Math 310: Proofs and Questions

C4E7: Prove that if x,y > 0, then (1/2)(x+y) ≥ √(xy). For which x,y does this equality hold?

(1/2)(x+y) ≥ √(xy) == x+y ≥ 2√(xy) == (x+y)^2 ≥ 4xy == x^2 + y^2 + 2xy ≥ 4xy == (x-y)^2 ≥ 0 So this equality holds when x=y.

C3E1: Express the decimal 1.813813813....(bar) as a fraction (m/n) where m and n are integers.

1.813813813....(bar) = 1 + (8/10) + (13/10^3) * (100/99) = (1812/999)

C4E5a: Which is bigger: 100^10000 or 1000^100?

10000^100 = (100^2)^100 = 100^200, so 100^10000 is bigger

C3E4: Without using a calculator, find the cube roots of 2 correct to one decimal place.

1^3 = 1 < 2 < 8 = 2^3 So we know that ∛(2) is atleast 1. 1.728 = 1.2^3 < 2 < 1.3^3 = 2.193 So, ∛(2) is approximately 1.2

C4E5b: Which is bigger: The cube root of 3 or the square root of 2?

2^(1/2) ≥ 3^(1/3). Taking sixth powers, this implies 2^3 ≥ 3^2, which is false. Hence, the cube root of three is bigger (2^(1/2)<3^(1/3))

C4E4: Find an integer n and a rational t such that n^t = 2^(1/2)3^(1/3)

2^(1/2)3^(1/3) = (2^(3)3^(2))^(1/6) = 72^(1/6)

C4E2: Simplify 2^(1/2)5^(1/2)4^(-1/4)20^(1/4)5^(-1/4)10^(1/2)

Because x^p *y^p = (xy)^p: 2^(1/2)5^(1/2)10^(1/2) = (2*5*10)^(1/2) = (100)^(1/2) = 10 and 4^(-1/4)5^(-1/4) = 20^(-1/4) Because x^p *x^q = x^(p+q): 20^(1/4)20^(-1/4) = 20^0 = 1 So 1*10 = 10

C1E6a: Prove √(6)-√(2) > 1

Equivalent to: 1+√(2) < √(6) == (1+√(2))^2 < 6 ==3 + 2√(2) < 3 + 2√(9/4) = 3 + (6/2) = 6

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: {1, 3} ∈ A

False

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: {1, α} ∉ A

False

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: {1, α} ⊆ A

False

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: {{1, 3}} ⊆ A

False

C1E5: True or false: If n^2-2n-3=0, then n=3.

False

C1E5: True or false: n^2-3n-3=0 only if n=3

False

True or False: The non-prime number 8881 has a prime factor greater than 89.

False

True or false: √(3) is rational.

False

C1E5: True or false: For integers a and b, ab is a square only if both a and b are squares.

False (e.g. if a=b=2, ab is a square, but a and b are not)

C2E3c: Prove it is true or five a counterexample to show it is false: The product of two irrational numbers is always rational.

False: The product of √(2) and 1+√(2) is √(2)+2, which is irrational.

C2E3b: Prove it is true or five a counterexample to show it is false: The product of two irrational numbers is always irrational.

False: √(2) and -√(2) are irrational, but √(2)*-√(2) = -2 which is rational.

C1E6c: Prove if n=m^3-m for some integer m, then n is a multiple of 6.

First observe: m^3-m = m(m+1)(m-1) That is, this number is the product of 3 consecutive numbers. At least one of these terms is even --> either m or m+1 is divisible by 2. Every 3rd number in the real line is divisible by 3, so at least one of these consecutive integers is divisible by 3. Because we have a factor of both 2 and 3 in m^3-m, we know m^3-m is a multiple of 6.

C4E8: When we want to add 3 numbers, we say a+b+c, because (a+b)+c = a+(b+c). But with powers, this is not true: (a^b)^c need not be equal to a^(b^c). Find positive integers a, b, c such that (a^b)^x < a^(b^c), and positive integers d, e, f such that (d^e)^f > d^(e^f)

If a=2, b=2, and c=3, then (2^2)^3 = 2^6 < 2^8 = 2^(2^3) If e=1, d=2, and f=2, then (2^1)^2 = 2^2 > 2^1 = 2^(1^2)

C2E2b: Rational or irrational: 1 + √(2) + √(3/2)

Irrational If a = 1 + √(2) + √(3/2), then a-1 would be rational, but we know from part C2E2a that √(2) + √(3/2) is irrational.

C2E2d: Rational or irrational: √(2) + √(3) + √(5)

Irrational If n is a positive integer that is not square, then √(n) is irrational.

C2E2a: Rational or irrational: √(2) + √(3/2)

Irrational Prove by contradiction: √(2) + √(3/2) = (m/n). Square both sides, see that it is still irrational.

C3E3b: Rational or irrational? If rational, express in the form (m/n) with m, n ∈ Z .101001000100001000001.....

