Maths project

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ppt for 10 cbse maths

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  • Maths project

    1. 1. Euclid’s DivisionLemmaGiven positiveintegers a andb, there existuniqueintegers q andr satisfying a= bq + r, 0 ≤ r< b.
    2. 2. Muhammad ibn Musa al-KhwarizmiAn algorithm is a series of welldefined steps which gives aprocedure for solving a type ofproblem.The word algorithm comes fromthe name of the 9th centuryPersian mathematicianal-Khwarizmi. In fact, even theword ‘algebra’ is derived from abook, he wrote, called Hisabal-jabr w’al-muqabala.A lemma is a proven statementused for proving anotherstatement.
    3. 3. Let us state Euclid’sDivision algorithm clearly.To obtain the HCF of two positive integers, say cand d, with c > d, follow the steps below:Step 1 : Apply Euclid’s division lemma, to c and d.So, we find whole numbers, q and r such that c =dq + r, 0 ≤ r < d.Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0,apply the division lemma to d and r.Step 3 : Continue the process till the remainder iszero. The divisor at this stage will be the requiredHCF.
    4. 4. FundamentalTheorem ofArithmeticEvery composite number can beexpressed ( factorised ) as a productof primes, and this factorisation isunique, apart from the order in whichthe prime factors occur.
    5. 5. Carl Friedrich GaussAn equivalent version of Theorem 1.2 was probably firstrecorded as Proposition 14 of Book IX in Euclid’sElements, before it came to be known as the FundamentalTheorem of Arithmetic. However, the first correct proofwas given by Carl Friedrich Gauss in his DisquisitionesArithmeticae.Carl Friedrich Gauss is often referred to as the ‘Prince ofMathematicians’ and is considered one of the threegreatest mathematicians of all time, along withArchimedes and Newton. He has made fundamentalcontributions to both mathematics and science.
    6. 6. Carl Friedrich Gauss(1777 – 1855)
    7. 7. HCF LCMProduct ofthe smallestpower ofeachcommonprime factorin thenumbersProduct ofthe greatestpower ofeach primefactor,involved inthe numbers.
    8. 8. Theorem : Let p be a prime number. If p divides a2, then pdivides a, where a is a positive integer.Proof : Let the prime factorisation of a be as follows :a = p1p2 . . . pn, where p1,p2, . . ., pn are primes, not necessarilydistinct.Therefore, a2= (p1p2 . . . pn)(p1p2 . . . pn) = p21P22 . . . p2n.Now, we are given that p divides a2. Therefore, from theFundamental Theorem of Arithmetic, it follows that p isone of the prime factors of a2. However, using theuniqueness part of the Fundamental Theorem ofArithmetic, we realise that the only prime factors of a2arep1, p2, . . ., pn. So p is one of p1, p2, . . ., pn. Now, since a = p1,p2, . . ., pn, p divides a.
    9. 9. Theorem : √2 is irrational.Proof : Let us assume, to the contrary, that √2 isrational.So, we can find integers r and s (≠ 0) such that √2 =r/s.Suppose r and s have a common factor other than 1.Then, we divide by the common factor to get √2a/b where a and b are co prime.So, b √2 = a.Squaring on both sides and rearranging, we get 2b2= a2. Therefore, 2 divides a2.Now, by Theorem 1.3, it follows that 2 divides a.So, we can write a = 2c for some integer c.
    10. 10. Theorem 1.5Let x be a rational number whosedecimal expansion terminates.Then x can be expressed in the formp/q where p and q are coprime, and theprime factorisation of q is of the form2n5m, where n, m are non-negativeintegers.
    11. 11. Theorem 1.6Let x = p/q be a rational number, suchthat the prime factorisation of q is ofthe form 2n5m, where n, m arenon-negative integers. Then x has adecimal expansion which terminates.
    12. 12. Theorem 1.7Let x = p/q be a rational number, suchthat the prime factorisation of q isnot of the form 2n5m, where n, m arenon-negative integers. Then, x has adecimal expansion which is non-terminating repeating (recurring).

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