Quiz 5 note

What is the best choice of proving the following statement? The sum of any two rational numbers is rational.

Proof: Suppose r and s are rational numbers. [ We must show that r + s is rational. ] Then, by definition of rational, r = a / b and s = c / d for some integers a, b , c, and d with b ≠≠ 0 and d ≠≠ 0. Thus r + s = abab + cdcd by substitution = ad+bcbdad+bcbd by basic algebra. Let p = a d + b c and q = b d . Then p and q are integer s because products and sums of integers are integers and because a, b, c, and d are all integers. Also q ≠≠ 0 by the zero product property. Thus r + s = pqpq where p and q are integers and q ≠≠ 0 . Therefore, r + s is rational by definition of a rational number. [ This is what was to be shown. ]

If n and d are integers and d ≠≠ 0 then n is divisible by d if, and only if, n equals d times some integer. What is the best choice that includes all equivalences of " n is divisible by d," from the following list? a. n is a multiple of d, b. n % d c. d is a factor of n d. n/d e. d is a divisor of n f. d|n

a, c, e and f

Proof by contradiction

- There Is No Greatest Integer - No Integer Can Be Both Even and Odd - The Sum of a Rational Number and an Irrational Number

Indirect proofs

-identify statement you want to prove -assume that it is false (opposite is true) -point out desired conclusion must be true since the contradiction proves it false

What is the best choice to fill in the following blanks ? According to the quotient-remainder theorem, if an integer n is divided by a positive integer d , the possible remainders are _____________________________________. This implies that n can be written in one of the forms _______________________________________________ for some integer q.

0, 1, 2, ..., ( d−1 ) ; dq , dq + 1 , dq + 2 , ..., dq + ( d − 1 )

proof by contra-diction

1. Suppose the statement to be proved is F. that is, suppose that the negation of the statement is T. 2. Show that this supposition leads logically to contradiction 3. Conclude that the statement to be proved is T

Choose the best answer for the following. Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the following equations: xyx+yxyx+y = a , xzx+zxzx+z = b , and yzy+zyzy+z = c. Is x rational ? If so, express it as a ratio of two integers.

Yes.Proof : Suppose a, b, and c are integers and x, y, and z are nonzero real numbers, where xyx+yxyx+y = a ............(1) xzx+zxzx+z = b .............(2) , and yzy+zyzy+z = c ................(3)Note that because a, b, and c are real numbers, none of the denominators x + y, z + x or y + z can equal zero.[ The strategy will be to express x in terms of the integers a, b, and c in hopes of showing that x can be written as a ratio of integers with a nonzero denominator. ]First observe that, by the zero product property,a ≠≠ 0 because from (1) x y = a ( x + y ) and x y ≠≠ 0 b ≠≠ 0 because from (2) z x = b ( z + x ) and z x ≠≠ 0c ≠≠ 0 because from (3) y z = c ( y + z ) and y z ≠≠ 0.i. Solve equation (1) for y in terms of a and x:x y = a ( x + y ) ⇔⇔ x y = a x + a y ⇔⇔ x y - a y = a x ⇔⇔ ( x - a ) y = a x. Now because a x is a product of nonzero real numbers, a x ≠≠ 0, and so ( x - a ) ≠≠ 0 . Thus we may divide by x - a to obtain y = axx−aaxx−a ii. Similarly, solve equation (2) for z in terms of b and x:z x = b ( z + x ) ⇔⇔ z x = b z + b x ⇔⇔ z x - b z = b x ⇔⇔ ( x - b ) z = b x.And because b x is a product of nonzero real numbers, b x ≠≠ 0. Thus ( x -b ) ≠≠ 0, and we may divide by x — b to obtain z = bxx−bbxx−b iii. Substitute the results of i and ii into equation (3) ( axx−aaxx−a ) ( bxx−bbxx−b ) = c ( axx−aaxx−a + bxx−bbxx−b ) and solve for x in terms of a, b, and c by first multiplying both sides ( x - a) ( y — b ) to obtain ( a x ) ( b x ) = c a x ( x - b) + c b x ( x - a ). Because x ≠≠ 0, both sides may be divided by x to yielda b x = c a x — c a b + c b x — c b a,or, equivalently (by putting all the terms involving x on the right-hand side and all the other terms on the left-hand side), 2 a b c = a c x + b c x — a b x = x ( a c + b c — a b).Because 2 a b c ≠≠ 0, by the zero product property, a c + b c — a b cannot be zero either. Thus we may divide both sides by a c + b c — a b to obtainx = 2abcac+bc−ab2abcac+bc−ab Finally, because products and sums of integers are integers, we see that x has been expressed s a ratio of integers with a nonzero denominator.

