1. ## Practice questions

Hey all, I have a test next week after Spring Break and would like some extra help with these problems in order to make sure I am prepared for the actual test!

1. Use a proof by cases to show that for all real numbers x and y, the absolute value of X+Y is less than or equal to the absolute value of X + the absolute value of Y.

For this I am thinking simply breaking into cases using X is + or -, and Y is + or -. That seems fairly simple. If X is - and Y is - the absolute value of X+Y is equal to the absolute value of X + the absolute value of X. The same applies for X and Y both being +. If one is + and the other -, the absolute value of X + Y is going to be less than the absolute value of X + the absolute value of Y. Am I missing anything here? Once I actually write out my proof I'll be sure to post it to make sure it makes sense.

2. Using problem 1, prove that for all real numbers A and B, the absolute value of A-B is greater than or equal to the absolute value of A - the absolute value of B.

Is this not essentially the exact same thing as 1?

3. Use a contradiction argument to show that for all positive real numbers X,Y, and Z, if X>Z and Y^2=XZ, then X>Y>Z. Hint: X>Y>Z is equivalent to saying X>Y and Y>Z.

Contradiction arguments use the general form of p and not q implies r and not r, correct? So for this problem it would be if X>Z and Y^2=XZ and X less than or equal to Y less than or equal to Z. Is this correct?

4. Prove that if D is an odd integer that divides both A+B and A-B, then D divides both A and B.

The definition of divides is that there A|B if there is some integer K we can multiply A by to get B, correct? So for this problem it would be if K_1D=A+B and K_2D=A-B then K_3D=A and K_4D=B correct? (The underscores represent subscripts).

5.Prove that N is odd if and only if N^2 is odd.

The definition of odd is that a number N can be represented by 2K_1+1 yes? So for an if and only if argument we prove if N is odd then N^2 is odd, then we prove the converse of if N^2 is odd, N is odd?

6. Prove that the sum of two odd numbers is even.

So I simply add (2K_1 +1)+(2K_2+1) and simply from there?

7. Prove by way on contradiction that for integers A,B, and C, if A^2+B^2=C^2, then at least one of A and B is even. Hint, use problem 5 and 6.

This seems pretty overwhelming but I'm sure it will be easier once I have a better grasp of 5 and 6.

8. Prove that for all real numbers X and Y, if Y^3+YX^2 less than or equal to X^3+XY^2, then Y less than or equal to X. Hint, use a proof by contrapositive.

Contrapositive proofs reverse and negate the implication right? So for this it would be if Y>X, then Y^3+YX^2>X^3+XY^2?

9. Prove that the square root of 3 is irrational.

A rational number can be expressed as A/B where A and B are both integers. I'm not sure how to start this as I don't know how to express an irrational number.

10. 7^n -2^n is divisible by 5 for all natural numbers n. Hint, add a clever form of zero when doing the inductive step.

I can set the base case up for this fairly easily but I have no idea what the clever form of zero is during the inductive step.

11. (1+1/2)^n greater than or equal to 1+n/2.

This is another induction problem but I don't know how to deal with the inequality. Are there any particular tips I should use for these that don't apply to equalities?

12.Suppose x+1/x is an integer. Prove that x^2+1/x^2,...x^n+1/x^n are all integers.

Another induction problem. Once I really get into it I imagine I'll come back with some questions but nothing seems terribly awful about it just looking at it.

13. Let P and Q be positive integers with Q>P. Prove that Q-P divides Q-1 if and only if Q-P divides P-1.

I don't even know what process to use on this one. I know the if and only if means I will be using the converse. But where do I even start?

14. A student claims to have shown that 2=1. Critique the student's proof by finding the error in their argument. Explain your reasoning below.

Proof: Let a=b be an integer. Then we have the following.
1. a=b
2. a^2=ab
3. a^2-b^2=ab-b^2
4. (a-b)(a+b)=b(a-b)
5. a+b=b
6. 2b=b
7. 2=1

Things start to look strange around step 4/5. However, I'm not sure what is wrong. Obviously 2 doesn't equal 1...but I don't see the error.

Thanks for anyone willing to help me out, I would like to do well on the test

2. ## Re: Practice questions

Originally Posted by Zerrigan
Hey all, I have a test next week after Spring Break and would like some extra help with these problems in order to make sure I am prepared for the actual test!

14. A student claims to have shown that 2=1. Critique the student's proof by finding the error in their argument. Explain your reasoning below.

Proof: Let a=b be an integer. Then we have the following.
1. a=b
2. a^2=ab
3. a^2-b^2=ab-b^2
4. (a-b)(a+b)=b(a-b)
5. a+b=b
6. 2b=b
7. 2=1

Things start to look strange around step 4/5. However, I'm not sure what is wrong. Obviously 2 doesn't equal 1...but I don't see the error.

Thanks for anyone willing to help me out, I would like to do well on the test
if $a=b$ then what does $a-b$ equal?
In going from step 4 to step 5 they divide by $a-b$. Is this allowed?

3. ## Re: Practice questions

Thanks for the help! I'm slowly working through the packet.

4. ## Re: Practice questions

So here is what I have for 1-7.

