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Artist 52 [7]
3 years ago
7

Use strong mathematical induction to prove the existence part of the unique factorization of integers theorem (Theorem 4.4.5). I

n other words, prove that every integer greater than 1 is either a prime number or a product of prime numbers.
Mathematics
1 answer:
valentina_108 [34]3 years ago
5 0

Answer:

Lets say that P(n) is true if n is a prime or a product of prime numbers. We want to show that P(n) is true for all n > 1.

The base case is n=2. P(2) is true because 2 is prime.

Now lets use the inductive hypothesis. Lets take a number n > 2, and we will assume that P(k) is true for any integer k such that 1 < k < n. We want to show that P(n) is true. We may assume that n is not prime, otherwise, P(n) would be trivially true. Since n is not prime, there exist positive integers a,b greater than 1 such that a*b = n. Note that 1 < a < n and 1 < b < n, thus P(a) and P(b) are true. Therefore there exists primes p1, ...., pj and pj+1, ..., pl such that

p1*p2*...*pj = a

pj+1*pj+2*...*pl = b

As a result

n = a*b = (p1*......*pj)*(pj+1*....*pl) = p1*....*pj*....pl

Since we could write n as a product of primes, then P(n) is also true. For strong induction, we conclude than P(n) is true for all integers greater than 1.

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Please help me answer all parts of question 3 in the photo.
Nookie1986 [14]
(i.) CA = πrl
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Find the area of a circle with a circumference of 18.48 units.
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There is a photo showing the answer and the formula and stuff hope this helps

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D-5x%2B6%7D%7B2x%5E%7B2%7D-7x%2B6%20%7D" id="TexFormula1" title="\frac{x^{
il63 [147K]

Answer:

\frac{x-3}{2x-3}. hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

x^2-5x+6

Now let's find what ac is (it's just c since a=1...)

AC= 6

List factors of -6

\pm1, \pm2, \pm3, \pm6.

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

(x-2)(x-3)

Ok now let's factor the second equation:

2x^2-7x+6

Multiply a and c

AC = 12

List factors of 12:

\pm1, \pm2, \pm3, \pm4, \pm6, \pm12.

Factors that add up to -7 and multiply to 12:

-3\ and\ -4

Rewrite equation:

2x^2-4x-3x+6

Group terms:

(2x^2-4x)+(-3x+6)

Factor out GCF:

2x(x-2)-3(x-2)

Rewrite:

(2x-3)(x-2)

Now let's write out the equation using these factors:

\frac{(x-2)(x-3)}{(2x-3)(x-2)}.

Here we can factor out the x-2 and the simplified form is:

\frac{x-3}{2x-3}

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0
2 years ago
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