# SUMMARY OF LEARNING-ECI831

Hello Everyone,
Below is the summary of my journey in ECI831. I gained  lot of knowledge in this course and understood the part that social media can play in education. Thank you Dr.Alec Couros and  my fellow classmates, you made this journey memorable. Hope you enjoy my video.

Thanks for stopping by!

# Reaching the finish line by finding Fraction of digits Ending with 9

IMAGE VIA GYAN SANCHAR

This blog post marks the completion of my learning project. Finally, I have successfully learned and implemented Vedic Techniques  in solving problems on Arithmetic

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operations . It was indeed fun and my journey was mesmerizing. Vedic Mathematics is a great way to make Mathematics simpler and joyful for everyone and yes, it is Magic math.

In my first blog post ,we learned what is Vedic Mathematics and how it  can make Mathematics simpler and joyful. I started this journey of learning and knowledge sharing with a great interest in the area of Vedic mathematics. My goal was to implemented Vedic Technic’s in solving problems on Arithmetic operations. My second blog post concentrated on implementing Vedic tricks in multiplication, we witnessed how easily one can multiply complex numbers in a minute and that too without using a calculator. My third and fourth blog post focused on finding Square and Square root of any numbers quickly and am sure most of us start looking for calculators if we want to find square or square root of numbers more than three digits and now we can do it mentally and that too within few seconds.

In my Fifth and Sixth blog post we learned how cool Vedic math is when it comes to finding cube and cube root of any numbers. This awe-inspiring technique helps us to easily find cube and cube root of any given digits mentally. Further Vedic Techniques made my seventh blog post more interesting by simplifying the division  process, now we can do more than just painlessly split a check. In my Ninth blog post Vedic mathematics offered a neat solution to calculate percentage  in our mind to overcome our day to day challenges.

IMAGE VIA EDVIA

In this blog post  we are going to learn how to convert any fraction  ending with 9 and denominator into its decimal equivalent. Vedic mathematics helps us to shorten the complex fraction process into simpler way ,In Sanskrit  this technic is called  as “Ekadhikena Purvena
To solve  fractions ending with 9 there is a Sutra or formula i.e.
BY ONE MORE THAN THE ONE BEFORE ,by using this formula let me solve two examples

Below image is a simple  illustration of the difference between usual method and Vedic method

1)Find the fraction of 1/39
STEPS:
i)   In 1st step, i will find the divisor ,here  the number 9 is the one before, previous             .    number of is 3 and the number 3  will be increase by 1, so we will get
.                                                       3+1=4
ii)   In 2nd step, on every example, i will start with zero point,  and  i will divide 1 by 4 i.e. 1/4 = 0.25 ,i  will  write 0.25 as 025
i.e                                              1/39=0.025
iii)   In 3rd step, i will divide 25/4, i will get remainder as 1 and quotient as 6
i.e                                             1/39=0.0256
.                                                                          1                                           (REMAINDER)
iv)  In 4th step, i will combine remainder and quotient, it will become 16, then i will             .      divide again 16 by 4 which gives me 4
i.e.                                            1/39=0.02564
(REMAINDER)
v)  In 5th step, i am going to divide 4 by 4 which gives me the value as 1
i.e.                                          1/39=0.02564 1
.                                                                         1                                           (REMAINDER)
vi) In 6th step, last digit is 1 ,if we divide 1 by 4, i will get as 0.25 again i will  write it          .      as 025,it became same as in 1st step 25 divide it by 4 gives me 1 as remainder and     as  quotient
i.e.                                           1/39=0.02564 10256
.                                                                                        1                         (REMAINDER)
vii) In next step again we are going to combine remainder and quotient i.e.16, and i will        divide it by 4 which gives me 4 and again 4 by 4 gives me 1,i will stop up-to last           .      digit  1, if i go on solving the digit i will get recurring values ,we can stop up-to 5         .      decimal  values.
i.e.                                           1/39=0.02564 1025641
.                                                                          1                1                       (REMAINDER)

