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If the root of a rational number is an irrational number then the resulting number is called surds.
eg:\(\sqrt{4}\)
∴ Here, 2 is a rational number .When square root of 4 is taken out it becomes irrational.
Hence, \(\sqrt{4}\) is a surd.
Here we can learn the basic operations addition, subtraction, multiplication, division, finding square root and finding cube roots of irrational numbers .
Addition
We can add irrational numbers if they have the same order and the same radicand .
eg :\(\sqrt{2}\) + 3\(\sqrt{2}\) = 4\(\sqrt{2}\) [∴ a + 3a = 4a ]
Here,
\(\sqrt{2}\) is added with 3\(\sqrt{2}\) as the radicand is \(\sqrt{2}\) as well as the order is 3.
Again,
\(\sqrt{3}\) + \(\sqrt{2}\) = \(\sqrt{3}\) + \(\sqrt{2}\) [∴ a + b = a + b ]
Then, we can not add with add \(\sqrt{3}\) with \(\sqrt{2}\) as the radicand is different even though the order is same .
Simillary,
\(\sqrt[3]{5}\) + \(\sqrt{5}\) = \(\sqrt[3]{5}\) + \(\sqrt{5}\)
Here, addition is not positive as the order of \(\sqrt[3]{5}\) is 3 and \(\sqrt{5}\) is 2 even though the radicand are same .
Subtraction
Here, we subtract irrational numbers if they have the same order and the same radicand same as in addition .
eg :
Multiplication
Here, multiplication is is possible with irrational numbers if the order of irrational number is same \(\sqrt{a}\) \(\times\) \(\sqrt{b}\) = \(\sqrt{ab}\)
eg :
Division
We can divide the rationals if they have some order again
If p and q are real numbers .
\(\frac{\sqrt{p}}{\sqrt{q}}\) = \(\sqrt{\frac{p}{q}}\)
eg;
\(\sqrt{8}\)÷ \(\sqrt{2}\) = \(\sqrt{\frac{8}{2}}\) = \(\sqrt{4}\) = 2
Here the square and cube root of all irritational numbers are not possible .even we can find the square root and cube root of some numbers that may later turn to be rational or irrational.
eg;
Let's see the example of following ;
Two irrationals are said to be rationalizing factors of each other if the product of these two irrationals is a rational number.
For example;
As we know , for the surd of form a-\(\sqrt{b}\) we need to multiply it with any another surd a + \(\sqrt{b}\) so that it turns it intoa perfect square and the irrational turns into rational a - \(\sqrt{b}\) and a + \(\sqrt{b}\) or \(\sqrt{a}\) + b and \(\sqrt{a}\) - b are the rationalizing factors for each other. these terms are called conjugates.
eg:
(a - \(\sqrt{b}\)× (a + \(\sqrt{b}\)
= (a - \(\sqrt{b}\) (a + \(\sqrt{b}\)
= (a)^{2} - (\(\sqrt{b}\)^{2}
= a^{2} - b
The process of multiplying a given surd by its rationalizing factor to get a rational number as a product is known as a rationalization of a given surd.
eg :
\(\frac{1}{\sqrt{2}}\)
Solution:
= \(\frac{1}{\sqrt{2}}\)× \(\frac{\sqrt{2}}{\sqrt{2}}\) (multiplying both numerator and denominator by \(\sqrt{2}\)
= \(\frac{\sqrt{2}}{\sqrt{2×2}}\)
= \(\frac{\sqrt{2}}{\sqrt{4}}\)
= \(\frac{\sqrt{2}}{\sqrt{2^2}}\)
= \(\frac{\sqrt{2}}{2}\)
Hence, the denominatior is rational number.
.
Solution
\(\sqrt{9}\)
= \(\sqrt{3×3}\)
= \(\sqrt{(3)}\)^{2}
=3
Solution
= \(\sqrt{25}\)
= \(\sqrt{5×5}\)
= \(\sqrt{(5)}\)^{2}
= 5
Solution
\(\frac{6\sqrt{3}}{2\sqrt{3}}\)
= \(\frac{6}{2}\)×\(\frac{\sqrt{3}}{\sqrt{3}}\)
= 3
Solution
= ( 8 - 3 + 4 ) \(\sqrt{3}\)
= ( 12 - 3 ) \(\sqrt{3}\)
= 9 \(\sqrt{3}\)
Solution
\(\sqrt{50}\)
= \(\sqrt{5×5×2}\)
= \(\sqrt{(5)^2×2}\)
= 5\(\sqrt{2}\)
If we multiply 5\(\sqrt{2}\) by \(\sqrt{2}\) we will get the rational number. So, \(\sqrt{2}\) is the simplest rationalizing factor of \(\sqrt{50}\).
Solve
(sqrt{27})
Solve
(sqrt{100})
Solve
(sqrt{3}) × (sqrt{3})
Solve
(sqrt{8}) ( imes) (sqrt{2})
Rationalize
(frac{1}{sqrt{5}})
The set of all rational and irrational numbers form _____________________ .
Solve
( 3 + (sqrt{5})) ( 3 - (sqrt{5}) )
Reduce into the simplest form.
(sqrt{9})
Solve
(sqrt{6}) ( imes) (sqrt{5})
Subtract
10(sqrt{5}) - (sqrt{5})
ASK ANY QUESTION ON Surds
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