ISE-112 - Fall 1999

Homework Assignment 3

Date Due: in lecture October 12, 1999
Late Deadline: in lecture October 19, 1999

1.

Prove the following, or disprove by finding a counterexample

(a)

There are real numbers a and b such that $\sqrt{a+b}=a+b$.

One such example is a=0 and b=0, therefore the statement is true

(b)

There is a real number x so that 2^x > x¹⁰

x=0 is such a number since 2⁰=1 and 0¹⁰=0.

(c)

Definition: an integer n is called a perfect square iff n=k² for some integer k. Prove there is a perfect square that can be written as a sum of two other perfect squares.

from the pythagorean theorem 3² + 4² = 5².

(d)

The sum of any two odd integers is an even integer

write the odd integers as n=2j+1 and m=2k+1. Then their sum is n+m = 2j+1+2k+1 = 2(j+k+1) which is obviously divisible by two.

(e)

For all positive integers n, if n is prime then n is odd.

This is not true, because 2 is a prime number and a positive integer

2.

Find the mistakes in the proof:

Theorem: The product of an even integer and an odd integer is even.
proof: ``Suppose m is an even integer and n is an odd integer. If $m\cdot n$ is even, then by definition of even there exists an integer r such that $m \cdot n = 2r$. Also, since m is even, there exists an integer p such that m = 2p and since n is odd there exists an integer q such that n=2q+1. Thus

$$mn=(2p)\cdot (2q+1) = 2r$$

where r is an integer. By definition of even, then, mn is even, as was to be shown.''

the mistake is that the proof begs the question: it starts out by supposing that mn is even, which is exactly what you're supposed to be proving.

3.

For each of the following, answer the question, and prove why that is so. Assume all variables are integers.

(a)

Is 52 divisible by 13?

yes: $52 = 13\cdot 4$.

(b)

Is 51 divisible by 17?

yes: $51 = 17 \cdot 3$.

(c)

Is $(3k+1)\cdot (3k+2)\cdot (3k+3)$ divisible by 3?

yes: $(3k+1)\cdot (3k+2)\cdot (3k+3) = (3k+3)\cdot (3k+2)\cdot (3k+1) = 3\cdot(k+1)\cdot (3k+2)\cdot (3k+1)$ which is clearly divisible by 3.

(d)

Is 2m (2m + 2) divisible by 4?

yes: 2m (2m + 2) = 4m(m+1) = 4(m² +m) which is divisible by 4.

(e)

Is 6a (a + b) a multiple of 3a?

yes: $6a(a+b) = 2 \cdot 3a(a+b) = 3a(2a+2b)$ which is a multiple of 3a.

(f)

If a is divisible by b and a is divisible by c then a is divisible by bc

no: if a=10, b=5, c=5 then a is divisible by b and a is divisible by c but a is not divisible by bc

(g)

If a is divisible by b and a is divisible by c then a is divisible by b+c

no: if a=15, b=3, c=5 then a is divisible by b and a is divisible by c but a is not divisible by b+c

(h)

If a is divisible by b and a is divisible by c then a is divisible by b-c

no: if a=10, b=5, c=2 then a is divisible by b and a is divisible by c but a is not divisible by b-c

(i)

If a is divisible by b and a is divisible by c then a is divisible by b+c

no: if a=10, b=5, c=2 then a is divisible by b and a is divisible by c but a is not divisible by b+c

(j)

if $2 \cdot 3 \cdot 4 \cdot 5 \cdot n = 29 \cdot 28 \cdot 27 \cdot 26 \cdot 25$, then is n divisible by 29? Why?

yes: 29 is a prime number. Yet $2 \cdot 3 \cdot 4 \cdot 5 \cdot n$ is divisible by 29 since it's equal to $29 \cdot 28 \cdot 27 \cdot 26 \cdot 25$, since 29 is a prime number you can't multiply two integers together to get it, which means that 29 must be a factor of n which is the same thing as saying n is divisible byu 29. Another (simpler) way is like this:

$$2 \cdot 3 \cdot 4 \cdot 5 \cdot n = 29 \cdot 28 \cdot 27 \cdot 26 \cdot 25$$

$$3 \cdot 4 \cdot 5 \cdot n = 29 \cdot 14 \cdot 27 \cdot 26 \cdot 25$$

$$4 \cdot 5 \cdot n = 29 \cdot 14 \cdot 9 \cdot 26 \cdot 25$$

$$2 \cdot 5 \cdot n = 29 \cdot 7 \cdot 9 \cdot 26 \cdot 25$$

$$5 \cdot n = 29 \cdot 7 \cdot 9 \cdot 13 \cdot 25$$

$$n = 29 \cdot 7 \cdot 9 \cdot 13 \cdot 5$$

which clearly shows n is divisble by 29.

(k)

Use the unique factorization theorem to write 4851 and 8925 in standard factored form.

$4851 = 3^2 \cdot 7^2 \cdot 11, 8925=3 \cdot 5^2 \cdot 7 \cdot 17$