2)     Let us solve one more example.
2/39
i)    In 1st step,i will find the divisor ,here  the number 9 is the one before, previous                   number 9 is 3  and the  number 3 we will be increase by 1,we will get.
.                                            3+1=4
ii)   In 2nd step,on every example, i will start with zero point  and  we will divide 2 by 4 i.e.  2/4 = 0.5, i will write 0.5 as 05.
i.e                                   2/39=0.05
iii)  In 3rd step, i will divide 5/4, we will get remainder as 1 and quotient as 1.
i.e                                2/39=0.051
(REMAINDER)
iv)  In 4th step, i will combine remainder and quotient, it will become 11, then i will             .      divide 11 by 4, which gives me remainder as 3 and quotient as.
i.e.                               2/39=0.0512
1 3                                                       (REMAINDER)
v)   In 5th step, if i combine remainder and quotient, it will become 32 .then i will divide  .      32 by 4 which gives me .
i.e.                               2/39=0.05128
1 3                                                        (REMAINDER)
vi)  In 6th step last digit is 8 ,i will  divide 8 by 4, i will get the value as 2.
i.e.                               2/39=0.051282
1 3                                                        (REMAINDER)
vii)
In 7th step last digit is 2, if i divide 2 by 4, i will get the values as  0.5, i will write 0.5   .       as  05  then if i divide 5 by 4 it gives me remainder 1 and quotient 1 .
i.e.                                2/39=0.051282051
1 3           1                                          (REMAINDER)
vii)
In next step again i am going to combine remainder and quotient i.e. 11 and i will             divide it by 4 which gives me 3 as remainder and 2 as quotient and again if i                  .       combine remainder and quotient and divide by   4 gives me 32/4=8,i will stop                    up- to last digit 8, if i go on solving the digit i will get recurring values .we can stop          up- to  5  decimal values.
i.e.                              2/39=0.05128 205128
1 3             1 3                                    (REMAINDER)

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# THOUGHTS ON USING PREZI AS AN INSTRUCTIONAL TOOL

Image via purdue.edu

I love stories and I consider myself a good story teller. As a math teacher I tell stories about mathematics  , mathematicians and about doing mathematics. Firstly, I do this because I love it and secondly because my students like it and I think it is a most effective instructional tool in teaching mathematics . Instruction via storytelling creates more realistic, powerful and unforgettable images in the student mind than any other means of delivering the same material. Teaching students by storytelling is an effective way to keep the learners occupied and PREZI offers a great platform to achieve it. In this week’s blog post I am reviewing Prezi, an awesome presentation tool which has a potential to make teaching moments memorable.

Prezi is a visual storytelling software presentation tool  which is a great alternative to traditional slide-based presentation programs which we use in classrooms. Prezi uses a large canvas that allows us to move and approach to different parts of the canvas and

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underline the ideas presented there, it also supports the use of text, images and videos and also offers a collection of models to choose from and help new users get used to the interface.

Let me acknowledge a fact that as an educator I invest a lot of my time in creating presentations  and use it in my classroom but most of the time it will be boring and I end up having a dull presentation that will not keep my students interested and in a math class, as a math teacher I cannot afford this. It takes a lot of creativity and in-depth knowledge of using software’s  to create interactive and informative presentation but Prezi offers a great solution for educators with its unique platform of making and delivering presentations. As an alternative of slides and boring animation, Prezi infuse fun and creativity into our presentation.

After reviewing Prezi, I found it  very straightforward presentation and easy to use, even for a first timer like me. We have total control over the details such as font size, adding audio files and images, importing PowerPoint slides, Embed PDF’s, sharing in social media etc. It also comes with a number of templates which we can use and still gives us flexibility to relate it to our topic and make it interesting. Prezi creates splendid visual impact in the classroom with its cinematic experience of the zooming function  and students will feel ecstatic with the topic which we have designed. The presentation can also be easily shared on the internet and can be accessed by our students at home and they can navigate with Prezi and play around with it observing the connections of ideas and visualize concepts.

The strength’s and advantages of Prezi are enormous yet it has some weaknesses too. The first issue which caught my attention is that it lacks auto correction feature  which can be a hassle for users. Prezi is completely web based, even thought it is an advantage but in few circumstances, it can be disadvantage when we have a limited internet access. It is not printer friendly so printing out our presentation may become a challenge to students.

Overall, Prezi is a great tool that most of us would find very useful in the classroom, especially for Math teachers like me. It is a visual storytelling software alternative which will be  great advantage to educators and embeds cinematic experience in classrooms which makes teaching moments memorable. I believe that storytelling in mathematics is an excellent approach for creating a safe-learning environment where a student might openly appreciate, comprehend and enjoy mathematics. I feel that storytelling in mathematics classroom creates an atmosphere of imagination, excitement, and thinking which make math fun and memorable for students and by the help of  Prezi we can create such an atmosphere in our classroom.

Emerging new tools and multimedia can definitely make storytelling easier, especially if the concept we want to share is complicated like mathematics. What do you guys think? will it be another distraction to students?  I fell storytelling is an exceptional tool in mathematics education. How could you see this used in your own subject area?

Until next time, happy blogging!

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# PERCENTAGE MADE EASY BY USING VEDIC MATHEMATICS

IMAGE via Aksharshilp

Hi ECI831,

There are situations in our day to day life when we need to calculate percentage in our mind and these scenarios are quite common. When was the last time we talked to our friends, colleagues or students about percentage? probably in restaurants, parting things with friends or while discussing with students about their grades, paying rent, mortgage these real-life situations are common and what if we can do this quicker and in our mind easily. Vedic mathematics offers a neat solution to these problems.

