Does vedic math trick of squaring numbers have proof of correctness?

Question

Vedic math tricks are usually verifiable, but an empirical proof may be bogus. Vedic squaring method (using duplex), for instance, Does it have a proof of correctness? Trick goes as follows:

Definition of Duplex: For a single digit number, the duplex is simply the square of the number. Thus the duplex of 2 is 4, the duplex of 6 is 36 and so on
For a 2-digit number, the duplex is simply twice the product of the 2 digits of the number. Thus, the duplex of 16 is 2x1x6 = 12, the duplex of 90 is 2x9x0 = 0, the duplex of 43 is 2x4x3 = 24, and so on
For n-digit numbers, the duplex is calculated as the sum of several individual duplexes. Pair up the first digit with the nth digit of the number and find the duplex of the resulting 2-digit number. Similarly pair up the second digit with the (n-1)th digit and find the duplex of the resulting 2-digit number. Continue this process until no more 2-digit pairs can be formed. If a middle digit (that could not be paired with anything else) exists at the end of the process, find its duplex also individually. Then add up all the duplexes found. The resulting sum of the duplex of the n-digit number

Suppose you have a four digits number, Find duplex of 1st digit,
Find duplex of 1st 2 digits,
Find duplex of 1st 3 digits,
Find duplex of all digits,
Find duplex of last 3 digits,
Find duplex of last 2 digits,
Find duplex of last digit,   
Jot down everything from top to bottom, if you get a number which is not a one digit, carry forwards.
Refer to link for more information and examples: http://mathlearners.com/vedic-mathematics/squares/dvanda-yoga/
Example: Function D() is duplex

Answer 1

Yuvadunum states that:

If $n = 10^k \pm m$ then $n^2 = (n\pm m)10^k + m^2$

Which is just algebra: $n^2 = (10^k \pm m)^2 = 10^{2k} \pm 2*10^km + m^2$ and $(n\pm m)10^k + m^2 = (10^k \pm 2m)10^k + m^2 = 10^{2k} \pm 2*10^km + m^2$.

Ekadhikena Purvena states that:

If $n = 10*a + 5$ then $n^2 = a(a+1)*100 + 25$

Which is just algebra:

$n^ = (10a + 5) = 100a^2 + 2*5*10a + 25 =100a^2 +100a + 25=100a(a+1) + 25$.

Finally:

Says that if $n = \sum_{k=0}^n b_k*10^k$ then

$n^2 = (a_0 + 10a_1 + 100a_2 + .... +10^ka_k)^2 =$

$(a_0)^2 + 10(a_0a_1 + a_1a_0) + 100(a_0a_2 + a_1a_1 + a_2a_0) + .......$

which followes directly by distribution.

Answer 2

I'll see.

For the first, it says $(10+a)^2 =(10(10+2a))+a^2 =100+20a+a^2 $ and that is simple algebra.

See what you can work out for the others.

Answer 3

The techniques for squaring numbers in Vedic Mathematics are based on mathematical principles and can be proven mathematically. These techniques use algebraic identities and patterns to simplify the squaring process. However, they are not universally accepted or used in mainstream mathematics education. Some people find these Vedic Math tricks helpful for mental calculations and as a way to approach squaring numbers, while others prefer traditional methods. Whether or not you find Vedic Math squaring techniques useful may depend on your personal preference and comfort with the method, but they are indeed based on mathematical principles and can be proven to work.Refer to the Vedic Maths information and knowlege books.