What is the best choice to fill in the following blanks? If a and b are positive integers and a | b, then _____ is less than or equal to _____.

a ; b

What is the best choice for the following statements? a. 32 div 9 b. 32 mod 9

a. 3 b. 5

What would be the best choice to proof the following statements? a. There are real numbers a and b such that a+b−−−−−−√a+b = a−−√a + b√b b. There is an integer n > 5 such that 2n − 1 is prime. c. There is a real number x such that x > 1 and 2x > x10 .

a. For example, let a = 1 and b= 0. Then a+b−−−−−−√a+b = 1-√1 = 1 and also a−−√a + b√b = 1-√1 + 0-√0 = 1 . Hence for these values of a and b, a+b−−−−−−√a+b = a−−√a + b√b In fact, if a is any nonzero integer and b = 0, then a+b−−−−−−√a+b = a+0−−−−−−√a+0 = a−−√a = a−−√a + 0 = a−−√a + 0-√0 = a−−√a + b√b b. For example, let n = 7. Then n is an integer such that n > 5 and 2n − 1 = 127, which is prime. c. For example, let x = 60. Note that to four significant digits 260 ≅≅ 1.153 x 1018 and6010 ≅≅ 6.047 x 1017 and so 2x ≥≥ x10 Examples can also be found in the approximate range 1 < x < 1.077. For instance, 21.07 ≅≅ 2.099 and 1.0710 ≅≅ 1.967,and so 21.07 > 1.0710

Choose the best choice for the following questions. a. Is 10/3 a rational number? b. Is − 539539 a rational number? c. Is 0.281 a rational number? d. Is 7 a rational number ? e. Is 0 a rational number?

a. Yes b. Yes c. Yes d. Yes e. Yes

What is the best choice that answers the following questions? a. If a and b are integers, is 3 a + 3 b divisible by 3 ? b. If k and m are integers, is 10 k m divisible by 5 ?

a. Yes. By the distributive law of algebra, 3 a + 3 b = 3 ( a + b ) and a + b is an integer because it is a sum of two integers. b. Yes. By the associative law of algebra, 10 k m = 5 · ( 2 k m ) and 2 k m is an integer because it is a product of three integers.

direct proof

start with the hypothesis of a statement and make one deduction after another until you reach the conclusion.

Choose the best answer to the following questions . a. Is 1 prime? b. Write the first six prime numbers. c. Write the first six composite numbers.

a. No. A prime number is required to be greater than 1. b. 2, 3, 5, 7, 11, 13 c. 4, 6, 8, 9, 10, 12

Choose the best answer for the following. I. The zero product property, says that if a product of two real numbers is 0, then one of the numbers must be 0. a. Write this property formally using quantifiers and variables. b. Write the contrapositive of your answer to part (a). c. Write an informal version (without quantifier symbols or variables) for your answer to part (b). II. Assume that a and b are both integers and that a ≠≠ 0 and b ≠≠ 0. Explain why ( b − a ) / ( a b2 ) must be a rational number.

I. 8. a. real numbers x and y, if xy = 0 then = 0 or y = 0.c. If neither of two real numbers is zero, then their product is nonzero. b. ∀realnumbersxandy,ifx̸=0andy̸=0thenxy̸=0. II. Because a and b are integers, b − a and ab2 are both integers (since differences and products of integers are integers). Also, by the zero product property, ab2 ̸= 0 because neither a nor b is zero. Hence (b − a)/ab2 is a quotient of two integers with nonzero denominator, and so it is rational.