In my previous blog ,we  learned division method when the divisor is 9 and also Nikhilam formula,today in this post we are going to learn how to find percentage of any number using vedic maths.

Below image is a simple  illustration of the difference between usual method and Vedic method

To find the percentage we are going to apply the sutra or formula i.e. Vertical and cross wise multiplication .which we have learned in multiplication method .
By using above formula i am  going to solve some examples.
1)Lets us find the percentage of  36⸓  of   85
STEPS
i) Write the given values in two rows
6
5
ii)Now we will apply the technic of Vertical and cross wise multiplication , consider the 1st pair where cross multiplication is not possible ,we will  solve it by vertical multiplication.
3      6
↓                                 (3×8=24)
8      5
24
iii)Consider 2 pairs, here cross multiplication is possible so we will apply the same rule.
3              6
.       Χ
8              5                (1548=63)
6                             (CARRIED FORWARD)
(24
iv) Next consider the last pair there is only one pair left , so we will apply vertical multiplication   method.
3              6
↓                       (6×5=30)
8              5
6      3                (CARRIED FORWARD)
(24 3   0
6
5
6        3                (CARRIED FORWARD)
(24)   3   0
vi) Divide the Answer by 100 we will get the final answer
3060÷100=30.60
36⸓  of   85 = 30.60

2)Lets us find the percentage of  52⸓  of   640
STEPS
i) Write the given values in two rows
0       5      2
4      0
ii)Now we will  apply  the technic of Vertical and cross wise multiplication, consider the 1st pair,cross multiplication is not possible , we will solve it by vertical multiplication.
5      2
↓                                 (0×6=0)
4      0
0
iii)Consider 2 pairs,  here cross multiplication is possible so we will apply the same rule.
5         2
.     Χ                                           (0+30=30)
6         4         0
3                                          (CARRIED FORWARD)
0  0
iii)Consider 3 pairs, here cross multiplication is possible,Multiply extreme digits by cross multiplication and middle digit by vertical method.
5         2
Χ                                     (0+12+20=32)
6         4        0
3   3                                      (CARRIED FORWARD)
0   0   2
iv) Next consider the last 2 pairs, cross multiplication is possible so we will apply same rule.
0       5       2
.              Χ                                     (0+8=8)
6       4       0
3     3                           (CARRIED FORWARD)
0     0    2    8
v)  Next consider the last pair there is only one pair left , so we will apply vertical multiplication   method.
0           5          2
.                         ↓                         (0×2=0)
6          4           0
3         3                        (CARRIED FORWARD)
0        0    2   8  0
5        2
6        4        0
3        3                      (CARRIED FORWARD)
0        0     2  8   0
3        3     2  8   0      (ANSWER)

vi) Divide the Answer by 100 we will get the final answer
33280÷100=332.80
52⸓  of   640 = 332.80

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# EDUCATORS IN ONLINE SPACES

Image via Thinkstock

Hello Everyone,

Nowadays, it appears as if everyone is on internet, writing and reading blogs, especially in our teaching community blogs are extremely popular platform where we find information posing new ideas, instruction and motivation.for educators blogging offers the greatest opportunity to share their work and ideas with the world. Just blogging doesn’t suffice, but doing it well is totally a different story, one that involves a lot of time and commitment, yet sometime on several occasions we will be provoked with some thoughtful questions and criticisms about the whole blogging thing. In this blog post I am  sharing one of my thoughts and concerns on sharing and blog post.

The first thing that pinches a relatively young, novice teacher and blogger like me is that critic of no one reads your blog or just cares on what you have to say? Or our work and ideas, plans will not be of benefit to others. Surely many of us might have heard of this one. I guess it is easily the very first criticism against the idea of blogging and sharing resources online and I think that this is the fear which keeps some teachers away from sharing their work online. In Sharing: The Moral Imperative. Dean Shareski talks that we all seek recognition’s for our contributions but the moment we focus on protecting our work we are in some ways antithesis of a teacher, I totally agree with it. No matter even if nobody is paying attention, the course of blogging and sharing our work, thoughts and ideas helps us find and evolve our inner voice and ideas. As Dean Shareski agrees even I feel that it is our work as a teacher to share online and not confine our teaching practices and ideas in a building. We teachers as a prolific member of the society must not only teach our students in a confined space but have to go beyond it and by sharing our work, ideas online we can accomplish this task . After hearing  what Dean Shareski says I agree that if there is no sharing then there is no education. The purpose of a teaching and a teacher is to educate irrespective of who the audience is.

