(1a) (4 pts) The domain is integers. Write the following using logical connectiv

Question

(1a) (4 pts) The domain is integers. Write the following using logical connectives, quantifiers, and required mathematical symbols: The difference between the square of an even integer and the square of an odd integer is an odd integer.

(1b) (8 pts) Write a direct proof of the proposition in (1a).

(2a) (4 pts) The domain is integers. Write the following using logical connectives, quantifiers, and required mathematical symbols: If the sum of the squares of two integers is even then either both the integers are even or both the integers are odd.

(2b) (8 pts) Prove the proposition in (2a) by contraposition.

(3) (6 pts) The domain is integers. Consider the following proposition. For every positive integer n, n n+2 < n n+1 . I have given the start of a proof by contradiction. Complete it and provide a complete proof as your answer. Proof by contradiction. Suppose the statement is false. Then, there exists a specific positive integer n such that n n+2 n n+1

Explanation / Answer

(1a)

E(x) : x is even integer

O(x): x is odd integer

SD(x, y): difference of squares of x & y

x y (E(x) O(y) -> O(SD(x,y)) )

(1b)

Suppose:

Consider two integers x even & y odd. The difference between squares of x & y is suppose z. Then

          z = x2 – y2

Proof statement: Prove z is an odd integer.

Proof:

          z = x2 – y2

z = (x+y)(x-y) => z * 1 = (x+y)(x-y)

Assume 1= x-y then z = x+ y, Solving these two equations gives:

          2x = z + 1

          x = (z+1)/2 => y = (z-1)/2

Now, since x & y are integers both z+1 and z-1 must be evaluated to an even number so that when divided by it produce an integer. For both of these to be evaluated as even, z should be odd.

Hence proved, difference between squares of an even and an odd number produces an odd number.

(2a)

E(x) : x is even integer

O(x): x is odd integer

SS(x, y): sum of squares of x & y

x y z ( E(SS(x,y)) -> (E(x) E(y)) (O(x) O(y)))

(2b)

Suppose:

Consider an integer z which is some of squares of two integers x &y. Then

z = x2 + y2

Proof statement: If one integer say x is even and y is odd then sum of squares of x & y is odd. (Proof by contraposition)

Proof:

          z = x2 + y2

z = (x+y)2 – 2x.y

Now since x is even and y id odd, sum x+y would be odd. Square of an odd number is also odd thus (x+y)2 is odd. Now 2 x.y is even since anything multiplied by 2 is even. So the (x+y)2 – 2x.y is difference between an odd minus an even number which is always an odd number. Thus z is an odd number. Hence proved, sum of squares of one odd and one even number is odd number. Thus, if the sum of the squares of two integers is even then either both the integers are even or both the integers are odd.

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