### Mathematical Proof

#### Disproving a Conjecture

In order to disprove a conjecture, you just need one example for which the conjecture is not true.

**E.g.1**

n^{2} + n + 41 is a *prime number* for any value of n.

Enter the value of 'n' into the text box and calculate.

It seems to be true for all the initial values of 'n'. We, however, cannot check it indefinitely by substituting integers.

Suppose n=41. Then the formula becomes

41^{2} + 41 + 41

Take 41 out to factorize:

41(41 + 1 + 1)

41(43)

So, when n = 41, the expression is no longer a prime number as there is a factor, 43, in addition to 1 and 43.

So, the conjecture is not true for all the values of n.

**E.g.2**

n^{2} + n + 1 is a *prime number* for any value of n.

Enter the value of 'n' into the text box and calculate.

It seems to be true for all the initial values of 'n'. We, however, cannot check it indefinitely by substituting integers.

Suppose n=4. Then the formula becomes

4^{2} + 4 + 1

21

So, when n = 4, the expression is no longer a prime number.

So, the conjecture is not true for all the values of n.

**E.g.3**

If n is prime, then n^{2} + n + 1 is a *prime number* for any value of n.

Enter the value of 'n' into the text box and calculate.

It seems to be true for all the initial values of 'n'. We, however, cannot check it indefinitely by substituting integers.

Suppose n=7. Then the formula becomes

7^{2} + 7 + 1

57, which is a product of 3 and 19.

So, when n = 7, the expression is no longer a prime number.

So, the conjecture is not true for all the values of n.

#### Proof by Exhaustion

As the name suggests, we have to experiment with all the possibilities to prove that the conjecture holds true.

**E.g.**

If n is an integer and 2 ≤ n ≤ 7, then p = n^{2} + 2 is not a multiple of 4

n | p | Y / N |
---|---|---|

2 | 6 | N |

3 | 11 | N |

4 | 18 | N |

5 | 27 | N |

6 | 38 | N |

7 | 51 | N |

As you can see, if n was not restricted, we would have been forced to check for thousands, if not millions, of possibilities to make sure it worked for all the values of n. It is very exhausting, indeed!

#### Proof by Deduction

In this method, we are not resorting to numerical proof - substituting numbers to show that the conjecture holds true for all of them. Instead, we use algebra with a certain logical argument to prove it.

**E.g.1**

n^{2} - 4n + 5 is positive for any integer.

Enter the value of 'n' into the text box and calculate.

As you can see, it works for any integer. We can easily prove it by using simple algebra, which is as follows:

Use completing the square:

n^{2} - 4n + 5 = (n-2)^{2} + 1

(n-2)^{2} is always positive, being a square. Adding 1 to a square number does not change it. So the conjecture is true for any vale of n.

**E.g.2**

1 + 2 + 3 + .......+ n = (n/2)(n+1), for any value of n.

Let T = 1 + 2 + 3 + ........+ n

Now, write the same thing from back to front;

T = n + (n-1) + (n-2) + (n-3) +.....3 + 2 + 1

Add the two up

2T = (n+1) + (n+1) + (n+1) + ......+ (n+1)

2T = n(n+1)

T = (n/2)(n+1)

The conjecture is true for any value of n.

#### Proof by Contradiction

This is also known as proof by assuming the opposite. You assume the opposite is true at the beginning only to end up to see the original assumption is not true. That, in turn, proves the conjecture.

**E.g.1**

√2 is an irrational number.

Suppose √2 is rational and can be written as p/q where p and q are two integers in the *simplest* form.

√2 = p/q

Square both sides:

2 = p^{2}/q^{2} => p^{2} = 2q^{2}

So, p^{2} is an even number. Therefore, p is even too and can be written as, p = 2k where k is an integer.

2 = (2k)^{2} /q^{2}

2q^{2} = 4k^{2}

q^{2} = 2k^{} = > q^{2} is even and so is q.

Since p and q are even, they both have a common factor of 2. This *contradicts* our original assumption that p and q were in the simplest form.
So, √2 is not a rational number. Therefore, it is an irrational number.

**E.g.2**

√3 is an irrational number.

Suppose √3 is rational and can be written as p/q where p and q are two integers in the *simplest* form.

√2 = p/q

Square both sides:

3 = p^{2}/q^{2} => p^{2} = 3q^{2}

So, p^{2} is a multiple of 3. Therefore, p is a multiple of 3 too and can be written as, p = 3k where k is an integer.

3 = (3k)^{2} /q^{2}

3q^{2} = 9k^{2}

q^{2} = 3k^{} = > q^{2} is a multiple of 3 and so is q.

Since p and q are multiples of 3, they both have a common factor of 3. This *contradicts* our original assumption that p and q were in the simplest form.
So, √3 is not a rational number. Therefore, it is an irrational number.

**E.g.3**

For all integers, if n^{2} is even, so is n.

Assume that n^{2} is even and n is not even, i.e. odd

So, n = 2k + k where k is any integer.

n^{2} = (2k+1)^{2}

n^{2} = 4k^{2} + 4k + 1 - an odd number

This is a contradiction; the left is *even*; the right is *odd.*

Therefore, our original assumption, is not true.

So, n^{2} is even, n is even too.

Now, it's time for some mathematical fun!

Let a = b

Multiply both sides by a

a^{2} = ab

Add a^{2} to both sides

2a^{2} = a^{2} + ab

Take away 2ab from both sides

2a^{2} - 2ab = a^{2} + ab - 2ab

2a^{2} - 2ab = a^{2} - ab

Factorize both sides,

2(a^{2} - ab) = a^{2} - ab

2 = 1!

Spot the contradiction!!