The benefits of sharing work online are bountiful. I believe by sharing our work Motivates learners and readers to do their very best. In his video Dean Shareski
states few awesome examples and as a math teacher the one which attracted me was Dan Meyer’s Math Stories. A relatively young mathematics teacher Dan Meyers from California who created some outstanding math resources which he shared for free, these resources are phenomenal for every math teacher and has helped numerous educators across the globe. Sharing our work online will not only help other educators to  improve their teaching and learning practices but it also helps us to build a strong writing, editing, and revision skills. It also gives other learners the opportunity to acquire experience and helps to build their confidence and self-esteem.

sharing the work of our students gives them good motivation to improve their work. Seeing their work published online can inspire a sense of self-esteem and confidence in students, it help’s them develop their self-esteem because they have the opportunity to be recognized for their hard work and commitment. Students will be encouraged by the fact that their work is online for everyone to see, which can give them the confidence boost they need to give the best of themselves.

Sharing the student’s work online also works for the benefit of other students by offering a valuable experience. If students can see a good example, they are more likely to improve their skills and knowledge base. In addition, students can better understand how to perform a well-qualified task and what mistakes to avoid seeing the work of others and use this knowledge for their future projects.

Finally, it leaves me with a question that how many of us truly believe in the culture of sharing? Most of us are harvesting the fruit of sharing culture and are we giving it back? How do we get teachers to share their work online? Do you agree with the idea of sharing our students work online? And will it help them?

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# “FISHING” FOR CONTENT IN OPEN EDUCATIONAL RESOURCES

Image via SCUS.EDU,

Open Educational resources (OER) are an exceptional way for educators to utilize digital content in classroom. Open Educational resources  are learning and teaching materials which are freely accessible for everyone online. As I discussed in my previous blog post  these material can be remixed, reorganized and revised and can be effectively used to deliver the knowledge to students. These collections of scrutinized, grade-level, content specific resources will be an asset for teachers and elevate our teaching standards and effectively use the digital content in our classrooms. New forms of communications are illuminating new opportunities for educators and students, this opportunity is why open education resources matter.

As a mathematics lecturer one of the big challenge is to keep our students engaged and excited about math. I must say that advances in technology has made our life easier and I must acknowledge that Open Educational resources has made math lessons fun in classroom and made my job as a teacher easier. It provides pre-made activities, options for distinguishing and even aiding with pre- and post- valuations, as well as grading.

In mathematical education , a manipulative is an object that is designed so that a student can perceive a mathematical concept by manipulating it, so its name. The use of manipulative provides a way for students to learn concepts through a practical experience appropriate to development. Manipulative ‘s aid learning math concepts

Image via Clip Art

exciting and engaging for students but for teachers buying these supplies for classroom can be expensive. OER provide us a free solution, by using it we can create manipulative’s ourselves.
Instead of investing a lot of time trying to format everything the resources below provide an excellent platform for teachers to get pre-made math goodies.

Black Line Masters , Provides a range of printable tools, materials and resources which is mainly focused for children’s in their Pre-K to ninth grade. The resources which are available here are extremely helpful for teachers, it provides an ocean of materials which can be easily downloaded and printed as per our needs and requirement.

SEN Teacher
has a load full of printable materials which are focused towards elementary level. It has polygons, clocks, coins, Nets (3D Models), arithmetic makers, number squares, Mazes etc.

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I AmHomeschooling.com   also offers ample of ready to use manipulative’s  and  tools which will help teachers for Pre-K to High school. It provides abundant resources for every student needs for different levels. We can print materials right from algebra tiles to cuisenaire style rods and fraction circles.
These Open educational resource repositories are very easy to access and extremely user friendly, the mathematical contents and material are very well organized and has high printing quality. It aids teachers prepare resourceful course contents for students and make mathematics fun.

ONLINE LESSONS AND VIDEOS
Our classrooms today’s are embedded with remarkable technology and it has helped us deliver course content efficiently and has enhanced classroom learning. There are many resources especially for mathematics teachers by which we can eradicate the long standing math fear phobia from students. The web resources below have some great resources for math teachers and most of them are free, these sites offer best lesson plans videos and cover entire course.

Khan Academy  Is an excellent pit stop for teacher’s and it is completely free. It contains exceptional collections of videos presenting difficult mathematics topics in a way which can be easily understood. There are numerous drill questions which can be used in conjunction with the videos. The resources are very well organized and it also

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easy to play around with the website. The grade level mathematics resources in Khan academy address all most all core mathematics standards and helps focused learning and the videos aid vibrant instruction.

Discovery Education  Provides a series of new and remarkable pre-made lesson plans for math teachers. It is regularly updated with new materials which covers levels from pre-K to K-12.

Where as the National Council of Teachers of Mathematics   Showcases that the math lessons can be fun and more than just a “drill and kill!. It has the outstanding accumulation of lesson plans and activities. They are free and accelerated towards different levels, it works to aid us by collective access to quality standards-based resources for learning and teaching mathematics. It also provides collaborative tools for students and instructional support for educators.

