Prove that $$6+\sqrt{2}$$ is irrational.
Let us assume $$6+\sqrt{2}$$ is rational. Then it can be expressed in the form $$\dfrac{p}{q}$$, where $$p$$ and $$q$$ are co-prime
Then, $$6+\sqrt{2}=\dfrac{p}{q}$$
$$\sqrt{2}=\dfrac{p}{q}-6$$
$$\sqrt{2}=\dfrac{p-6q}{q}$$ -----($$p,q,-6$$ are integers)
$$\dfrac{p-6q}{q}$$ is rational
But, $$\sqrt{2}$$ is irrational.
This contradiction is due to our incorrect assumption that $$6+\sqrt{2}$$ is rational
Hence, $$6+\sqrt{2}$$ is irrational
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.
Classify the following numbers as rational or irrational: (i) $$\sqrt{23}$$ (ii) $$\sqrt{225}$$ (iii) $$0.3796$$ (iv) $$7.478478...$$ (v) $$1.101001000100001...$$
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?
Prove that $$3+2\surd{5}$$ is irrational.
Prove that $$\sqrt{3}$$ is irrational.
Show that $$3\sqrt{2}$$ is irrational.
The fact that $$3+2\sqrt{5}$$ is irrational is because
Prove that the following are irrational. $$\displaystyle\frac{1}{\sqrt{2}}$$.
Prove that $$\sqrt{5}-\sqrt{3}$$ is not a rational number.
Show that the square of $$\dfrac{(\sqrt{26 - 15\sqrt{3}})}{(5\sqrt{2} - \sqrt{38 + 5\sqrt{3}})}$$ is a rational number.