Real Numbers Questions

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. …

The following real numbers have decimal expansions as given below. Read More »

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)  \(13\over 3125\) …

Write down the decimal expansion of these given rational numbers which have terminating decimal expansions. Read More »

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)  …

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 : Read More »

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 the following are irrationals : Read More »

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 \(3 + 2\sqrt{5}\) is irrational. Read More »

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 = \(p^2\over q^2\)   \(\implies\)  \(p^2\) = \(5q^2\)              …

Prove that \(\sqrt{5}\) is an irrational number by contradiction method. Read More »

Solution : They will be again at the starting point at least common multiples of 18 and 12 minutes. To find the L.C.M of 18 and 12, we have : 18 = \(2 \times 3\times 3\) and  12 = \(2 \times 2 \times 3\) L.C.M of 18 and 12 = \(2 \times 2 \times 3 …

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point ? Read More »

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 …

Explain why \(7 \times 11 \times 13\) + 13 and \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\) + 5 are composite numbers. Read More »

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

Check whether \(6^n\) can end with the digit 0 or any n \(\in\) N. Read More »

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.