Learn Zillion  is an excellent stop for math teachers if they are looking to improve classroom discussions with the help of videos and lesson plans complex mathematical subjects can be easy to understand in engaging way. The website has to offer math educational contents in fun and exciting way which is suitable for every level of students. Teachers can use these videos to start a lesson and introduce new ideas which is very important to engage students in math class.

With Mathalicious, Math can be fun and exciting, Using this resource Math teacher can spice up the middle and high school students by making math lesson a whole lot easier. These remarkable assortments of lessons will not only get the students thinking and problem solving, but they will have a lot fun while solving it ! The lessons are classically geared towards using math in practical applications, which really helps students see the significance of learning and understating math

Even though we were supposed to choose one major Open education resource repositories and evaluate its resource but mathematics has such a wide scope and enormous availability resources made me explore quite a few repositories. These resources are very well organized and is extremely user friendly and a great asset for educators to elevate their teaching practices.

The efficient use of these open educational resources embeds a new style to learning which can be implemented to every level of the education sector., letting learning and teaching materials to be produced and then used and reused, remixed by everyone. It will be a boon to novice educators to guide them into perfection. The mathematics teaching fraternity has been extremely enriched by these resources, it has simplified the instructional process and made math class fun and engaging for students. OER’S has increased flexibility in how we use the course content and emphasizes learning communities, learner engagement. The advantages of OER’s are bountiful yet it has some challenges to offer. If not properly trained and educated on using OER’s and finding the right course content teachers tend to invest a lot of time planning their lessons and selecting alternate readings and other materials for class.

With that being said finally it leaves me with questions on how many of us are efficient utilizing OER’s? and contributing to updated the current content? how do we deliver better learning experiences to more students? And How do we get the most up-to-date content when we want it? How to overcome the accessibility issues when file formats that are not always compatible with our computer’s software or version? What do you guy’s think?

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# DIVISION MADE EASIER BY VEDIC MATHEMATICS

Image via Nikhi

Hello Everyone,
Vedic math is an awesome technique for solving arithmetic problems in quicker and simpler way. The cool thing about this technic is that every math checks out. Whether you want to painlessly split a check, to amaze your friends, or to learn a different way to quickly divide numbers, this easy method can be learned within no time.
In this week blog post let us learn how to find division  of 3 or more digits using Vedic mathematics.

Below image is a simple  illustration of the difference between usual method and Vedic method

Finding the division of 3 or more digit number when the divisor is 9.
1)Let us solve example on  3 digit number
Let us consider the number 102÷9

102÷9
STEP 1:
Splits the digit 102 into 2 parts as a divisor has only one digit as 2 goes in                                remainder part and rest all digits will goes in Quotient part i.e.
Q       R
1 0 ǀ  2

STEP 2:
To find the 1st digit of a quotient put 1 as it is i.e.
Q       R
1 0 ǀ  2

1
STEP 3:
i)To find the 2nd digit of a quotient ,add quotient and divident  digit i.e.                                         (1+0=1)
ii)To find the remainder ,add remainder and quotient i.e.(2+1=3) i.e.
Q        R
1 0 ǀ  2

1 1 ǀ  3
Quotient     = 11
Remainder =   3

2) Let us solve examples of 4 digit number
Let us consider the 1232÷9
1232÷9
STEP 1:
Split the digit 1232 into 2 parts as a divisor has only one digit as 2 goes in                                 remainder part and rest all digits will goes in Quotient part i.e.
Q          R
123 ǀ 2
STEP 2:

Q           R
123 ǀ  2

1
STEP 3:

i)To find the 2nd digit of a quotient add quotient and the divident i.e.(1+2=3) ,                      ii) To  find the  3rd digit of a quotient add 2nd digit of quotient with the divident                        digit i.e. (3+3=6)
iii)To find the remainder add 3rd digit of quotient with  remainder i.e. (6+2=8) i.e.
Q            R
1 2 3  ǀ     2

1 3  6  ǀ    8
Quotient = 136
Remainder=8

Division method when divisor is closer and smaller than power of 10.
By understanding the  Concept of Nikhilam formula i.e.all from 9 & last from 10,(i.e. Subtract all digits from 9 & last digit from 10)

Below image is a simple  illustration of the difference between usual method and Vedic method

Lets us consider the value
234÷87
STEP 1:
Splits the digit 234 into 2 parts as a divisor has two  digit as 34 goes in                                 remainder part and rest all digits will goes in Quotient part i.e.
Q          R
2 ǀ 34
STEP 2:

Apply Nikhilam formula on 87 and we will get the compliment 13 (i.e. 9-8=1, 10-7=3)
Q           R
2 ǀ  34
13
2
STEP 3:

i)Multiply 2 with individual digit of 13  i.e.
Q      R
2   ǀ  3 4
13               ↓   ǀ  2 6
2  ǀ  6 0
Quotient    =   2
Remainder= 60

Lets us consider the value
113401÷997
STEP 1:
Splits the digit 113401 into 2 parts as a divisor has three digit so remainder will also have 3 digit ,as 401 goes in remainder part and rest all digits will goes in Quotient part i.e.
Q          R
113 ǀ 401
STEP 2:

Apply Nikhilam formula on 997 and we will get the compliment 003
(i.e. 9-9=0, 9-9=0,10-7=3)
Q           R
113 ǀ  401
003
1
STEP 3:

i)Multiply 1 with individual digit of 003  i.e.
Q      R
1 1 3  ǀ  401
003         ↓ 0 0 ǀ  3
1 1     ǀ
ii)Multiply 2nd quotient with individual digit of 003 ,and add the third column,to get 3rd quotient
Q      R
1 1 3  ǀ  401
003         ↓ 0 0  ǀ   3
0  ǀ  0 3
1 1 3  ǀ
iii)
Multiply 3rd quotient with individual digit of 003 ,and add the 4th,5th and 6th column,to get the  remainder.
(Note: Have to repeat the function till we get the number in last column)
Q      R
1 1 3  ǀ  4  0   1
003         ↓ 0 0  ǀ  3
ǀ  0  3
ǀ  0  0    9
1 1 3 ǀ   7  3  (10)  (1 is carried forward)
1 1 3 ǀ   7  4   0
Quotient    = 113
Remainder= 740

A quick update on answering questions from my previous blog post.
One of my fellow classmate Kara Tylor had a question on one of my blog post on how to find cube of two numbers using vedic technic. She asked me how to find cube of more than 2 digit using Vedic math, below I have solved few examples based on the method and my video will further aid the understanding the concept.

Finding cube of 3 digit number using vedic maths.
Lets us consider the number
(153)3

STEPS:
i)Split the given digits into two part
15     3
ii)consider the 1st part i.e. 15 from left to right go on reducing power i.e
153    152    15
iii) Next consider the 2nd part i.e. 3 from right to left go on reducing the power i.e.
153    152.3   15.32   33
iv) Find the values and write in 2nd Row
153          152.3         15.32           33
3375        675           135             27
v) Multiply middle terms by 3 i.e.
153         152.        15.32         33
3375     675           135             27
×3            ×3
3375       2025       405            27
vi)To get the final answer   we are going to keep unit digit as it is rest other digit are carried forward
153           152.3         15.32          33
3375        675           135             27
×3            ×3
3375       2025          405             27
(206)        (40)             (2)                              (CARRIED FORWARD)
3581         5                   7                 7
(153)3
= 35,815,77

I hope you like this post and found this technic interesting .

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# FINDING CUBE ROOT USING VEDIC METHOD

Hi Eci831,
In My Previous blog , we learned how to find cube of any number using Vedic maths. In this Post we are going to study how to find cube root . of any perfect cube easily by using Vedic maths. This is an remarkable trick and is always quite easy and appreciated by many This awe inspiring technique helps us to easily find cube root of 5 or 6  digits number mentally

Before going ahead on this method to find cube root, please make a note and memorize the below table.

To calculate cube root of any Perfect Cube easily, we need to Memorize the cubes, cube of unit digit and its relations from 1 to 10.

 Number Cube Cube of unit digit Relations 13 1 1 1 ⇒1 23 8 8 8⇒2 33 27 7 7⇒3 43 64 4 4⇒4 53 125 5 5⇒5 63 216 6 6⇒6 73 343 3 3⇒7 83 512 2 2⇒8 93 729 9 9⇒9 103 1000 0 0⇒0

Mathematics is very interesting and easy to learn when we use tricks to make it simple,now let’s see how easily we can solve cube root of perfects cube by using  Vedic method and also by viewing the table above.

Example 1: Find cube root of 19683
Check the last digit number & make a groups of three, three digits
from right side i.e. 19683  Can be written as
19   ,    683
STEP 1

1. Take the last group which is 683, in this case the last digit of 683 is 3
2. Now compare the last digit number i.e. 3 in the above table in the  relation we can see that if last digit of perfect cube is 3 then last digit of cube root is 737
3. So we replace 3 by 7 at the right most digit of the cube root
19   ,    683

STEP  2
1. Now check the remaining digit which is left i.e. 19
2. We have to check the cube which is less then or equal to the digit 19
3.  In this case the number 23=8 which is less than 19
4. We will replace 19 by 2

i.e.   19  ,   683
2          7
Hence the answer for cube root of 19683 is 27

Example 2: Find cube root of 551368
Check the last digit number & make a groups of three, three digits from                                      right side i.e. 19683  Can be written as
551 ,     368

