Question

Part I: Constructing proofs.

You must write down all proofs in acceptable mathematical language: make sure you mark the beginning and end of the proof, define all variables, use complete, grammatically correct sentences, and give a justification for each assertion (e.g., by definition of…).

Definitions:

• An integer ? is even if and only if there exists an integer ? such that ? = 2?.

• An integer ? is odd if and only if there exists an integer ? such that ? = 2? + 1.

• Two integers have the same parity when they are both even or when they are both odd. Two integers have opposite parity when one is even and the other one is odd.

• An integer ? is divisible by an integer ? with ? ≠ 0, denoted ? | ?, if and only if there exists an integer ? such that ? = ??.

• A real number ? is rational if and only if there exist integers ? and ? with ? ≠ 0 such that ? = ?/?.

• For any real number ?, the absolute value of ?, denoted |?|, is defined as follows: |?| = { ? if ? ≥ 0; −? if ? < 0 4.

Prove each of the following statements using a direct proof, a proof by contrapositive, a proof by contradiction, or a proof by cases. For each statement, indicate which proof method you used, as well as the assumptions (what you suppose) and the conclusions (what you need to show) of the proof.

a. If ? is divisible by ? and ? is divisible by ?, where ?, ?, and ? are positive integers, then ? +? is divisible by ?.

b. The difference of any rational number and any irrational number is irrational.

c. There is no integer that is both even and odd.

d. Any two consecutive integers have opposite parity.

e. For all real numbers ? and ?, ???(?, ?) = ?+?−|?−?| ? and ???(?, ?) = ?+?+|?−?| ? .

Answer 1

a) Since x is divisible by z and y is divisible by z, then there exists integers c and d such that

x = cz and y = dz.

Now, x+y = cz+dz = (c+d)z.

Since c and d are integers then c+d is also an integer.

Therefore (c+d)z is divisible by z ,i.e., x+y is divisible by z.

b) Assume x is an irrational number and the difference of x and a rational number is a rational number , where a,b,c,d are integers (b,d 0).

Then,

By addition,

i.e.,

Since integers are closed under multiplication and addition, ad, bc, bd, bc+ad are integers, making a rational number by definition.

This is a contradiction to the given fact that the difference of x and is a rational number.

The assumption is wrong.

Therefore, the difference of a rational number and an irrational number is an irrational number.

c) Let us assume that there exists a number A that is both even and odd.

Since A is even, then there exists an integer k such that A = 2k.

Snce A is odd, then there exists an integer m such that A = 2m+1.

By condition we have, 2k = 2m+1.

i.e., 2k-2m = 1

i.e., 2(k-m) = 1

Since k and m are integers, then k-m is an integer and 2(k-m) is an even number.

But in other side we have an odd number 1.

This is a contradiction to the given fact that an integer can be both even and odd.

The assumption is wrong.

Therefore, an integer can't be both even and odd.

d) Let us assume that x and y be any two consecutive integers that have same parity.

Case I : both of x and y are even.

Then there exists integers c and d such that x = 2c and y = 2d.

Since x and y are consecutive integers, then x+1 = y.

i.e., 2c+1 = 2d

Here, in one side we have an odd integer and in other side we have an even integer.

This is a contradiction to the given fact that Any two consecutive integers have same parity.

The assumption is wrong.

Case II : both of x and y are odd.

Then there exists integers c and d such that x = 2c+1 and y = 2d+1.

Since x and y are consecutive integers, then x+1 = y.

i.e., (2c+1)+1 = 2d+1

i.e., 2c+1+1 = 2d+1

i.e., 2c+2 = 2d+1

i.e., 2(c+1) = 2d+1

Here, in one side we have an even integer and in other side we have an odd integer.

This is a contradiction to the given fact that Any two consecutive integers have same parity.

The assumption is wrong.

Therefore, Any two consecutive integers have opposite parity.

e) Case I : x > y.

Here, x-y > 0 ,i.e., x-y is positive.

Then, |x-y| = x-y.

Then, [x+y-|x-y|]/2 = [x+y-(x-y)] = [x+y-x+y]/2 = [2y]/2 = y = min(x,y).

And, [x+y+|x-y|]/2 = [x+y+(x-y)]/2 = [x+y+x-y]/2 = [2x]/2 = x = max(x,y).

Case II : x < y.

Here, x-y < 0 ,i.e., x-y is negative.

Then, |x-y| = -(x-y) = y-x.

Then, [x+y-|x-y|]/2 = [x+y-(y-x)] = [x+y-y+x]/2 = [2x]/2 = x = min(x,y).

And, [x+y+|x-y|]/2 = [x+y+(y-x)]/2 = [x+y+y-x]/2 = [2y]/2 = y = max(x,y).

Therefore, for all real numbers x and y, min(x,y) = [x+y-|x-y|]/2 and max(x,y) = [x+y+|x-y|]/2.