Choose the best choice to proof the following. I. Given any integer n , if n > 3, could n , n + 2 , and n + 4 all be prime ? Prove or give a counterexample. II. The fourth power of any integer has the form 8 m or 8 m + 1 for some integer m.

I. Given any integer n , the numbers n, n + 2 , and n + 4 cannot all be prime. Proof : Suppose n is any integer with n > 3 . By the quotient-remainder theorem with d = 3 , we know that n = 3 q, or n = 3 q + 1, or n = 3 q + 2 for some integer q. Note that because n is greater than 3 , either q is greater than 1 or q = 1 and n = 4 = 3 q + 1 or q = 1 and n = 5 = 3 q + 2. Case 1 ( q > 1 and n = 3 q ) : In this case, n is not prime because it is a product of 3 and q and both 3 and q are greater than 1. Case 2 ( q > 1 and n = 3 q + 1 ) : In this case, n + 2 = ( 3 q + 1 ) + 2 by substitution = 3 q + 3 = 3 ( q + 1 ) by algebra. So n + 2 is not prime because it is a product of 3 and q + 1 and both 3 and q + 1 are greater than 1. Case 3 ( q > 1 and n = 3 q + 2 ) : In this case, n + 4 = ( 3 q + 2 ) + 4 by substitution = 3 ( q + 2 ) by algebra.So n + 4 is not prime because it is a product of 3 and q + 2 and both 3 and q + 2 are greater than 1. Conclusion : In all three cases, at least one of n or n + 2 or n + 4 is not prime. II. Proof : Suppose n is any integer. By the quotient-remainder theorem with d =2 , n = 2 q or n = 2 q + 1 for some integer q. Case 1 ( n = 2 q for some integer q ) : In this case, by substitution, n4 = ( 2 q )4 = 16 q4 = 8 ( 2 q4 ). Let m = 2 q4 . Then m is an integer because it is a product of integers. Hence n8 = 8 m where m is an integer. Case 2 ( n = 2 q + 1 for some integer q ) : In this case, by substitution, n4 = ( 2 q + 1 )4 by substitution = ( 2 q + 1 )2 ( 2 q + 1 )2 = ( 4 q2 + 4 q + 1 ) ( 4 q2 + 4 q + 1 ) = 16 q4 + 16 q3 + 4 q2 + 16 q3 + 16 q2 + 4 q + 4 q2 + 4 q + 1 = 16 q4 + 32 q3 + 24 q2 + 8 q + 1 = 8 ( 2 q4 + 4 q3 + 3 q2 + q ) + 1 by algebra. Let m = 2 q4 + 4 q3 + 3 q2 + q . Then m is an integer because products and sums of integers are integers. Hence n 4 = 8 m + 1 where m is an integer. Conclusion : In both cases n4 = 8 m or n4 = 8 m + 1 for some integer m.

Choose the best proof for the following. I. For all real numbers r and c with c ≥ 0 , if − c ≤ r ≤ c, then | r | ≤ c . II. For all real numbers r and c with c ≥ 0 , if | r | ≤ c, then − c ≤ r ≤ c .

