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# सूत्र ३. ऊर्ध्वतिर्यग्भ्यां

## (Sūtra 3. ūrdhva tiryagbhyāṃ) - Vertically and crosswise.

The Sūtra: ūrdhva tiryagbhyāṃ (Vertically and crosswise) is used for multiplication (x × y) of numbers. It is considered as the easiest multiplication method in Vedic Mathematics amongst practitioners, and thus, is most popular.

As an illustration, let us use this Sūtra for:
14 × 12
 Steps 14 × 12 1. Arrange the numbers vertically. For non-equidigit numbers, prefix as many Zeroes in front. In this case, × 14 × 12 ----- 2. Consider sets of vertical columns, preferably from left to right, in ascending order, and then in descending order. In this case, the sets are 1   14   4 1 , 12 , 2 Sets are obtained by including one vertical column at a time from left to right (ascending), and then again excluding one vertical column at a time from right to left (descending) 3. Find the sum of the cross-products - of the digits equidistant from the edges, to represent each set. In this case, (1 × 1), (1 × 2) + (4 × 1), (4 × 2) 4. Re-format the number in Base10 (Normalized) form, if required. In this case, [1,6,8] = 168, which is the answer!
Most discussions on the Sūtra: ūrdhva tiryagbhyāṃ, uses diagrams to explain the crosswise products of horizontal columns - as provided below for 2-digit, 3-digit and 4-digit multiplications: However, while finding cross-products, diagrams of steps do not do justice to the underlying concept of taking the vertical rows, and their sums of cross-wise multiplications. The sum of cross-products of the digits, is simplified in the following rule:
 For n-digit set, the Cross-Product is: 1. For even-number of digits: Sum of the cross-wise product of the digits, that are equidistant from the ends. 2. For odd-number of digits: Sum of the cross-wise product of the digits, that are equidistant from the ends - summed with central-digits multiplied with each other.  Then, for a single-digit set, m1 m2Cross-Product is is m1 × m2  For a double-digit set, m1n1 m2n2Cross-Product is (m1 × n2)+(n1 × m2)  For a 3-digit set, m1n1o1 m2n2o2Cross-Product is (m1 × o2)+(o1 × m2)+(n1 × n2)  For a 4-digit set, m1n1o1p1 m2n2o2p2Cross-Product is (m1 × p2)+(p1 × m2)+(n1 × o2)+(o1 × n2)  For a 5-digit set, m1n1o1p1q1 m2n2o2p2q2Cross-Product is (m1 × q2)+(q1 × m2)+(n1 × p2)+(p1 × n2)+(o1 × o2)  And so on.
So, for a practitioner of Vedic Mathematics, for something like:
873 × 234
 × 873 × 234 ------  Clearly, the sets are: 8   87   873   73   3 2 , 23 , 234 , 34 , 4  Which means, [8 × 2, (8 × 3)+(7 × 2), (8 × 4)+(3 × 2)+(7 × 3), (7 × 4)+(3 × 3), (3 × 4)] = [16, 24+14, 32+21+6, 28+9, 12] = [16, 38, 59, 37, 12] By śūddhikaran, we get: = [20,4,2,8,2] = 204282  Thus, 873 × 234 = 204,282  Note: For starters, the śūddhikaran in the final stages may get confusing. This has been discussed here »
Most practitioners, identify the sets, and sum the cross-products from right to left - including the śūddhikaran with the carry-overs as part of the calculations. Not that we advice it, but practitioners of Vedic Mathematics find the steps so simple, with some practice, that they tend to execute all the steps mentally - within a few seconds. Taking it ahead, for something like:
7196 × 4671
 × 7196 × 4671 --------  Clearly, the sets are: 7   71   719   7196   196   96   6 4 , 46 , 467 , 4671 , 671 , 71 , 1  Let us take a professional approach, from right to left. Considering the right-most set, gives 6 × 1 = 6, as the last digit of the answer.  Taking it ahead, for the set: 96 71, we get (9 × 1)+(6 × 7) = 51 So, our answer becomes 516  Again, for the set: 196 671, we get (1 × 1) + (6 × 6) + (9 × 7) = 1 + 36 + 63 = 100 Considering the carry-over 5, this becomes 100 + 5 = 105 So, our answer becomes 10516  For the next set: 7196 4671, we get (7 × 1)+(6 × 4)+(1 × 7)+(9 × 6) = 7 + 24 + 7 + 54 = 92 Considering the carry-over 10, this becomes 91 + 10 = 102 So, our answer becomes 102516  Now, for the set: 719 467, we get (7 × 7)+(9 × 4)+(1 × 6) = 49 + 36 + 6 = 71 Considering the carry-over 10, this becomes 71 + 10 = 81 So, our answer becomes 812516  For the next set: 71 46, we get (7 × 6)+(1 × 4) = 42 + 4 = 48 Considering the carry-over 8, this becomes 48 + 8 = 56 So, our answer becomes 5612516  Finally, for the set: 7 4, we get 7 × 4 = 28 Considering the carry-over 5, this becomes 28 + 5 = 33 Which, gives us: 33612516  Thus, 7196 × 4671 = 33,612,516
The above examples should clarify the underlying technique. One can take any (n-digit) number and multiply it with another (m-digit) number by iterating the steps above to obtain the desired calculated value.

Note that, the number of sets also provide an indicator of the number of digits of the answer.

But, why does it work? For this Sūtra (ūrdhva tiryagbhyāṃ), let us consider the following:
 Assuming 2 two-digit numbers, mn and op, Let N1 = 10m + nAnd, N2 = 10o + p N1 × N2 = (10m + n)(10o + p) = 100om + 10mp + 10on + np = 100om + 10(mp + on) + np This is exactly what this Sūtra makes us do.  Similarly, for 2 three-digit numbers, mno and pqr, Let N1 = 100m + 10n + oAnd, N2 = 100p + 10q + r N1 × N2 = (100m + 10n + o)(100p + 10q + r) = 10000mp + 1000mq + 100mr + 1000np + 100nq + 10nr + 100op + 10oq + or = 10000mp + 1000(mq + np) + 100(mr + nq + op) + 10 (nr + oq) + or This is exactly what this Sūtra makes us do.  Note that, as the number of digits increase - so does the number of steps that this Sūtra makes us perform - in the same pattern as above.

Lastly, please remember that, as in any other form of mathematics, the mastery of Vedic Mathematics require practice and the judgement of applying the optimal method for a given scenario - a guideline of which, is presented in Applications »

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