__Proofs__
==========

__Direct Proof__
================

### _Section 4.1, #27_

By _Joram Stith_ and _Noah Soto_

Prove the following statment

- The sum of any two odd integers is an even integer.

## Direct Proof:

Suppose:

$a$ is an odd integer
$b$ is an odd integer

Therefore, by the definition of an odd integer:

$a=2c+1$ and $b=2d+1$ for some integers c and d.

By substiting a and b with the values defined above:
$$a+b = (2c+1)+(2d+1)$$

$$(2c+1)+(2d+1) = 2c+2d+2$$

Factor out a 2:
$$2c+2d+2=2(c+d+1)$$

By definition of adding integers:

$e=(c+d+1)$ for some integer e, as the sum of integers will always yield another integer.

Thus:

$$a+b=2e$$

where $2e$ is even by the definition the of an even number.

### Therefore, the sum of any two odd integers is an even integer.

__Direct Proof #2__
================

### _Section 4.3, #21_

Prove the following statment

- Prove that the product of any two even integers is a multiple of 4.

## Direct Proof:

Suppose:

$a$ is an even integer
$b$ is an even integer

Therefore, by the definition of an even integer:

$$a=2d$$ and $$b=2e$$

where $d,e$ are integers.

By substiting a and b with the values defined above:
$$
ab = (2d)(2e)
$$
$$
ab=4de
$$
By definition on an integer, $g=de$ where g is an integer because it is the product of two integers.

Therefore:
$$ab=4g$$

$ab$ must be a multiple of 4 becuase the expression equals $4g$ which is a multiple of 4.

### Therefore, the product of any two even integers is a mutiple of 4.

__Proof by Contradiction__
================

### _Section 4.6, #12_

- If $a$ and $b$ are rational numbers, $b ≠ 0$, and $r$ is an irrational number, then $a+br$ is __irrational__

## Proof by Contraction

__Suppose Not:__

- Contradicting if __P__ _then_ __Q__ becomes __P__ _and_ __!Q__

Suppose $a$ and $b$ are rational numbers, $b ≠ 0$, $r$ is an irrational number, and $a+br$ is __rational__.

Because a and b are both rational, by the definition of a rational number,
$$a=\frac cd$$ and $$b=\frac ef$$
where $c$, $d$, $e$, and $f$ are integers, and $d ≠ 0$ and $f ≠ 0$.

Becuase $a+br$ is rational, by defintion of a rational number:
$$a+br=\frac gh$$
For some integers $g$ and $h$ where $g ≠ 0$

By subsituting a and b with the values defined above:

$$(\frac cd) + br=\frac gh$$
$$br=\frac gh - \frac cd$$

$$ br = {gd\over hd} -{ch\over dh} $$
$$ b r = {gd-ch\over hd} $$
$$ r = {gd-ch\over hdb} $$

And $k$ is an integer as the product and difference of integers will always yield an integer, such that:

$$k=gd-ch$$

And $l$ is an integer as the product of integers will always yield an integer, such that:

$$l=hdb$$

Thus:

$$r=\frac kl$$

Where r is an integer, as the quotient of two integers will always yield another integer.

__This is a contradiction__

This proves the original statment.

### Therefore, if $a$ and $b$ are rational numbers, $b ≠ 0$, and $r$ is an irrational number, then $a+br$ is __irrational__

__Proof By Contraposition__
======================

### _Section 4.6 problem #20_

By _Joram Stith_ and _Noah Soto_

Prove the following statement using contraposition:

- If the sum of 2 real numbers is less than 50, then at least one number is less than 25.

## Proof by Contraposition

### Contrapositive:

- If __P__ _then_ __Q__ becomes If __~Q__ _then_ __~P__

- If two real numbers are greater than or equal to 25, their sum must be at least 50.

### Solve:

Suppose $m,n \in$ Real Numbers and $m,n \ge 25$.

Thus, $m = 25 + a$ and $n = 25 + b$ for $a,b \in$ Real Numbers where $a,b \ge 0$

So, $m + n = a + b + 50$

$a + b + 50 \ge 50$

Thus, $m+n \ge 50$

### Which was to be proven

---

- Therefore, if the sum of 2 real numbers is less than 50, then at least one number is less than 25.

__Proof By Induction__
======================

### _Section 5.2 problem #14_

By _Joram Stith_ and _Noah Soto_

Prove the following statement by mathematical induction _for all integers n greater than or equal to 0_:

$$
\sum_{i=1}^{n+1} i*2^i = n*2^{n+2}+2
$$

## Proof by Induction

### Basis Step:

Let n = 0

$$
\sum_{i=1}^{(0)+1} i*2^i = (1)*2^{(1)} = 2
$$

$$
(0)*2^{(0)+2}+2 = 2
$$

Thus, for n = 0

$$
\sum_{i=1}^{n+1} i*2^i = n*2^{n+2}+2
$$

### Inductive Step:

Suppose for integer $K \ge 0$

$$
\sum_{i=1}^{k+1} i*2^i = k*2^{k+2}+2
$$

Show that for $k+1$:

$$
\sum_{i=1}^{(k+1)+1} i*2^i = (k+1)*2^{(k+1)+2}+2
$$

or

$$
\sum_{i=1}^{k+2} i*2^i = (k+1)2^{k+3}+2
$$

### Solve:

$$
\sum_{i=1}^{k+2} i*2^i = \sum_{i=1}^{k+1} i*2^i + [(k+2)*2^{k+2}]
$$

and

$$
\sum_{i=1}^{k+1} i*2^i = k*2^{k+2}+2
$$

So,

$$
\sum_{i=1}^{k+1} i*2^i + [(k+2)*2^{k+2}] = [k*2^{k+2}+2] + [(k+2)*2^{k+2}]
$$

$$
= k*2^{k+2} + k*2^{k+2} + 2*2^{k+2} + 2
$$

$$
= 2k*2^{k+2} + 2^{k+3} + 2
$$

$$
= k*2^{k+3} + (1)*2^{k+3} + 2
$$

$$
= (k+1)*2^{k+3} + 2
$$

### Which was to be proven