years, 3 months ago
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Nitya Mishra 7 years, 3 months ago
Posted by Bhumi Sharma 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let a and d be the first term and common difference respectively of the given A.P. Then
an = a + (n - 1)d
{tex}\frac { 1 } { n } ={/tex} mth term
{tex}\Rightarrow \frac { 1 } { n } {/tex}= a + ( m - 1 ) d ...(i)
{tex}\frac { 1 } { m }{/tex}= nth term
{tex}\Rightarrow \frac { 1 } { m } {/tex}= a + ( n - 1 ) d ...(ii)
On subtracting equation (ii) from equation (i), we get
{tex}\frac { 1 } { n } - \frac { 1 } { m } = {/tex} [a+ (m-1) d] -[ a+ (n -1)d]
= a + md - d - a - nd + d
{tex}= ( m - n ) d{/tex}
{tex} \Rightarrow \frac { m - n } { m n } = ( m - n ) d {/tex}
{tex}\Rightarrow d = \frac { 1 } { m n }{/tex}
Putting d = {tex}\frac { 1 } { m n }{/tex} in equation (i), we get
{tex}\frac { 1 } { n } = a + \frac { ( m - 1 ) } { m n } {/tex}
{tex}\Rightarrow \frac { 1 } { n } = a + \frac { 1 } { n } - \frac { 1 } { m n } {/tex}
{tex}\Rightarrow a = \frac { 1 } { m n }{/tex}
{tex}\therefore{/tex} (mn)th term = a + (mn - 1) d
= {tex}\frac { 1 } { m n } + ( m n - 1 ) \frac { 1 } { m n } {/tex}{tex}\left[ \because a = \frac { 1 } { m n } = d \right]{/tex}
= {tex}\frac { 1 } { m n } + \frac { mn } { m n } - \frac { 1 } { m n }{/tex}
= 1
Posted by Shiv Shankr 7 years, 3 months ago
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Posted by Karan Thirpola 7 years, 3 months ago
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Posted by Diksha Rathor 7 years, 3 months ago
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Mohit Singh. 7 years, 3 months ago
Posted by Soumya Prabhunatti 7 years, 3 months ago
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