Can someone please help me with this

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

valentina_108 [34]

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.

5 0

2 years ago

A triangle has side lengths of 9 in, 13 in, and 20 in. What is the measurement of this triangle’s largest angle?

zvonat [6]

1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

$a^{2}=b ^{2}+c ^{2}-2bc(cosA)$

2.

$20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)

400=81+169-234(cosA) 150=-234(cosA)

cosA=150/-234= -0.641$

3. m(A) = Arccos(-0.641)≈130°,

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc

4 0

3 years ago

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For right triangle PQR what is the measure of PQ to the nearest hundredth of a centimeter?

antoniya [11.8K]

Angle PQR=90 degrees.
cos 52=PQ/14.72
PQ=14.72 x (cos 52)
PQ= 9 cm to the nearest cm

6 0

3 years ago

Read 2 more answers

Find the distance between xequals[Start 2 By 1 Matrix 1st Row 1st Column 8 2nd Row 1st Column negative 5 EndMatrix ]and yequals[

kherson [118]

Answer:

Distance =√(x₁ - y₁)²+ (x₂ - y₂)² = √97 = 9.85

Step-by-step explanation:

The Matrix X and Y could also be referred to as vectors in Rⁿ dimensions.

if Vector X = ( x₁ , x₂) and Vector Y = (y₁ , y₂)

then, Distance (X-Y) = ||X-Y|| = √(x₁ - y₁)²+ (x₂ - y₂)²

where, x₁ = 8, x₂ = -5 and y₁ = -1 , y₂ = -9

Distance = √(8 - (-1))²+ (-5 - (-9))² = √9² + 4² =√97 = 9.85

4 0

3 years ago

Amanda needs to make a garden plot which has an area less than 18 sq. feet. The length should be 3 feet longer than the width. W

SCORPION-xisa [38]

Answer:

Any width less than 3 feet

Step-by-step explanation:

Inequalities

The garden plot will have an area of less than 18 square feet. If L is the length of the garden plot and W is the width, the area is calculated by:

A = L.W

The first condition can be written as follows:

LW < 18

The length should be 3 feet longer than the width, thus:

L = W + 3

Substituting in the inequality:

(W + 3)W < 18

Operating and rearranging:

$W^2 + 3W - 18 < 0$

Factoring:

(W-3)(W+6)<0

Since W must be positive, the only restriction comes from:

W - 3 < 0

Or, equivalently:

W < 3

Since:

L = W + 3

W = L - 3

This means:

L - 3 < 3

L < 6

The width should be less than 3 feet and therefore the length will be less than 6 feet.

If the measures are whole numbers, the possible dimensions of the garden plot are:

W = 1 ft, L = 4 ft

W = 2 ft, L = 5 ft

Another solution would be (for non-integer numbers):

W = 2.5 ft, L = 5.5 ft

There are infinitely many possible combinations for W and L as real numbers.

6 0

2 years ago