Subjective Type

Recall, $$\pi$$ is defined as the ratio of the circumference(say c) of a circle to its diameter(say d). That is, $$\pi=\displaystyle\frac{c}{d}$$. This seems to contradict the fact that $$\pi$$ is irrational. How will you resolve this contradiction?
Solution
Here,
π=227π=227
Now, it is in the form of pqpq , which is a rational number, but this is an approximate value of ππ .
If you divide 2222 by 77 , the quotient (3.14⋯)(3.14⋯) is a non-terminating number i.e. it is irrational.
Approximate fractions include (in order of increasing accuracy)
227,333106,355113,⋯227,333106,355113,⋯
If we divide anyone of these we get, the quotient (3.14⋯)(3.14⋯) is a non-terminating number. i.e. it is irrational.
So,
There is no contradiction as either cc or dd irrational and
hence ππ is a irrational number.
Number Systems
State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form $$\sqrt{m}$$, where $$m$$ is a natural number.
(iii) Every real number is an irrational number.
Number Systems
Classify the following numbers as rational or irrational:
(i) $$\sqrt{23}$$
(ii) $$\sqrt{225}$$
(iii) $$0.3796$$
(iv) $$7.478478...$$
(v) $$1.101001000100001...$$
Number Systems
Prove that $$3+2\surd{5}$$ is irrational.
Number Systems
Prove that $$\sqrt{3}$$ is irrational.
Number Systems
Show that $$3\sqrt{2}$$ is irrational.
Number Systems
The fact that $$3+2\sqrt{5}$$ is irrational is because
Number Systems
Prove that the following are irrational.
$$\displaystyle\frac{1}{\sqrt{2}}$$.
Number Systems
Prove that $$6+\sqrt{2}$$ is irrational.
Number Systems
Prove that $$\sqrt{5}-\sqrt{3}$$ is not a rational number.
Number Systems
Show that the square of $$\dfrac{(\sqrt{26 - 15\sqrt{3}})}{(5\sqrt{2} - \sqrt{38 + 5\sqrt{3}})}$$ is a rational number.