Your Algebra Homework Can Now Be Easier Than Ever!

Math 111 Practice Midterm II Solutions

For each problem, prove the statement. Indicate what type of proof (trivial, vacuous, direct, by
contrapositive, or by contradiction) you are using.

1. Let x be a real number .

(a) If x >-7, then -5-x^2 < 0.

For any real number x , x≥0. Therefore -x^2≤0, and-5-x^2 ≤-5 + 0 = -5 < 0.
(This is a trivial proof.)

(b) If |x| = 5, then x^2 + x + 1 > 20.

If  |x| = 5, then either x = 5 or x = -5. Thus we can consider the following two
cases:

Case 1. x = 5.

Then x^2 + x + 1 = 5^2 + 5 + 1 = 31 > 20.

Case 2. x = -5.

Then x^2 +x+1 = (-5)^2 +(-5)+1 = 21 > 20. (This is a direct proof by cases.)

(c) If 2x > x^2 + x^3, then x < 1.

We will prove this statement by contrapositive. Suppose x≥1. Then x^2≥x and
x^3≥x. Adding these two inequalities gives x^2 + x^3≥2x; thus

2. Let n and m be integers.

(a) If 3n^2 + 5n is odd, then n≥10.

We will show that for any integer n, the number 3n^2 +5n is even. To do this, we will
consider two cases :

Case 1. n is even.

Then n = 2k for some k ∈ Z. Therefore 3n^2+5n = 3(2k)^2+5(2k) = 12k^2+10k =
2(6k^2 + 5k). Since 6k^2 + 5k ∈ Z, the number 3n^2 + 5n is even.

Case 2. n is odd.

Then n = 2k + 1 for some k ∈ Z, and 3n^2 + 5n = 3(2k + 1)^2 + 5(2k + 1) =
12k^2+12k+3+10k+5 = 12k^2+22k+8 = 2(6k^2+11k+4). Since 6k^2+11k+4 ∈ Z,
the number 3n^2 + 5niseven.

Since 3n^2 + 5n is never odd, the implication follows. (This is a vacuous proof.)

(b) If n is even, then 3n^2-2n -5 is odd.

Suppose n is even. Then n = 2k for some k ∈Z. Therefore

Since 6k^2-2k-3 ∈ Z, the number 3n^2-2n-5 is odd. (This is a direct proof.)

(c) If 7n^2 + 4 is even, then n is even.

We will prove this statement by contrapositive. Suppose n is odd. Then n = 2k + 1
for some k ∈ Z. Therefore 7n^2 + 4 = 7(2k + 1)^2 + 4 = 7(4k^2 + 4k + 1) + 4 =
28k^2 +28k +11 = 2(14k^2 +14k +5)+1. Since 14k^2 +14k +5 ∈ Z, 7n^2 +4 is odd.

(d) If n-5m is odd, then n and m are of opposite parity.

We will prove the statement by contrapositive. Thus we will prove that if n and m
are of the same parity, then n-5m is even.

Case 1. n and m are both even.

Then n = 2k and m = 2l for some k,l ∈ Z. Therefore n-5m = 2k-5(2l) =
2k-10l = 2(k-5l). Since k-5l ∈ Z, the number n -5m is even.

Case 1. n and m are both odd.

Then n = 2k + 1 and m = 2l + 1 for some k,l ∈ Z. Then n -5m = 2k + 1 -
5(2l +1) = 2k +1-10l-5 = 2k-10l-4 = 2(k-5l-2). Since k-5l-2 ∈ Z,
the number n -5m is even.

(e) If 5 | (n-1), then 5 | (n^3 + n-2).

Suppose 5 | (n-1). Then (mod 5). Therefore (mod 5).
This implies that 5 | (n^3 + n-2). (This is a direct proof.)

Another proof: Suppose 5 | (n-1). Then n-1 = 5k for some k ∈ Z. Therefore
n = 5k+1, and we have n^3 +n-2 = (5k+1)^3 +(5k+1)-2 = 125k^3 +75k2 +15k+
1+5k+1-2 = 125k^3+75k^2+20k = 5(25k^3+15k^2+4k). Since 25k^3+15k^2+4k ∈ Z,
we have 5 | n^3 + n-2. (This is also a direct proof.)

(f) 3 | mn if and only if 3 | m or 3 | n

We have two implications to prove.

