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To prove: √3+√5 is irrational
To prove it let us assume it to be a rational number
Rational numbers are the ones which can be expressed in p/q form where p,q are integers and q isn't equal to 0
√3+√5=p/q
√3=(p/q)-√5
Squatting on both sides
3=p²/q²-(2√5p)/q+5
(2√5p)/q=5-3-p²/q²
2√5p/q=(2q²-p²)/q²
√5=(2q²-p²)/q²*q/2p
√5=(2q²-p²)/2pq
As p and q are integers RHS is rational
As RHS is rational LHS is also rational i.e √5 is rational
But this contradicts the fact that √5 is irrational
This contradiction arose because of our false assumption
So √3+√5 is irrational.
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Gaurav Seth 6 years, 3 months ago
nth term, an=a+nb
n-1 th term, an-1=a+(n-1)b
Differenece of two consecutive terms
= an-an-1= a+nb-a-nb+b = b
which is constant
Hence the sequence is an AP and the common difference is b
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Sia ? 6 years, 3 months ago
7 × 11 × 13 + 13
Take 13 common there we get
= 13(7 x 11 +1 )
= 13(77 + 1 )
= 13(78)
It is the product of two numbers and both numbers are more than 1. So, it is a composite number.
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Posted by Epic Boy Rahul 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
The given pair of equations is
px + qy = p - q .....(1)
qx - py = p + q ....(2)
Multiplying equation (1) by p and equation (2) by q, we get
p2x + pqy = p2 - pq....(3)
q2x - pqy = pq + q2.....(4)
Adding equation (3) and equation (4), we get
(p2 + q2)x = p2 + q2
{tex}\Rightarrow \;x = \frac{{{p^2} + {q^2}}}{{{p^2} + {q^2}}} = 1{/tex}
Substituting this value of x in equation (1), we get
p(1) + qy = p - q
{tex}\Rightarrow{/tex} qy = -q
{tex}\Rightarrow \;y = \frac{{ - q}}{q} = - 1{/tex}
So, the solution of the given pair of linear equations is x = +1, y = -1.
Verification, Substituting x = 1, y = -1,
We find that both the equations (1) and (2) are satisfied as shown below:
px + qy = p(1) + q(-1) = p - q
qx - py = q(1) - p(-1) = q + p = p + q
This verifies the solution.
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