The GCD(Ax,By) by definition divides Ax ;as a result Ax must be some multiple of

Question

The GCD(Ax,By) by definition divides Ax ;as a result Ax must be some multiple of GCD(Ax,By) i.e M*GCD(Ax,By) where M is some integer The GCD(Ax;By) by definition evenly divides By; As a result; By must be some multiple of GCD(Ax;By); i:e:NGCD(Ax;By) = By whereN is someinteger case 1 M*GCD(Ax;By) + N*Gcd(Ax;By) (M + N)*GCD(Ax;By) = D So we can see that GCD(Ax;By) evenly dividesD: case 2 M*gcd (Ax;By) - N * gcd(Ax;By) = EdividesE case 3 M*gcd (Ax;By)/N * gcd(Ax;By) = F results 1dividesF case 4 M*gcd (Ax;By) * N *gcd(Ax;By) = G evenly dividesG Ax;By multiple of D/E/F/G; D,E,F,G < A;D,E,F,G < Bcondition as D/E/F/G divisor then A;B multiple of D/E/F/G/,Ax+By=Cz results Cy multiple of D results C multiple of D/E/F/G induction D/E/F/G Think of it this way: Let P(n) be the statement that n can be written as the product of prime numbers. Then the proposition says: P(n) is true for each integer greater or equal to 2. Proof (by strong induction). (1) Base case: 2 is a prime, so it is the product of a single prime. (2) Strong inductive step: Suppose for some k 2 that each integer n with 2 n k may be written as a product of primes. We need to prove that k + 1 is a product of primes. Case (a): Suppose k + 1 is a prime. Then we are done. Case (b): Suppose k + 1 is a not prime. Then by the fact stated above, there exist integers a and b with 2 a, b k such that k + 1 = a _ b. By the strong inductive hypothesis, since 2 a, b k, both a and b are the product of primes. Thus k + 1 = a *b is the product of primes. then exist a subgroup of primes .is it correct as a proof for beal conjecture?

Explanation / Answer

{Beal’s conjecture states that if A^x + B^y = C^z where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.}

The strong induction principle says that you can prove a statement of the form:

P(n) , for each positive integer n,
as follows:
Base case:P(1) is true.
Strong inductive step:Suppose k is a positive integer such that

P(1),P(2)......P(k) are all true. Prove that P(k+ 1) is true.

So the key step is to show:

P(1),P(2).....and P(k) =)P(k+ 1):So to speak, the statement is true if you can prove that:

(1) The rst domino has fallen.

(2) If k is such that the rst k dominos have fallen,then the (k+ 1)th domino has fallen.

Thus the above proof of beal conjecture is complete by strong induction principle

hope this helps!!

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.