please try to solve these questions.

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- Dec 15th 2006, 09:31 PMm777induction,irrational and proves......
please try to solve these questions.

- Dec 16th 2006, 04:01 AMtopsquark
1a) What the heck is the definition of Q?

1b) Any even integer is of the form where n is some integer. (This is how an even number is defined.) Now square this:

Now, since n is an integer, so is . Again, since 2 and [tex]n^2 are integers so is . Thus is an integer it is also an even integer.

-Dan - Dec 16th 2006, 04:20 AMSoroban
Hello, m777!

Here's some help . . .

Quote:

Use mathematical induction to prove:

. . for any positive integer .

Verify . . . true!

Assume

Add to both sides:

. .

. . This is the left side of

The right side is: .

. . which is the right side of . . . We're done!

- Dec 16th 2006, 04:37 AMSoroban
Hello again, m777!

Quote:

3) Prove by induction that is divisible by 3.

Verify is divisible by 3 . . . true!

Assume is divisible by 3.

. . Then: . for some integer .

Add to both sides:

. .

We have: .

Therefore: . is divisible by 3 . . . We're done!

- Dec 16th 2006, 04:44 AMm777reply to topsquark
Hello,

Topsquark.

In part(a) of question no 1 i have done this so far.

Q(a,b)= o if a<b.

Q(a-b,b)+1 if b is less than equal to a.

so

Q(3,4)=0 SINCE 3<4

GIVEN Q(a,b)= Q(a-b,b)+1) if b is less than equal to a

NOW

Q(14,3)=Q(11,3)+1

=Q(8,3)+1

=Q(5,3)+1

=Q(2,3)+1

SO AT THAT POINT 2 IS LESS THAN 3.

THEREFORE

=0+4

=4 IS THE ANSWER - Dec 16th 2006, 02:00 PMThePerfectHacker
To prove is irrational.

Assume,

Thus,

Thus,

.

Prime decompose .

Prime decompose .

Note,

has an even number of prime factors.

Note,

has an even number of prime factors.

But,

has an odd number of primes factors.

Thus, we have a contradiction because both sides are note equal by the uniquess part of the fundamental theorem of arithmetic. - Dec 16th 2006, 03:05 PMPlato
This is a very general proof.

If p is a positive integer that is not a perfect square then is irrational.

Suppose that it is rational, where a & b are integers.

Let

By well ordering there is a first term in T, call it K.

Using the floor function and because p is not a square we have the following:

But is an integer less than K.

But is an integer which contradicts the minimumality of K. - Dec 18th 2006, 10:30 AMJudi
YOu guys are awesome!!!