Irrational Prove by contradiction: Suppose this is rational. Then this has to be periodic, so for some n there is a sequence a1a2a3...an such that it repeats itself. If we go far enough along the decimal, there will be a sequence of at least 2n consecutive zeroes. This means the decimal ends 0000..... (goes on forever), which is false. This is irrational.

C3E3c: Rational or irrational? If rational, express in the form (m/n) with m, n ∈ Z 1*b1b2b3.... where bi = 1 if i is a square, and bi=0 if i is not a square.

Irrational This would be 1.100100001000001..... There are an ever increasing sequence of zeroes, so the argument for C3E3b shows that this number is irrational.

C1E10: Prove by contradiction that a real number that is less than every positive real number cannot be positive.

Prove by contradiction: Here we interpret "less than" as "strictly less than" (<). In this case, if such a number existed, it would be strictly less than itself, which it cannot be.

C1E8: Given that the number 8881 is not a prime number, prove that it has a prime factor that is at most 89.

Prove by contradiction: Suppose no prime factor of 8881 is less than or equal to 89. Let p1 and p2 be the two smallest prime factors of 8881. The next prime number after 89 is 97. By assumption, p1 and p3 must be at least as big as 97, so: 8881≥97^=9409 This is a contradiction, therefore 8881 must have a prime divisor less than or equal to 89.

C2E1a: Prove that √(3) is irrational.

Prove by contradiction: Suppose √(3) is rational, so √(3) = (m/n). Squaring this, we get m^2=3n^2. Thus, m^2 is a multiple of 3 and that means that m is a multiple of 3. We write m=3k, for some integer k. 3n^2 = m^2 = 9k^2, so n^2 = 3k^2. Therefore, n^2 is a multiple of 3. We have shown that m and n are multiples of 3, but (m/n) is in its lowest terms, so this is a contradiction. Therefore, √(3) is irrational.

C2E1b: Prove that there are no rationals r, s such that √(3) = r+s√(2).

Prove by contradiction: Suppose √(3) = r+s√(2) with r, s rational. Square both sides: 3=r^2+2s^2+2rs√(2). If rs≠0, then √(2) = (3-r^2-2s^2)/(2rs). Since res, are both rational, this implies √(2) is rational, which is a contradiction.

C1E6b: Prove if n is an integer such that n^2 is even, then n is even.

Prove contrapositive: Prove if n is odd, then n^2 must also be odd n=2k+1 for some integer k. n^2 = (2k+1)^2 = 4k^2+4k+1 = 2(2k^2 +2k)+1 This shows that n^2 must also be odd if n is odd.

C2E2e: Rational or irrational: √(2) + √(3) - √(5+2√(6))

Rational (√(2) + √(3))^2 = 5+2√(6), so √(2) + √(3) - √(5+2√(6)) = 0

C2E2c: Rational or irrational: 2√(18) - 3√(8) + √(4)

Rational 2√(18) - 3√(8) + √(4) = 6√(2) - 6√(2) + 2 = 2

C3E3a: Rational or irrational? If rational, express in the form (m/n) with m, n ∈ Z 0*a1 a2 a3 ... where for n=1,2,3.... the value of an is the number 0,1,2,3 or 4 which is dividing n by 5

Rational: All periodic decimals are rational. The number is .12340123401234... which is periodic (.01234bar) so 12340/99999

C4E6: Find all real solutions x of the equation: x^(1/2) - (2-2x)^(1/2) = 1

Squaring both sides of the equation and simplifying gives: 1-x = 2x^(1/2)(2-2x)^(1/2) Squaring again gives: 9x^2 - 10x + 1 = 0 Solving the quadratic equation gives x=1, x=(1/9), but (1/9) doesn't work so x=1 is the only solution.

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: {3, {3}} ⊆ A

True

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: {{1, α}} ⊆ A

True

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: {α} ∉ A

True

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: Ø ⊆ A

True

C1E1: Let A be the set A={α, {1,α}, {3}, {{1,3}}, 3}. True or False: α ∈ A

True

C1E5: True or false: n=3 only if n^2-2n-3=0

True

True or False: If n is an integer such that n^2 is even, then n is even.

True

True or false: A real number that is less than every positive real number cannot be positive

True

True or false: Between any two different real numbers, there is a rational number and an irrational number.

True

True or false: If n is a positive integer that is not square, then √(n) is irrational.

True

True or false: Prove if n=m^3-m for some integer m, then n is a multiple of 6.

True

C1E5: True or false: For integers a and b, ab is a square if both a and b are squares.

True (e.g. if a,b are squares then a=m^2, b=n^2, so ab=(mn)^2)

C2E3a: Prove it is true or five a counterexample to show it is false: The product of two rational numbers is always rational.

True: If x=(m/n), and y=(p/q) are rational, then xy = (mp/nq) which is rational.

C2E3d: Prove it is true or five a counterexample to show it is false: The product of a non-zero rational and an irrational is always irrational.

True: Prove by contradiction: Suppose there is a rational a≠0, and an irrational b such that c=ab is rational. Then b=(c/a) and a and c are both rational, therefore b is rational. This is a contradiction.