I. Proof : Let c be any positive real number and let r be any real number. Suppose that - c ≤≤ r < c. ( * ) By the trichotomy law ( see Appendix A, T 17 ), either r ≥≥ 0 or r < 0. Case 1 ( r ≥≥ 0 ) : In this case | r | = r, and so by substitution into ( * ) , - c ≤≤ | r | ≤≤ c. In particular, | r | ≤≤ c . Case 2 ( r < 0 ): In this case | r | = - r, and so r = - | r | . Hence by substitution into ( * ) , - c ≤≤ - | r | ≤≤ c. In particular, - c ≤≤ - | r | . Multiplying both sides by -1 gives c ≥≥ | r | , or equivalently, | r | ≤≤ c . Therefore, regardless of whether r ≥≥ 0 or r < 0, | r | ≤≤ c [ as was to be shown ]. II. Proof : Suppose that | r | ≤≤ c. ( ** ) By the trichotomy law, either r ≥≥ 0 or r < 0. Case 1 ( r ≥≥ 0 ): In this case | r | = r, and so by substitution into ( ** ) , r ≤≤ c. Since r ≥≥ 0 and c ≥≥ r , then c ≥≥ 0 by transitivity of order ( Appendix A, T 18 ) . Then, by property T 24 of Appendix A, 0 ≥≥ - c, and, again by transitivity of order, r ≥≥ - c . Hence - c ≤≤ r ≤≤ c. Case 2 ( r < 0 ) : In this case | r | = - r, and so by substitution into ( ** ) , - r ≤≤ c. Multiplying both sides of this inequality by —1 gives r ≥≥ - c . Also since r < 0 and c ≥≥ 0 , then c ≥≥ r . Thus - c ≤≤ r ≤≤ c. Conclusion : Regardless of whether r ≥≥ 0 or r < 0, we have that - c ≤≤ r ≤≤ c [ as was to be shown ].

Choose the best choice to proof the following. I. For any integer n, n2 + 5 is not divisible by 4. II. The sum of any four consecutive integers has the form 4 k + 2 for some integer k.

I. Proof : Let n be any integer. [ We must show that n2 + 5 is not divisible by 4 for any integer n . ] By the quotient-remainder theorem, n = 2 q or n = 2 q + 1 for some integer q. Case 1 ( n is even ) : In this case, n = 2 q for some integer q, and so n2 + 5 = ( 2 q )2 + 5 by substitution = 4 q2 + 4 + 1 = 4 ( q2 + 1 ) + 1 by algebra.Let k = q2 + 1. Then k is an integer because it is a product of integers, and thus n2 + 5 has the form 4 k + 1 for some integer k. So, by the uniqueness part of the quotient-remainder theorem, the remainder obtained when n2 + 5 is divided by 4 is 1, not 0, and so n2 + 5 is not divisible by 4. Case 2 ( n is odd ): In this case, n = 2 q + 1 for some integer q, and so n2 + 5 = ( 2 q + 1 )2 + 5 by substitution = ( 4 q2 + 4 q +1 ) + 5 = ( 4 q2 + 4 q + 4 ) + 2 = 4 ( q2 + q + 1 ) + 2 by algebra. Let k = q2 + q + 1 . Then k is an integer because it is a product of integers, and thus n2 + 5 has the form 4 k + 2 for some integer k. So, by the uniqueness part of the quotient-remainder theorem, the remainder obtained when n2 + 5 is divided by 4 is 2 , not 0 , and so n2 + 5 is not divisible by 4. Conclusion : In both possible cases n2 + 5 is not divisible by 4. II. Proof : Consider any four consecutive integers. Call the smallest n. Then the sum of the four integers is n + ( n + 1 ) + ( n + 2 ) + ( n + 3 ) = 4 n + 6 = 4 ( n + 1 ) + 2. Let k = n + 1 . Then k is an integer because it is a sum of integers. Hence n can be written in 4 k + 2 form for some integer k .

Choose the best answer for the following. Find the mistakes in the "proofs" shown in below. I. Theorem: The product of an even integer and an odd integer is even. " Proof : Suppose m is an even integer and n is an odd integer. If m · n is even, then by definition of even there exists an integer r such that m · n = 2 r . Also since m is even, there exists an integer p such that m = 2 p , and since n is odd there exists an integer q such that n = 2 q + 1. Thus m n = ( 2 p ) ( 2 q + 1 ) = 2 r, where r is an integer. By definition of even, then, m · n is even, as was to be shown." II. Theorem: The sum of any two even integers equals 4 k for some integer k. " Proof: Suppose m and n are any two even integers. By definition of even, m = 2 k for some integer k and n = 2 k for some integer k. By substitution, m + n = 2 k + 2 k = 4 k. This is what was to be shown."

