Proof

Question 1

How to proofe infinite prime numbers by contradiction

Answer 1

Assume that there is a finite list of prime numbers p1, p2, p3, … pn

Now say a number q is a number not on the list is the product of all numbers in the list , +1

Q= p1p2p3p4p5…pn +1

As a fact Q is prime or not prime !
If prime , then it’s a prime NOT on the “finite list”, which CONTRADICTS STATEMENT, proving there are infinite

If Q is not prime, then it must be divisible by a prime number by number theory

• since we are saying there are only these set primes, it must be divisible by a combination of these set primes
• HOWEVER, these set combination of primes will also have to be wholly divisible into 1 .
• == WHICH IS IMPOSSIBLE
• as a result the number is divisible by a prime number NOT ON THE LIST
• = which is a CONTRADICTION TO THE ORIGINAL STATMENT

thus proving by contradiction that there are an infinite number of primes !

Question 2

How to prove any irrational number is irrational

Say root 5

Answer 2

Assume that root 5 is rational , so that it can be written in the form m/n , where m and n are integers and have no common multiple and n is not 0

Thus root 5= m/n
5= m2/n2
5n2 = m2

Thus m2 a multiple of 5 and m too, m can be written as 5p

5n2 = (5p)2
5n2 = 25p2
N2 = 5p2
Thus n2 is a multiple of 5 and so is n, can be written as 5q

Thus m/n = 5p/5q
And as there is a common factor this contradicts original statement and shows root 5 is irrational !

Question 3

What are properties of 0 (in terms of odd and even)

Answer 3

0 has to be even, so if you can prove left hand side is odd you can disprove

Question 4

What form can all numbers be written like

Thus how prove all saaure numbers don’t end inn3?

Answer 4

In the form 10x +y, where y part of positive integers 0<= x <= 9

Here (10x +y)2 = 100x2 + 20xy + y2
Anything multipler by 100 and 20 ends in 0, so last Didier of any square number ends in last digit of y

Find all possible last digit numbers of y by exhausting 0 to 9

Now can conclude no sqaures end in a 3

Question 5

How to show there are no rational solutions to like x3 + x +1 = 0by contradiction

Answer 5

Assume there is a rational in form p/q solution in simplest terms etc

Now sub in to equation

And test for every situation 
- p odd q odd
- p even q odd
- p odd q even
- p even a even
For first three this will all lead to an odd number which CANT BE =0, thus it contradicts statement in terms of being a valid solution 

For last statement both can’t be even as this means p and q are not in their lowest terms, contradicting Orginal solution

Thus there is no rational solution to this!