1. By definition, the absolute value of a number k is as follows. |k|=k when k>=0 and |k|=-k when k<0.

Case 1. Let X and Y both be >=0. Thus, |X|=X and |Y|=Y. Since X+Y>=0, both |X|+|Y|>=0 and |X+Y|>=0. Therefore, |X|+|Y|=|X|+|Y| and |X|+|Y|<=|X+Y|.

Case 2. Let X and Y both be <=0. Thus, |X|=-X and |Y|=-Y. Hence, |X|+|Y|=-X-Y=-(X+Y). However, since the absolute value of a negative number is the opposite of the number, for X,Y<0, -(X+Y)=|X+Y|. Therefore, |X|+|Y|=-(X+Y)=|X+Y| and |X|+|Y|<=|X+Y|.

Case 3. Let X>=0 and Y<0. Thus, |X|=X and |Y|=-Y. Hence |X|+|Y|=X-Y. Not that |X+Y|=X+Y when X+Y>=0 and |X+Y|=-(X+Y) when X+Y<0. Since X>=0 and Y<0, X-Y>=X+Y and -X-Y=-(X+Y)>=-X+Y. Either way, |X+Y|<=|X|+|Y|.

Case 4. Let X<0 and Y>=0. Using case 3, |X|=-X and |Y|=Y. Hence |X|+|Y|=Y-X. Since X<0 and Y>=0, Y-X>=Y+X and -Y-X=-(Y+X)>=-Y+X. Either way, |X+Y|<=|X|+|Y|.

2. Using problem 1,

Case 1. Let |A|>=|B|. Also, let A=A-B+B. Thus, |A|=|A-B+B|. Using problem one, |A|=|A-B+B|<=|A-B|+|B|. Therefore, |A|-|B|<=|A-B|.

Case 2. Let |B|>=|A|. Also, let B+B-A+A. Thus, |B|=|B-A+A|. Using problem one, |B|=|B-A+A|<=|B-A|+|A|. Therefore, |B|-|A|<=|B-A|.

Assume if X>Z and Y^2=XZ then X>Y>Z and X<=Y<=Z. Thus, we have Z>Y>X. Hence, Z>X. However, this contradicts the hypothesis of X>Z.

Is that really all there is to 3?

4. If D is odd then D=2N+1. Also, if D is even then D=2N. Note, if D|A+B then there is some integer K_1 we can multiply by D to get A+B. This means there is also some integer K_2 we can multiply by D to get A-B.

Knowing D=2N+1 and that D|A+B and D|A-B, we know that there are some integers K_1 and K_2 such that (A+B)=(2N+1)K_1 and (A-B)=(2N+1)K_2.

Note that (A+B)+(A-B)=2A, which by definition of even numbers, is even. Also, note that 2A=(2N+1)(K_1+K_2). Therefore, since 2A is even, (2N+1)(K_1+K_2) must be even. Hence, let K_1+K_2=2K_3. Thus, 2A=4NK_3+2K_3 and A=K_3(2N+1). Note that D=2N+1 so A=K_3(2N+1). Therefore, D|A.

Now, note that (A+B)-(A-B)=2B, which by definition of even numbers, is even. Also, note that 2B=(2N+1)(K_2-K_1). Therefore, since 2B is even, (2N+1)(K_2-K_1) must be even. Hence, let K_2-K_1=K_4. Thus, 2B=4NK_4+2K_4 and B=K_4(2N+1). Note that D=2N+1 so B=K_4(2N+1). Therefore, D|B.

5. If a number K is odd, then K=2J+1 for some integer J.

If N is odd, then N=2K_1+1 for some integer K_1. Thus, N^2=(2K_1+1)^2=4(K_1)^2+4K_1+1. Let 4(K_1)^2+4K_1=2K_2 for some integer K_3. Thus, N^2=2k_2+1. Therefore, if N is odd, N^2 is odd.

By contraposition, if N is even, then N=2K_3 for some integer K_3. Thus, N^2=(2K_3)^2=4(K_3)^2. Let 4(K_3)^2=2K_4 for some integer K_4. Hence, N^2=2k_4. Therefore, if N is even, N^2 is even, and by contraposition, if N^2 is odd, then N is odd.

6. Let X and Y both be odd integers. Thus, X=2K_1+1 and Y=2K_2+1 for some integer K_1 and K_2. Hence, X+Y=2K_1+1+2K_2+1=2K_1+2K_2+2=2(K_1+K_2+1) where K_1+K_2+1=K_3 for some integer K_3. Hence, X+Y=2K_3, which is even.

7. By contradiction, if A^2+B^2=C^2 and both A and B are odd, then by problem 5 if A^2 is odd, then A is odd and if B^2 is odd, then B is odd. Also, A^2+B^2=C^2 which is even by 6. By 5, if C^2 is even, then C is even. Hence, C=2K_1 and C^2=4(K_1)^2=4M for some integers K_1 and M.

Note that A^2+B^2=(2K_2+1)^2+(2K_3+1)^2 for some integers K_2 and K_3. Thus, A^2+B^2=4((K_2)^2+K_2+(K_3)^2+K_3)+2=C^2. Hence, C^2=4K_4+2 for some integer K_4=(K_2)^2+K_2+(K_3)^2+K_3. However, 4M doesn't equal 4K_4+2 for integers K_4 and M. Therefore, by contradiction, A or B must be even.

Thanks again for all the help!