-> Suppose 3 | mn. Show that 3 | m or 3 | n.
We will prove this statement by contrapositive. Suppose . Then
m = 3k + c and n = 3l + d for some k, l, c, d ∈ Z where c and d are equal to
either 1 or 2. We have

Case 1. c = d = 1

Then cd = 1, and mn = 3(3kl + k + l) + 1. Thus

Case 2. WLOG c = 2, d = 1.

Then cd = 2, and mn = 3(3kl + 2k + l) + 2. Thus

Case 3. c = d = 2

Then cd = 4, and mn = 3(3kl+2k +2l)+4 = 3(3kl+2k +2l+1)+1. Thus

<-Suppose 3 | m or 3 | n. Show that 3 | mn.

We will prove this statement directly. Without loss of generality, suppose 3 | m.
Then m = 3k for some k ∈ Z. Therefore mn = 3kn is divisible by 3.

3. The number log 32 is irrational.

Suppose log 32 is rational. Then log32 = m/n for some m,n ∈ Z, n > 0. Therefore 3m/n = 2,
so 3m = 2n. Since n > 0, 3m = 2n > 1. We know that 3m is odd since it is a product
of odd integers (or you can prove a little lemma here that says, If a is odd and m is a
positive integer , then am is odd.). But we also know that 2n is even since it is a product
of even integers (or you can prove another similar lemma if you are not convinced). This
is a contradiction, since an odd number cannot be equal to an even number. Therefore
log 32 is irrational. (This is a proof by contradiction.)

4. The product of a nonzero rational number and an irrational number is irrational.

Suppose there exist a nonzero rational number x and an irrational number y such that xy
is rational . Then x = k/l for some k,l ∈ Z, k, l ≠ 0 and xy = m/n for some m, n ∈Z, m≠0

We haveSince ml,nk ∈Z and nk ≠ 0, y is rational . Contradiction.
(This is a proof by contradiction.)

5. Let A and B be sets. Thenif and only if

We have two implications to prove.

-> SupposeShow that

We will prove this statement by contrapositive. Suppose

Then there exists an elementThereforeand

Thus x = (y, z) for some(and It follows that)

<-SupposeShow that

We will prove this statement by contrapositive as well. SupposeThen
there existsSince x ∈ A and x ∈ B, we haveand

Thus

Prev Next

Start solving your Algebra Problems in next 5 minutes!

Algebra Helper
Download (and optional CD)

Only $39.99

Click to Buy Now:


OR

2Checkout.com is an authorized reseller
of goods provided by Sofmath

Attention: We are currently running a special promotional offer for Algebra-Answer.com visitors -- if you order Algebra Helper by midnight of November 23rd you will pay only $39.99 instead of our regular price of $74.99 -- this is $35 in savings ! In order to take advantage of this offer, you need to order by clicking on one of the buttons on the left, not through our regular order page.

If you order now you will also receive 30 minute live session from tutor.com for a 1$!

You Will Learn Algebra Better - Guaranteed!

Just take a look how incredibly simple Algebra Helper is:

Step 1 : Enter your homework problem in an easy WYSIWYG (What you see is what you get) algebra editor:

Step 2 : Let Algebra Helper solve it:

Step 3 : Ask for an explanation for the steps you don't understand:



Algebra Helper can solve problems in all the following areas:

  • simplification of algebraic expressions (operations with polynomials (simplifying, degree, synthetic division...), exponential expressions, fractions and roots (radicals), absolute values)
  • factoring and expanding expressions
  • finding LCM and GCF
  • (simplifying, rationalizing complex denominators...)
  • solving linear, quadratic and many other equations and inequalities (including basic logarithmic and exponential equations)
  • solving a system of two and three linear equations (including Cramer's rule)
  • graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions)
  • graphing general functions
  • operations with functions (composition, inverse, range, domain...)
  • simplifying logarithms
  • basic geometry and trigonometry (similarity, calculating trig functions, right triangle...)
  • arithmetic and other pre-algebra topics (ratios, proportions, measurements...)

ORDER NOW!

Algebra Helper
Download (and optional CD)

Only $39.99

Click to Buy Now:


OR

2Checkout.com is an authorized reseller
of goods provided by Sofmath
Check out our demo!
 
"It really helped me with my homework.  I was stuck on some problems and your software walked me step by step through the process..."
C. Sievert, KY
 
 
Sofmath
19179 Blanco #105-234
San Antonio, TX 78258
Phone: (512) 788-5675
Fax: (512) 519-1805
 

Home   : :   Features   : :   Demo   : :   FAQ   : :   Order

Copyright © 2004-2024, Algebra-Answer.Com.  All rights reserved.