**(pg. 42 : 2, 6, 7, 10, 12 )2. (a) intend that**

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## 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.You are watching: If a divides b then a divides bc

**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**.