I. This incorrect " proof " assumes what is to be proved. The second sentence states a conclusion that follows from the assumption that m • n is even. The next-to-last sentence states this conclusion as if it were known to be true. But it is not known to be true. In fact, it is the main task of a genuine proof to derive this conclusion, not from the assumption that it is true but from the hypothesis of the theorem. II. The mistake in the " proof " is that the same symbol, k, is used to represent two different quantities. By setting both m and n equal to 2 k, the " proof " specifies that m = n, and, therefore, it only deduces the conclusion in case m = n. If m ≠≠ n, the conclusion is often false.For instance, 6 + 4 = 10 but 10 ≠≠ 4 k for any integer k.

Choose the best choice for the following. I. a. Use the quotient-remainder theorem with d = 3 to prove that the square of any integer has the form 3 k or 3 k + 1 for some integer k. b. Use the mod notation to rewrite the result of part a. II. a. Use the quotient-remainder theorem with d = 3 to prove that the product of any two consecutive integers has the form 3 k or 3 k + 2 for some integer k. b. Use the mod notation to rewrite the result of part a.

I. a. Proof : Suppose n is any integer. By the quotient-remainder theorem with d = 3, we have that n = 3 q , or n = 3 q + 1, or n = 3 q + 2 for some integer q.Case 1 ( n = 3 q for any integer q ) : In this case, n2 = ( 3 q )2 by substitution= 3 ( 3 q2 ) by algebra.Let k = 3 q2. Then k is an integer because it is a product of integers. Hence n2 = 3 k for some integer k.Case 2 ( n = 3 q + 1 for some integer q ): In this case, n2 = ( 3 q + 1 )2 by substitution = 9 q2 + 6 q + 1 = 3 ( 3 q2 + 2 q )+ 1. by algebra. Let k = 3 q2 + 2 q . T hen k is an integer because sums and products of integers are integers. Hence n2 = 3 k + 1 for some integer k.Case 3 ( n = 3q + 2 for some integer q ): In this case, n2 = ( 3 q + 2 )2 by substitution = 9 q2 + 12 q + 4 = 9 q2 + 12 q + 3 +1= 3 ( 3 q2 + 4 q + 1 ) + 1 by algebra. Let k = 3q2 + 4 q + 1. Then k is an integer because sums and products of integers are integers.Hence n2 = 3 k + 1 for some integer k. Conclusion : In all three cases, either n2 = 3 k or n2 = 3 k + 1 for some integer k [ as was to be shown ]. b. Given any integer n, n2 mod 3 ≠≠ 1. II. a. Proof: Suppose n and n + 1 are any two consecutive integers. By the quotient-remaindertheorem with d = 3 , we know that n = 3 q, or n = 3 q + 1 , or n = 3 q + 2 for some integer q.Case 1 ( n = 3 q for some integer q ) : In this case,n ( n + 1 ) = 3 q ( 3 q + 1 ) by substitution = 3 [ q ( 3 q + 1) ] by algebra.Let k = q ( 3 q + 1 ). Then k is an integer because sums and products of integers are integers. Hence n ( n + 1 ) = 3 k for some integer k. Case 2 ( n = 3 q + 1 for some integer q ) : In this case,n ( n + 1 ) = ( 3 q + 1 ) ( 3 q + 2 ) by substitution = 9 q2 + 9 q + 2 = 3 [ ( 3 q2 + 3 q ) + 2 by algebra.Let k = 3 q2 + 3 q. Then k is an integer because sums and products of integers are integers. Hence n ( n + 1 ) = 3 k + 2 for some integer k. Case 3 ( n = 3 q + 2 for some integer q ) : In this case, n ( n + 1 ) = ( 3 q + 2 ) ( 3 q + 3 ) by substitution = 3 [ ( 3 q + 2 ) ( q + 1 ) ] by algebra. Let k = ( 3 q + 2 ) ( q + 1 ) . Then k is an integer because sums and products of integers are integer. Hence n ( n + 1 ) = 3 k for some integer k. Conclusion : In all three cases, the product of the two consecutive integers either equals 3 k or it equals 3 k + 2 for some integer k [as was to be shown]. b. Given any integer n, m n mod 3 ≠≠ 1.

