# Real Numbers Questions

## The following real numbers have decimal expansions as given below.

Question : The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational and of the form $$p\over q$$, what can you say about the prime factors of q ? (i)  43.123456789 (ii)  0.120120012000120000….. (iii)  43.$$\overline{123456789}$$ Solution : (i)  43.123456789 is terminating. …

## Write down the decimal expansion of these given rational numbers which have terminating decimal expansions.

Question : Write down the decimal expansion of these given rational numbers which have terminating decimal expansions. (i) $$13\over 3125$$ (ii)  $$17\over 8$$ (iii)  $$64\over 455$$ (iv)  $$15\over 1600$$ (v)  $$29\over 343$$ (vi)  $$23\over {2^3 5^2}$$ (vii)  $$129\over {2^2 5^7 7^5}$$ (viii)  $$6\over 15$$ (ix)  $$35\over 50$$ (x)  $$77\over 210$$ Solution : (i)  $$13\over 3125$$ …

## Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :

Question : Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion : (i) $$13\over 3125$$ (ii)  $$17\over 8$$ (iii)  $$64\over 455$$ (iv)  $$15\over 1600$$ (v)  $$29\over 343$$ (vi)  $$23\over {2^3 5^2}$$ (vii)  $$129\over {2^2 5^7 7^5}$$ (viii)  $$6\over 15$$ (ix)  …

Question : Prove that the following are irrationals : (i)  $$1\over \sqrt{2}$$ (ii)  $$7\sqrt{5}$$ (iii)  $$6 + \sqrt{2}$$ Solution : (i)  Let us assume, to the contrary, that $$1\over \sqrt{2}$$ is rational. That is, we can find co-prime integers p and q (q $$\ne$$ 0) such that $$1\over \sqrt{2}$$ = $$p\over q$$  or   $$1\times \sqrt{2}\over … ## Prove that \(3 + 2\sqrt{5}$$ is irrational.

Solution : Let us assume, to the contrary, that $$3 + 2\sqrt{5}$$ is an irrational number. Now, let $$3 + 2\sqrt{5}$$ = $$a\over b$$, where a and b are coprime and b $$ne$$ 0. So, $$2\sqrt{5}$$ = $$a\over b$$ – 3  or  $$\sqrt{5}$$ = $$a\over 2b$$ – $$3\over 2$$ Since a and b are integers, …

## Prove that $$\sqrt{5}$$ is an irrational number by contradiction method.

Solution : Suppose that $$\sqrt{5}$$ is an irrational number. Then $$\sqrt{5}$$ can be expressed in the form $$p\over q$$ where p, q are integers and have no common factor, q $$ne$$ 0. $$\sqrt{5}$$ = $$p\over q$$ Squaring both sides, we get 5 = $$p^2\over q^2$$   $$\implies$$  $$p^2$$ = $$5q^2$$              …

Solution : We have,  $$7 \times 11 \times 13$$ + 13 = 1001 + 13 =1014 1014 = $$2 \times 3 \times 13 \times 13$$ So, it is the product of more than two prime numbers. 2, 3 and 13. Hence, it is a composite number. $$7 \times 6 \times 5 \times 4 \times 3 … ## Check whether \(6^n$$ can end with the digit 0 or any n $$\in$$ N.

Solution : If the number $$6^n$$ ends with the digit zero. Then it is divisible by 5. Therefore the prime factorisation of $$6^n$$ contains the prime 5. This is not possible because the only primes in the factorisation of $$6^n$$ are 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that …

## Given H.C.F. (306, 657) = 9, find L.C.M. (306, 657).

Solution : We have, H.C.F. (306, 657) = 9. We know that, Product of L.C.M and H.C.F = Product of two numbers. $$\implies$$  L.C.M $$\times$$ 9 = $$306 \times 657$$ $$\implies$$  L.C.M = $$306 \times 657 \over 9$$ = 22338 Hence, L.C.M is 22338.