l>M3000 Homework #4a

## 2175forals.com 3000 Homework answers #5

From Smith, Eggen, & St. Andre, A transition to progressed 2175forals.comematics, 5th Ed.Answers to * troubles are given in the back of the book and also will not be reproduced here.

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(pg. 42 : 2, 6, 7, 10, 12 )2. (a) intend that AB is no an invertible matrix. ... Thus, one of two people A or B is not an invertible matrix. Therefore, if A and B are invertible matrices, then AB is an invertible matrix.(b) suppose that either A or B is no an invertible matrix. ... Thus, AB is not an invertible matrix. Therefore, if AB is an invertible matrix, then A and B room invertible matrices.(c) expect that A and also B are invertible matrices and that AB is no an invertible matrix. ... Thus, some Q is true. ..... Therefore, ~Q is true. Contradiction. So, if A and B space invertible matrices, then AB is one invertible matrix.

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(d) mean that AB is an invertible matrix and that one of two people A or B is no an invertible matrix. ... Thus, part Q is true. ..... Therefore, ~Q is true. Contradiction. So, if AB is one invertible matrix, then A and B are invertible matrices.(e) First, intend that A and B room invertible matrices. ... Thus, AB is one invertible matrix. Therefore, if A and B space invertible matrices, climate AB is one invertible matrix. Next, intend that AB is an invertible matrix. ... Thus, A and B room invertible matrices. Therefore, if AB is an invertible matrix, climate A and B room invertible matrices. Combining this statements we have that A and also B room invertible matrices if and also only if AB is one invertible matrix.6. (a) Assume the a divides b and also that a > b. Since a divides b, there exists an essence k so the b = ak. Due to the fact that a is hopeful it is not 0 for this reason we deserve to divide by a to acquire that k = b/a. Now, because a > b, we have the right to divide by a to obtain 1 > b/a = k. This is a contradiction since no integer have the right to be less than 1. Hence, our assumption is false and also so we have shown that if a divides b then a b.(c) Assume that a is odd and also that a+1 is odd. Because a is odd, there exists an creature k so that a = 2k+1. Then a+1 = (2k+1) + 1 = 2k+2 = 2(k+1). So, a+1 is even, a contradiction. Thus, our assumption is false and if a is odd then a+1 have to be even.(d) Assume that a - b is odd and also that a+b is even. Since a-b is odd, there is an essence k so the a-b = 2k+1. Now, a+b = (a-b) + 2b = 2k+1 + 2b = 2(k+b) + 1. Due to the fact that k+b is an integer, a+b is odd. This is a contradiction, so if a-b is odd then a+b is odd.7. (a) First, expect that ac divides bc. Climate there exist an essence k so that bc = (ac)k. Due to the fact that c is not 0, we deserve to divide by that to obtain b = ak. Therefore, a divides b. Next, expect that a divides b. Climate there exists an creature k so the b = ak. Main point both sides by the optimistic number c to gain bc = (ak)c = (ac)k. Thus, ac divides bc. Putting these 2 statements together we have that ac divides bc if and only if a divides b.(b) First, suppose that a+1 divides b and b divides b+3. Because a+1 divides b there is an essence k so that b = (a+1)k, and since b divides b+3 there is an integer m so that b+3 = bm. Individually b in this critical equation offers 3 = b(m-1). This method that b is a optimistic integer i m sorry divides 3, i.e., b is one of two people 1 or 3. If b = 1 then us would obtain a contradiction indigenous the very first statement which would say that a+1 (an integer bigger than 1) divides 1. So us must have actually b = 3. Currently the first statement claims that a+1 have to divide 3, therefore a+1 is one of two people 1 or 3. If a+1 = 1, climate a = 0 contradicting the assumption that a is positive. Hence we must have actually a+1 = 3, i.e., a = 2. Next, assume that a = 2 and b = 3. Because a+1 = 3 and also 3 divides 3, we have actually that a+1 divides b. Also, b+3 = 6 and also 3 divides 6, so b divides b+3. Putting these two statements together we have that a+1 divides b and b divides b+3 if and only if a = 2 and b = 3.(c) If a is odd then a+1 is even (Exercise 6c). Next, mean that a+1 is even. Over there exists an integer k so that a+1 = 2k. Now, a = 2k -1 = 2(k-1+1) -1 = 2(k-1) + 1 and k-1 is one integer, for this reason a is odd. Putting these 2 statements together we have actually that a is weird if and also only if a+1 is even.10. We will certainly prove this through contradiction. BWOC assume that sqrt(5) is a reasonable number. Climate there exist integers p and q through q no 0 so the sqrt(5) = p/q. We might reduce this fraction and so deserve to assume the p and q have no typical factor. Through squaring we acquire 5q2 = p2. This way that 5 divides p2 (since q2 is one integer). Because 5 is a prime number, if 5 divides p2 then 5 divides p. We might now compose p = 5k for part integer k. Return to the equation we now have 5q2 = p2 = (5k)2, or 5q2 = 25k2. Thus, q2 = 5k2 and 5 divides q2. Again, since 5 is prime this indicates that 5 divides q. This is the desired contradiction due to the fact that we have that p and also q have actually no typical factor, yet have shown that 5 divides both p and q. Hence our presumption is false and so, sqrt(5) should be an irrational number.12.(a) F. The proof given is a correct proof that the converse that the claim (by the contrapositive method), not the claim. (c) F. The fourth sentence "The left side ...." that the proof is just the claim that is being proved. This is an example of one reasoning and also doesn"t prove anything.(d) A.