Q.1. Based on Euclid’s algorithm: a = bq + r ; Using Euclid’s algorithm: Find the HCF of 825 and 175.
Ans:Since 825>175, we apply division lemma to 825 and 175 to get825 = 175 x 4 + 125. Since r ≠ 0, we apply division lemma to 175 and 125 to get 175 = 125 x 1 + 50 Again applying division lemma to 125 and 50 we get, 125 = 50 x 2 + 25. Once again applying division lemma to 50 and 25 we get. 50 = 25 x 2 + 0. Since remainder has now become 0, this implies that HCF of 825 and 125 is 25. Q.2. Based on Showing that every positive integer is either of the given forms: Solved example: Prove that every odd positive integer is either of the form 4q + 1 or 4q + 3 for some integer q. Ans: Let a be any odd positive integer (first line of problem) and let b = 4. Using division Lemma we can write a = bq + r, for some integer q, where 0≤r<4. So a can be 4q, 4q + 1, 4q + 2 or 4q + 3. But since a is odd, a cannot be 4q or 4q + 2. Therefore any odd integer is of the form 4q + 1 or 4q + 3. Q.Find H C F (26,91) if LCM(26,91) is 182 Sol: We know that LCM x HCF = Product of numbers. or 182 x HCF = 26 x 91 or HCF = 26 x 91 = 13 Hence HCF (26, 91) = 13. Q. prove that √5 is irrational. Solution: let us assume on the contrary that √5 is rational. That is we can find coprimes a and b b (≠0) such that √5 = a/b. Or √5b = a. Squaring both sides we get 5b2 = a2. This means 5 divides a2. Hence it follows that 5 divides a. So we can write a = 5c for some integer c. Putting this value of a we get 5b2 = (5c)2 Or 5b2 = 25c2 Or b2 = 5b2. It follows that 5 divides b2. Hence 5 divides b. Now a and b have at least 5 as a common factor. But this contradicts the fact that a and b are coprimes. This contradiction has arisen because of our incorrect assumption that √5 is rational. Hence it follows that √5 is irrational. Q. prove that product of three consecutive positive integers is divisible by 6? Ans: Let three consecutive positive integers be, n, n + 1 and n + 2. Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2. ∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. ⇒ n (n + 1) (n + 2) is divisible by 3. Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1. ∴ n = 2q or 2q + 1, where q is some integer. If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2. Since, n (n + 1) (n + 2) is divisible by 2 and 3. ∴ n (n + 1) (n + 2) is divisible by 6. Q. Express HCF of F 65 and 117 in the form of 65m +117n Ans: Among 65 and 117; 117 > 65 Since 117 > 65, we apply the division lemma to 117 and 65 to obtain 117 = 65 x 1 + 52 … Step 1 Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain 65 = 52 x 1 + 13 … Step 2 Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain 52 = 4 x 13 + 0 … Step 3 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 From Step 2: 13 = 65 – 52 x 1 … Step 4 From Step 1: 52 = 117 – 65 x 1 Thus, from Step 4, it is obtained 13 = 65 – (117 – 65 x 1) x 1 ⇒13 = 65 x 2 – 117 ⇒13 = 65 x 2 + 117 x (–1) In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1 1oth maths term1
44 Comments
Veena
13/8/2013 04:03:42 am
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Tom
13/5/2012 05:18:11 pm
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Jb
19/7/2016 06:34:49 am
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28/5/2012 08:49:21 pm
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15/6/2012 05:24:05 pm
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3/4/2013 12:57:50 am
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6/4/2013 10:33:46 pm
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30/4/2013 09:36:58 pm
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varun
27/6/2013 05:29:56 pm
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29/6/2013 02:21:04 pm
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khushpriya
16/9/2013 03:44:57 pm
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nivetha
16/3/2014 10:28:39 pm
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16/3/2014 10:29:23 pm
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parinnethi
28/3/2014 08:14:44 pm
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Aaisha Thomas
9/4/2014 12:00:50 pm
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17/4/2014 04:40:02 pm
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Rahul
20/4/2014 10:34:12 pm
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14/6/2014 07:59:20 pm
very simple
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vs meghana
5/8/2014 10:03:35 pm
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10/9/2014 02:25:58 am
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7/4/2015 01:55:59 am
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Janaharshini R
18/6/2015 12:30:36 am
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Nameera
20/6/2015 05:31:43 pm
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14/5/2016 06:26:11 am
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abi
31/5/2016 06:25:15 am
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