practice problems pending

Q1. if p | ab then either p divides a or p divides b. 'p' is a prime.
S1.
Case 1: p | a: done
Case 2: p doesn't divide a:
gcd(a,p) = 1
=> au + pv = 1
Multiply by 'b':
bau + bpv = b
p divides both terms on LHS => p | RHS => p | b
H.P.

Corollary:
If p | a1a2....an then p divides some ai.

Q2.
Theorem: A composite number has at least one prime divisor.
Prove this.
S2.

A set of all its +ve divisors except 1 and itself.

Let (S) be the set.

As the no. must have some other divisors than 1 & itself, then

(S) is non-empty and S is subset of N(natural numbers).

So by well ordering property (S) has a least element (l).

If (l) is a prime then we are done.

But if (l) is not a prime then there exists d | l (d divides l).

C is the composite number we are talking about.
d | l and l | C => d | C
=> d is in S and d < l which is a contradiction.

Q3.
find all possible values of 'p' s.t. (p) and (p^2 + 8) both are prime.
S3.
Case 1: p mod 3 = 0 => p can only be 3 => 9+8 = 17 is prime
Case 2: p mod 3 = 1,4 => p^2 mod 3 = 1 => p^2 + 8 mod 3 = 0, no good
So only p = 3 works.

Q4. Prove Fundamental Theorem of Arithmetic:
Each positive integer is either 1 or a prime or a composite that can be expressed as a product of primes.
S4.
Prove by induction.
1 is in a category of itself.
Prove for n>= 2.
Base case: 2. It's a prime. So it holds.
Now assume it's true for 2,3,4.... k
Come to k+1
Case 1: it's a prime we are done.
Case 2: it's a composite number => k+1 = a*b where both a,b are > 1. 
And both a,b are strictly less than k+1.
By our inductive hypothesis, both a,b are made up by multiplication of prime numbers or they are prime numbers themselves.
In both cases k+1 can be expressed as product of primes.
H.P.

Q5.
Prove that there are infinitely many primes.
S4.
Let there be a finite number or primes:
p1,p2 ... pk

Let N = p1.p2....pk + 1
N is co-prime to each prime in our list.
So there are 2 cases(using fundamental theorem of arithmetic and N > 1):
Case 1: N is a new prime.
Case 2: N is a composite number made up of primes which are not in our list.

Both cases => that there are more primes than what we have in our list.









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