## Knowledge in Mathematics Proof

Real Numbers - Ncert Solutions for Class 10 Maths CBSE

Chapter 1 - Real Numbers Excercise Ex. 1.1Solution 1(i) 135 and 225&nbsp;Step 1:&nbsp;&nbsp;Since 225 &gt; 135, apply Euclid's division lemma, to a =225 and b=135 to find q and r&nbsp;&nbsp;such that 225 = 135q+r, 0&nbsp;&nbsp;r&lt;135&nbsp;On dividing 225 by 135 we get quotient as 1 and remainder as 90&nbsp;i.e 225 = 135 x 1 + 90&nbsp;Step 2:&nbsp;Remainder r which is 90&nbsp;&nbsp;0, we apply Euclid's division lemma to b =135 and r = 90 to find whole numbers q and r such that&nbsp;135 = 90 x q + r,&nbsp;0&nbsp;&nbsp;r&lt;90On dividing 135 by 90 we get quotient as 1 and remainder as 45i.e 135 = 90 x 1 + 45Step 3:&nbsp;&nbsp;Again remainder r = 45&nbsp;&nbsp;0 so we apply Euclid's division lemma to b =90 and r = 45 to find q and r&nbsp;&nbsp;such that&nbsp;90 = 90 x q + r,&nbsp;&nbsp;0&nbsp;&nbsp;r&lt;45On dividing 90 by 45 we get quotient as 2 and remainder as 0&nbsp;i.e 90 = 2 x 45 + 0Step 4:&nbsp;Since the remainder is zero, the divisor at this stage will be HCF of (135, 225).&nbsp;Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45.&nbsp;(ii)&nbsp;&nbsp;&nbsp;196 and 38220Step 1:&nbsp;&nbsp;Since 38220 &gt; 196, apply Euclid's division lemmato a =38220 and b=196&nbsp;to find&nbsp;whole numbers q and r&nbsp;&nbsp;such that&nbsp;38220 = 196 q + r, 0&nbsp;&nbsp;r &lt; 196On dividing 38220 we get quotient as 195 and remainder r as 0&nbsp;i.e 38220 = 196 x 195 + 0Since the remainder is zero, divisor at this stage will be HCF&nbsp;Since divisor at this stage is 196 , therefore, HCF of 196 and 38220 is 196.NOTE: HCF( a,b) = a if a is a factor of&nbsp;b. Here, 196 is a factor of&nbsp;38220 so HCF is 196.&nbsp;(iii)&nbsp;&nbsp;&nbsp;867 and 255&nbsp;Step 1:&nbsp;Since 867 &gt; 255, apply Euclid's division lemma, to a =867 and b=255 to find q and r&nbsp;&nbsp;such that 867 = 255q + r, 0&nbsp;&nbsp;r&lt;255On dividing 867 by 255 we get quotient as 3 and remainder as 102i.e 867 = 255 x 3 + 102&nbsp;Step 2:&nbsp;Since remainder 102&nbsp;&nbsp;0, we apply the division lemma to a=255 and b= 102 to find whole numbers q and r such that&nbsp;255 = 102q + r where 0&nbsp;&nbsp;r&lt;102On dividing 255 by 102 we get quotient as 2 and remainder as 51&nbsp;i.e 255 = 102 x 2 + 51Step 3:&nbsp;Again remainder 51 is non zero, so we apply the division lemma to a=102 and b= 51&nbsp;to find whole numbers q and r such that&nbsp;102 = 51 q + r where 0&nbsp;r &lt; 51&nbsp;&nbsp;On dividing 102 by 51 quotient is 2 and remainder is 0i.e 102 = 51 x 2 + 0Since the remainder is zero, the divisor at this stage is the HCFSince the divisor at this stage is 51,therefore, HCF of 867 and 255 is 51.&nbsp;Concept Insight: To crack such problem remember to apply the Euclid's division Lemma which states that "Given positive integers a and b, there exists unique integers q and r satisfying a = bq + r, where 0&nbsp;&nbsp;r &lt; b" in the correct order.&nbsp;Here, a &gt; b.&nbsp;Euclid's algorithm works since Dividing 'a' by 'b', replacing 'b' by 'r' and 'a' by 'b' and repeating the process of division till remainder 0 is reached, gives&nbsp;a number which divides a and b exactly.&nbsp;i.e&nbsp;&nbsp;&nbsp;HCF(a,b) =HCF(b,r)Note that do not find the HCF using prime factorisation in this question when the method is specified and do not skip steps.

Engineering mathematics unit 2 chapter 1 fourier series,explained, formula

Engineering mathematics subject is one to study hardly . Engineering mathematics second unit consist of unit namedd fourier series. In this pdf fourier series is explained with simplicity & numerically.

Engineering mathematics unit 2 chapter 1 fourier series,explained, formula 4

Engineering mathematics subject is one to study hardly . Engineering mathematics second unit consist of unit namedd fourier series. In this pdf fourier series is explained with simplicity & numerically.

Mathematics 1, engineering mathematics, fourier series numericals & formulae 2

Engineering mathematics is a subject in first year engineering it is in first semester. In this pdf we are going to learn mathematics unit 2 & chapter 2 ie. Fourier series numericals

Mathematics 1, engineering mathematics, fourier series numericals & formulae

Engineering mathematics is a subject in first year engineering it is in first semester. In this pdf we are going to learn mathematics unit 2 & chapter 2 ie. Fourier series numericals

Mathematics 1, engineering mathematics, fourier series numericals & formulae 2

Engineering mathematics is a subject in first year engineering it is in first semester. In this pdf we are going to learn mathematics unit 2 & chapter 2 ie. Fourier series numericals

Mathematics 1, engineering mathematics, fourier series numericals & formulae 1

Engineering mathematics is a subject in first year engineering it is in first semester. In this pdf we are going to learn mathematics unit 2 & chapter 2 ie. Fourier series numericals

Mathematics asssignment on l'hospital rule part 2

This pdf has handwritten notes of assignment on l'hospital rule