Section 46 number 8. Use number 6 and 7 to help and 46.11 6. By referring to Example 46.11, actually express the gcd 23 in the form 2(22,471) + µ(3,266) for 2, µ e Z.[Hint: From the next to the last line of the computation in Example 46.11, 23 = (138)3 – 391. From the linebefore that, 138 = 3,266 – (391)8, so substituting, you get 23 = [3,266 – (391)8]3 – 391, and so on. That is,work your way back up to actually find values for and u.]7. Find a gcd of 49,349 and 15,555 in Z.8. Following the idea of Exercise 6 and referring to Exercise 7, express the positive gcd of 49,349 and 15,555 inZ in the form 2(49,349) + µ(15,555) for 2, µ e Z. factórization of an integer into46.11 Example Note that the division algorithm Condition 1 in the definition of a Euclidean norm saysnothing about r being "positive." In computing a gcd in Z by the Euclidean algorithmfor | |, as in Example 46.10, it is surely to our interest to make |r¡| as small as possiblein each division. Thus, repeating Example 46.10, it would be more efficient to writea = 22,471b = 3,266a to22,471 = (3,266)7 – 3913,266 = (391)8 + 138391 = (138)3 – 23138 = (23)6+0ri = -391r2 = 138ai CS evar3 = -23r4 =0We can change the sign of r¡ from negative to positive when we wish since the divisorsof r; and -r; are the same.

Name: Section 46 number 8. Use number 6 and 7 to help and 46.11 6. By referr
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Description: Section 46 number 8. Use number 6 and 7 to help and 46.11 6. By referring to Example 46.11, actually express the gcd 23 in the form 2(22,471) + µ(3,266) for 2, µ e Z.[Hint: From the next to the last line of the computation in Example 46.11, 23 = (138