STEP 1

1. Take the last group which is 368, in this case the last digit of 368 is 8
2. Now compare the last digit number i.e. 8 in the above table in the  relation we can see that if last digit of perfect cube is 8 then last digit of cube root is 2 i.e 82
3. So we replace 8 by 2 at the right most digit of the cube root
551   ,    368
2

STEP 2
1. Now check the remaining digit which is left i.e. 551
2. We have to check the cube which is less then or equal to the digit 551
3. In this case the number 83=512 which is less than 551
4. We will replace 551 by 8
i.e.  551  ,   368
8           2
Hence the answer for Cube root of 551,368 =82

Example 3: Find cube root of 148,877
Check the last digit number & make a groups of three, three digits from right                         side i.e. 148,877 Can be written as
148     ,     877
STEP 1

1. Take the last group which is 877, in this case the last digit of 877 is 7
2. Now compare the last digit number i.e. 7 in the above table in the  relation we can see that if last digit of perfect cube is 7 then last digit of cube root is 3 i.e 73
3. So we replace 7 by 3 at the right most digit of the cube root
148 ,    877
3

STEP 2

1. Now check the remaining digit which is left i.e. 148
2. We have to check the cube which is less then or equal to the digit 148
3. In this case the number 53=125 which is less than 148
4. We will replace 148 by 5
i.e       148   ,   877
5      3

Hence the answer for Cube root of 19683 =53
In the below video i have solved few more problems based on the vedic technics

I hope you liked the video and  found the Technic interesting and it simplified the process of finding cube root in a faster way, I appreciate your time for reading my blog post .

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# OPEN EDUCATION AND REMIXED TEACHING

Image via Wikiversity

As the famous Quote states “Education is not the answer to the question. Education is the means to the answer to all questions.” Therefore, I feel that a well-educated individual has one of the life’s most valuable possessions that too as a result of one’s own hard work. Education is the only companion who will be by your side at all times and will never let you down. That is why; a concrete foundation is indispensable in shaping one’s future.

An educated and qualified person has always something to fall back upon. I strongly believe that one cannot perform better in any other field beside the field that deeply fascinates and captures one’s interest. Determination and persistent hard work follow one’s decision in undertaking a challenging task, there was a time where one was supposed to get this education/knowledge by enrolling to the educational institutions by meeting their academic requirement, it was reachable for only privileged class, but today the whole arena has changed ,the concept of open education has revolutionized the field of education it has replaced the traditional class room and made education free and accessible to all.

Web-based educative platforms have allowed us to remotely replace or facilitate many interactive processes such as purchases, booking appointments, and banking activities, buying flight tickets etc. Getting an education online is the next frontier for web-based convenience, and there are a number of reasons why this is an effective and attractive platform. Some of them are,

Accessibility: Open Education allows students to take courses at their own convenience, on their own time and place. Open learning systems are very effective and importantly Accessible when compared to its traditional counterparts. The rich resources in online communities help students to overcome hurdles. Open education has attracted many students who were very unlikely to gain higher education.

Affordability: The word free attracts everyone and in my view the concept of free education is best served by open education. There are hundreds of lecture hours which are available for free in reputed universities websites such as MIT, these are free and open for anyone and there are many online learning platform which are also cheap and affordable for students.

Flexibility: Open education offers students flexibility to choose what to learn and when to learn and how much to learn, in return putting them in control. It provides ocean of resources for students in-comparison with the traditional classrooms.

This week’s concept and videos were very interesting and covered the concept of open education and knowledge sharing. As a mathematics major I had a strong mathematical knowledge but teaching was something very new to me in my initial days of my teaching career. As Ashley mentions in her blog post We are not responsible for creating the material that we provide to students, I think I have exploited most of the resources online which helped me to effectively deliver knowledge to students and thanks to all those who believed in the concept of sharing culture , it made my initial days a lot easier. The resources like Karnataka education ,Open Educational Resources for Mathematics has helped me a lot in developing material to facilitate my course. It has improved my teaching material and delivery.

And Finally, as an educator I believe that the ultimate goal of teaching is to promote learning and education takes place in different circumstances and contexts. I totally agree with message of Kirby Ferguson  in his video series that we are all leaders of remix

Kirby Ferguson’s  popular videos illustrates that everything is remix and everything we have created takes motivation from something which is already in existence. The video showcases number of evidences that supports his claims. I was convinced and it kept me thinking that most of things I do in my teaching practices might be exactly the same.

I believe that in order to lay a strong foundation of knowledge we need to copy and when we relate this to our profession we take ideas, material, technic’s from resources and effectively deliver knowledge to students and I think it has made our life as a teacher much easier and effective. Just Imagine the amount of time required if we had to create everything original ..??

