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JaneFairfax wrote:

JaneFairfax, here is a basic proof of L4:

For all real a > 1, y = a^x is a strictly increasing function.

log(base 2)3 versus log(base 3)5

2*log(base 2)3 versus 2*log(base 3)5

log(base 2)9 versus log(base 3)25

2^3 = 8 < 9

2^(> 3) = 9

3^3 = 27 < 25

3^(< 3) = 25

So, the left-hand side is greater than the right-hand side, because

Its logarithm is a larger number.

iamaditya, it has been four and a half months since your last reply to bobbym, who has since died.

The answer to # 15 is log(12).

Samuel.lagodam wrote:

It's more appropriate to say 0^0 is 1 than saying its undefined

No, it is incorrect to say that 0^0 is 1, because it is undefined. For instance, in the graph of

y = x^x, there is a "hole" at (0, 1), because there is no function value associated with x = 0.

luca-deltodesco wrote:

we have:

you could also have:

These are incorrect. All of the forms involving the natural logarithm for the answer must have absolute value

bars where the logarithm of the quantity is being taken of in these examples, not just parentheses.

This goes for the rest of your posts in this thread.

Monox D. I-Fly wrote:

MathsIsFun wrote:But rather strangely there is disagreement on whether 1 should be included or not.

Like the disagreement on whether 0 should be integer or not?

There's no disagreement I know of about whether 0 should be an integer or not.

There's inconsistency of whether 0 is a called a counting number or not.

Zeeshan 01 wrote:

Two non parallel lines intersect at

A) 0 point

B) one point

C infinity points

Can any one show plz??

If these two lines are in the same plane, and they are not the same line, what can you state about them?

MisterMaths wrote:

I was to determine whether this sequence is convergent or divergent: Un=http://i.imgur.com/BOwr53L.png

That sequence is convergent.

As n approaches infinity, each expression on either side of the subtraction sign approaches 1. That answers the part of convergence/the question.

(If you were being asked about what value the whole expression is approaching, then you would know the value of the expression on the left-hand

side of the subtraction sign is greater for all (finite) values of n than the right-hand expression. So, the value of the whole sequence would be a

finite number greater than or equal to 0, but less than 1. I would expect the value to equal to 0.)

bob bundy wrote:

??? No, still don't get it. What's the difference between a circle and a disk? If two semicircles meet what does it matter if they're open?

Here's my solution.

http://i.imgur.com/sXx6Z60.gif

You cannot see the original circle (it is red) because it's covered.

Bob

You cannot see the circumference of the fill circle/disk even though it

is not covered. Again,

the semi-disks do not include their circumferences.

And you cannot see the diameter between the two semi-disks.

It is also not covered. Not being able to see with our eyes is not proof.

bob bundy wrote:

hi Mario23

So what's wrong with choosing the semicircles to be half the circle in which case it's two for both parts.

Bob

There's a circle (consisting of just the circumference), and then there's the disk containing the circle and the area inside of the circle.

From what the original poster stated, it must be an open disk, instead of an open circle.. If a full circle [really the disk] is going to be fully

covered using congruent open semicircles [really open semi-disks] with the same radius of that of the original full circle or even the original

full disk, then more than two of them will be needed to ensure that the circumference of the full circle (or disk) and a diameter (where the

open semi-disks would meet if there were only two of them) would also be covered.

.

Zeeshan 01 wrote:

By doing rationalization

1-cosx÷sin (x)^2 . . . . . .This isn't correct, because you didn't use grouping symbols.

[1 - cos(x)]÷[sin(x)]^2

mumba wrote:

2 Expand each expression into factored form. The rewrite it in the shortest exponent form.

or

**Write the second one with grouping symbols around the denominator:**

bobbym wrote:

That is also a problem, many online math classes do not like their students getting help on their problems. Your is okay with that?

Your next post to the OP after the one above makes no sense. You can't just take the word of a user whose own interest is to get answers on something

without disclosing proof of A) where these problems are really from and B) whether there is documented permission from a third source to use the problems.

The OP's reply "They do not mind" gives you zero assurance, and you could still be in trouble for discussing the answers. Why would you entertain trusting

a user's word on likely sensitive material!?

bobbym wrote:

As you know humans say

zetafunc wrote:

Go back to your previous expression and prove that the LHS is negative. The point I was trying to make was that you'll need to establish an ordering relation between p, c, r to do it, as per the below:

zetafunc wrote:

If a, b are negative but c is positive, then you can also establish that a < b < c. Now multiply your original expression by abc and derive a contradiction by proving the LHS is negative, whilst the RHS is positive.

You must have an error somewhere. I know I am asking about all integers, except zero.

But it should not make a difference if I were to ask about a, b being negative, c is positive and all three belonging

to the set of real numbers. And, I will make sure [in general, and in my example below] that a < b < c.

I came up with this counterexample for you:

If you work this out, you will see that the expression on the left-hand side simplifies to 1.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

In general where a and b negative and a < b:

Let r, p be positive integers

Let b = -r

To make sure a < b,

let a = -p - r

c = c

Substitute:

The sides do not contradict each other in signs.

If I were to multiply each side by cp(p + r), I would get:

You're not going to manipulate that and get a negative LHS versus a positive RHS.

zetafunc wrote:

greg1313 wrote:Here, I included all integers, except zero.

You can show that there aren't any solutions by considering the cases where one, two, or three of (a,b,c) are negative.

If a, b are negative but c is positive, then you can also establish that a < b < c. Now multiply your original expression

by abc and derive a contradiction by proving the LHS is negative, whilst the RHS is positive.

I isolated your second case to further discuss.

My work:

----------

Let r, p be positive integers.

Suppose a = -r

Suppose b = -p

Keep c positive.

Multiply each side by rpc:

Add two terms to each side:

I don't see any contradiction of signs here, because the LHS is positive and so is the RHS.

It just means this case is inconclusive.

If your third case is based on the technique of the second case, then that might be inconclusive, too.

bobbym wrote:

Hi;

Are a,b and c positive integers?

If I want a, b, c to be positive integers for other problems, I will state "positive integers" in those cases.

It seems some younger students aren't aware of negative integers, and I could see the clarification question

being asked here because of that.

Here, I included all integers, except zero.

So, it has been established that positive integers don't work, had that been the question. (I had already

researched that angle of the question some days prior.)

**greg1313**- Replies: 9

Does the equation

have a solution in nonzero integers a, b, c ?

___________________________________________________________________

The background would be intermediate/college algebra and number theory if

this could be proven or disproven.

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