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Inder Kumar Rana
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Set Theory-difficulties

Started Jul 22, 2014

Spoken tutorials on math topics
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Started this discussion. Last reply by ankit rawat Mar 2, 2014.

How romans multiplied

Started Jun 8, 2010

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Sep 25, 2016
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making math meaningful

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Profile Information

Profession
I am a faculty member at IIT Bombay
What interests you most in Math Education
Almost all aspects of Math education. More specifically: Teaching of Calculus, Role of Technology in Math Education, Professional Development of Teachers.
More details about you
I am academic advisor to www.math4all.in, do visit the site.

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Thanks

Thanks to our member Mr. M. S. Sloanki Director Acad&Training, Mahaishi Vidya Mandir Schools, Bhopal for contributing Rs 1000.00 towards iMEN fund.

Posted on August 18, 2010 at 4:35pm

Thanks

This is to thank Dr. Vivek Sinha, NIT Suratkal for contributing Rs. 500.00 towards iMEN

Posted on August 13, 2010 at 4:25pm

Comment Wall (28 comments)

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At 10:43am on March 1, 2016, dasari naga vijay krishna said…

sorry sir,

At 6:42pm on January 26, 2015, Devarakonda said…

When They Resisted The Booming
And Shelling Of The Colonial Guns,
Our Founder Fathers Wanted Nothing
But Sovereignty For This Nation,
Let Us Always Defend This Gift
From Our Predecessors.

Happy Republic Day.

At 12:15pm on January 11, 2013, trantancuong said…

Hi Monsieur Jean Yves Rolli.

I am so happy can talking to you continue.

Happy new year.

Pierre De Fermat 's last Theorem.
The conditions:
x,y,z,n are the integers and >0. n>2.
Proof:
z^n=/x^n+y^n.

We have;
z^3=[z(z+1)/2]^2-[(z-1)z/2]^2
Example;
5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125
And
z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2
Example;
5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189
And
z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2
Example
5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216
And
z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2
Example
5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224
General:
z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2

We have;
z^3=z^3+(z-m-1)^3 - (z-m-1)^3.
Because:
z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3]
So
z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3.
Similar:
z^3=z^3+(z-m-2)^3 - (z-m-2)^3.
So
z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3.
....
....
Suppose:
z^n=x^n+y^n
So
z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3.
So
z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3}
Similar:
z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-

At 12:14pm on January 11, 2013, trantancuong said…

Hi Monsieur Jean Yves Rolli. I am so happy can talking to you continue. Happy new year. Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.

At 8:24pm on July 8, 2012, MT Ansari said…

Hi Sir 

      We are working here with elementary classes and developing content for children. Could you suggest us what approach we have to adopt for content in math. 

        

At 9:09pm on November 11, 2010, Devarakonda said…
Advance Eid Mubarak.........!
At 9:08pm on November 11, 2010, Devarakonda said…

At 12:56am on July 10, 2010, N.Balasubramanian said…
It is disturbing to learn that we may have to close. I wish we are able to keep it going at some cost - But what cost, can be discussed if a concrete proposal is made.
At 4:09pm on May 13, 2010, G.S.Shravankumar said…
Dear sir
if i have any queries can i mail you ?
At 6:00pm on April 2, 2010, Manidipa Pal said…
thank you sir for accepting my friend request ...
 
 
 

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