After reading the blog post of Joe Marine his question kept me thinking and I wanted to share the same with everyone , Is there a need to reform intellectual property laws to better represent how ideas actually spread? Is it possible to have original work, or is everything truly a remix?

What do you guys think?

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# FINDING CUBE OF ANY NUMBER USING VEDIC MATH

Hello Everyone,
It was indeed fun and interesting journey in EC&I 831, I have learned a lot by reading my fellow classmates blog post, every week was so interesting and now we are almost halfway through the course. I have always been fascinated by mathematical studies and having a flair for the subject, there was never any doubt that I would choose  concepts in mathematics as my major learning project. I would call this as journey of learning, knowledge sharing. My project was focused on learning how to make mathematics simple, easy and fun for everyone and the tricks of Vedic mathematics

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would be perfect to answer to it. In my first blog post we learned what is Vedic mathematics  and its sutra’s(Formula’s), the idea behind this was to introduce the concept of Vedic math to my classmates. In my second blog post I focused on learning how to implement this technic in various mathematical concepts, applying Vedic math in solving multiplication problems of complex numbers made multiplication so easy, we could solve 4-5 digit problems in less than a minute and that too without using calculator. My next post focused on learning how to find square of any number using Vedic trick, followed by another week blog post where we learned to find square root of any number using this trick.

I feel happy regarding my progress after reading the comments about my journey by my fellow classmates. Initially it was not so easy as I thought it would be, I had to understand the whole technic first and solve problems related to it. After searching a lot over internet I understood the best way to showcase this concept and learn with my classmates is by making a video of my own where I solve problems, which in-return help everyone understands the Vedic tricks and it applications. Going forward in the second half of our course we learn few more applications of Vedic technic in mathematics and by the end of our journey in EC&I 831 we will make Mathematics simpler and joyful using the tricks of Vedic Mathematics.

Now that we have learned how to find square root of any numbers using Vedic math , we will go ahead and use this technic to find cube of numbers. Finding cube of the number plays a vital role in Mathematics, so it is useful to find quick ways to find the cube of the number. We will learn the Vedic trick which will help us finding the cube of numbers in quick way. To find the cube of numbers we are going to learn four types Vedic Tricks, In the first we are going to find the cube of a number which starts with 1 i.e. 11,12,13……. In second type we are going to find the cube of a number which ends with 1 i.e.21,31,41……In the third type we are going to find the cube of a number having same digit i.e.22,33,44……and Finally in the fourth type we are going to find the cube of a number having different digit.

Type 1: Numbers starting with 1
(16)3
1)   We consider 1 as 1st term and 6 as 2nd term, we will write the given term as it is  i.e.                               6
2)  Square the 2nd term i.e. 62=36 and also cube the 2nd term i.e.=63=216 and write the         values in 1st row  i.e.
1     6      36       216
3)  In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
1      6      36       216
12     72                       (Doubled the value)
4)  Add them vertically in column. carry forward the 10th places digit to next column
1      6       36       216
+         12       72
3   12       21                       (Carried forward)
(16)3=4,096

Type 2: Numbers Ending with 1
(61)3
1)  We consider 6 as 1st term and 1 as 2nd term, we will write the given term as it is but           in reverse order i.e.        6     1
2)  Square the 1st term i.e. 62=36 and also cube the 1st term i.e.=63=216 and write the
values in 1st row i.e.
216    36       6     1
3)  In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
216     36       6      1
72     12                                (Doubled the value)
4)  Add them vertically in column. carry forward the 10th places digit to next column
216      36            1
+                72     12
10        1                                 (Carried forward)
(61)3=226,981

Type 3: Numbers having same digit
(55)3
1)  We consider 5 as 1st term and 5 as 2nd term, here both the digit is same so we take             any one digit .Cube of 5 =125 and write 4 times
i.e.              125   125   125   125
2)  In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
125    125   125   125
250    250                 (Doubled the value)
4)  Add them vertically in column. carry forward the 10th places digit to next column
125    125    125    125
+               250    250
41        38      12                (Carried forward)
(55)3=166,375

Type 4: Numbers having different digit
(32)3
1)  We consider 3 as 1st term and 2 as 2nd term, Cube the 1st term and 2nd term
i.e.       (3)3 =27     and         (2)3=8
27                                  8
2)  Square the 1st   term i.e. 32=9 then multiply by 2nd term i.e.9×2=18
27         18                   8
3)  Square the 2nd term i.e. 22 =4 then multiply by 1st term i.e.4×3=12
27          18      12       8
4)  In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
27          18       12      8
36       24               (Doubled the value)
5)  Add them vertically in column. carry forward the 10th places digit to next column
27         18       12      8
+                     36       24
5           3                             (Carried forward)