What is the best choice for the following ? I. Consider the following statement : The negative of any multiple of 3 is a multiple of 3. a. Write the statement formally using a quantifier and a variable. b. Determine whether the statement is true or false and justify your answer. II. Show that the following statement is false : For all integers a and b, if 3 | ( a + b ) then 3 | ( a − b ).

I. a. ∀ integers n if n is a multiple of 3 then − n is a multiple of 3. b. The statement is true. Proof : Suppose n is any integer that is a multiple of 3. [ We must show that − n is a multiple of 3. ] By definition of multiple, n = 3 k for some integer k. Then − n = − ( 3 k ) by substitution = 3 ( − k ) by algebra. Hence, by definition of multiple, − n is a multiple of 3 [ as was to be shown ]. II. Counterexample : Let a = 2 and b = 1. Then a + b = 2 + 1 = 3, and so 3 | ( a + b ) because 3 = 3 · 1 . On the other hand, a − b = 2 − 1 = 1, and 3 ∤∤ 1 because 1 / 3 is not an integer. Thus 3 ∤∤ ( a − b ) . [ So the hypothesis of the statement is true but its conclusion is false. ]

What is the best choice for the following statement? Prove: The square of any odd integer has the form 8 m + 1 for some integer m

Proof: Suppose n is a [particular but arbitrarily chosen] odd integer. By the quotient-remainder theorem, n can be written in one of the forms 4 q or 4 q + 1 or 4 q + 2 or 4 q + 3 for some integer q . In fact, since n is odd and 4 q and 4 q + 2 are even , n must have one of the forms 4 q + 1 or 4 q + 3 . Case 1 ( n = 4 q + 1 for some integer q ) : [We must find an integer m such that n 2 = 8 m + 1 . ] Since n = 4 q + 1, n 2 = ( 4 q + 1 ) 2 by substitution = ( 4 q + 1 ) ( 4 q + 1 ) by definition of square = 16 q 2 + 8 q + 1 = 8 ( 2 q 2 + q ) + 1 by the laws of algebra. Let m = 2 q 2 + q . Then m is an integer since 2 and q are integers and sums and products of integers are integers. Thus, substituting, n 2 = 8 m + 1 where m is an integer. Case 2 ( n = 4 q + 3 for some integer q ) : [We must find an integer m such that n 2 = 8 m + 1. ] Since n = 4 q + 3, n 2 = ( 4 q + 3 ) 2 by substitution = ( 4 q + 3 ) ( 4 q + 3 ) by definition of square = 16 q 2 + 24 q + 9 = 16 q 2 + 24 q + ( 8 + 1 ) = 8 ( 2 q 2 + 3 q + 1 ) + 1 by the laws of algebra. [The motivation for the choice of algebra steps was the desire to write the expression in the form 8 · (some integer) + 1.] Let m = 2 q 2 + 3 q + 1 . Then m is an integer since 1, 2, 3, and q are integers and sums and products of integers are integers. Thus, substituting, n 2 = 8 m + 1 where m is an integer. Cases 1 and 2 show that given any odd integer, whether of the form 4 q + 1 or 4 q + 3, n 2 = 8 m + 1 for some integer m. [This is what we needed to show.]

Theorem 4.5.3

The sum of any rational number and any irrational number is irrational.

Argument by contradiction

if a man accused of holding up a bank can prove that he was some place else at the time the crime was committed, he will certainly be acquitted. >> Suppose I did commit the crime. Then at the time of the crime, I would have had to be at the scene of the crime. In fact, at the time of the crime I was in a meeting with 20 people far from the crime scene, as they will testify. This contradicts the assumption that I committed the crime since it is impossible to be in two places at one time. Hence that assumption is false.