P4.4.2 Give a rigorous proof, using strong induction, that every positive natural has at least one...

P4.4.2 Give a rigorous proof, using strong induction, that every positive natural has at least one factorization into prime numbers.

Expert Solution

Colby Messinger answered 1 year ago

Prove that the proof by mathematical induction and the proof by strong induction are equivalent

Use strong induction to show that every positive integer, n, can be written as a sum...

Use strong induction to show that every positive integer, n, can be written as a sum of powers of two: 20, 21, 22, 23, .....

Use strong induction to show that every positive integer n can be written as a sum...

Use strong induction to show that every positive integer n can be written as a sum of distinct powers of two, that is, as a sum of a subset of the integers 2^0 =1, 2^1 = 2, 2^2 = 4, and so on. [Hint: For the inductive step, separately consider the case where k + 1 is even and where it is odd. When it is even, note that (k + 1)/2 is an integer.]

Give a proof by induction that the number of nodes of a full binary tree with...

Give a proof by induction that the number of nodes of a full binary tree with a height of "h" is: (2^(h+1))-1.

Please be able to follow the COMMENT Use induction proof to prove that For all positive...

Please be able to follow the COMMENT Use induction proof to prove that For all positive integers n we have the inequality n<=2^n here is the step: base step: P(1)= 1<=2^1 inductive step: k+1<= 2^(k)+1 <= 2^(k)+k (since k>=1) <= 2^(k)+2^(k) = 2X2^(k) =2^(k+1) i don't understand why 1 can be replaced by k and i don't know why since k>=1

Regular => Context-Free Give a proof by induction on regular expressions that: For any language A,...

Regular => Context-Free Give a proof by induction on regular expressions that: For any language A, if A is regular, then A is also context-free. You may assume that the statements from the previous Closure under Context-Free Languages problem are proved. R is a regular expression if RR is: a for some a∈alphabet Σ, the empty string ε, the empty set ∅, R1∪R2, sometimes written R1∣R2, where R1 and R2 are regular expressions, R1∘R2, sometimes written R1R2, where R1 and...

1a. Proof by induction: For every positive integer n, 1•3•5...(2n-1)=(2n)!/(2n•n!). Please explain what the exclamation mark...

1a. Proof by induction: For every positive integer n, 1•3•5...(2n-1)=(2n)!/(2n•n!). Please explain what the exclamation mark means. Thank you for your help! 1b. Proof by induction: For each integer n>=8, there are nonnegative integers a and b such that n=3a+5b

5 Regular => Context-Free Give a proof by induction on regular expressions that: For any language...

5 Regular => Context-Free Give a proof by induction on regular expressions that: For any language A, if A is regular, then A is also context-free. You may assume that the statements from the previous Closure under Context-Free Languages problem are proved.

C. Prove the following claim, using proof by induction. Show your work. Let d be the...

C. Prove the following claim, using proof by induction. Show your work. Let d be the day you were born plus 7 (e.g., if you were born on March 24, d = 24 + 7). If a = 2d + 1 and b = d + 1, then an – b is divisible by d for all natural numbers n.

Let d1, d2, ..., dn, with n at least 2, be positive integers. Use mathematical induction...

Let d1, d2, ..., dn, with n at least 2, be positive integers. Use mathematical induction to explain why, if d1+ d2+…+dn = 2n-2, then there must be a tree with n vertices whose degrees are exactly d1, d2, ..., dn. (Be careful with reading this statement. It is not the same as saying that any tree with vertex degrees d1, d2, ..., dn must satisfy d1+ d2+...+dn = 2n-2, although this is also true. Rather, it says that if...

Question