CPhill

I can only do that with my "normal" answers........LOL!!!

It's right there in my eighth "sentence".......

"Thus,   ∞ - ∞ = ∞"..........

On first blush, we might consider the answer to be "0.'

But, it's not.....

Consider the set of integers on the number line. You will agree - I hope - that this set is infinite. Now, consider the set of even integers on the number line. Again, we'll consider this set to be infinite, too.

So, subtracting the second "infinity" from the first "infinity," we end up with another infinite set - the set of all odd integers !!

Thus,   ∞ - ∞ = ∞

Also, notice another odd property......I've divided one set in half and gotten another infinite set. Thus....

∞ / 2  = ∞

One more interesting idea, and then I'll shut up.  Which "infinity" is "greater?"  The set of all positive integers, or the set of all positive even integers??

Note, that I can "count" each positive even integer.....For example, 2 is the first one, 4 is the second one, etc. Thus, I can put each positive even integer into a one-to-one correspondence with a "counting" number. And since I can do so, then the set of positive even integers must be just as large as the set of all the positive integers !!!

Yes....you are correct, Melody.......I should have said it is undefined everywhere except for my answer.......

Good answer, GoldenLeaf !!!

Now...as to your question......in Zen practice, the answerer is not expected to always give an answer. In fact, there may BE no answer. Contemplation about the answer is sometimes more important that the "correct" answer itself.

I'm still contemplating...........

Your answer was fine, GoldenLeaf.  Let me see if I can add to it.

Note, that by a property of exponents,we can write √(100) as 100^1/2 

So we have.......log√(100) = log(100)^1/2

And by a property of logs we can write log (a)^b  = b log (a)

So we an write log(100)^1/2  as (1/2) log (100) ......which we also can write as (1/2) log₁₀ (100)

Don't get too frustrated by a log...it's just an exponent (power)...basically, we're asking ourselves what power (exponent) do we need to put on the "little" 10 to raise it to 100??  Answer........2.

So log₁₀ (100)  = 2

And...... (1/2) [log₁₀ 100]  =  (1/2) [2]  = 1

And there you go !!

Hope that helps !!

So we have

(y/15) = (7/3)

Using cross-product equality, we have

3y = 15*7

3y = 105 

divide by 3

y = 35

Here's another method that sometimes works well. What did I have to multiply 3 by in the denominator of the second fraction to get 15 in the denominator of the other??  Answer, 5.  So, if I multiply the 7 in the numerator of the second fraction by 5, I get 35 for "y" in the numerator of the first fraction.

So, we have the points (2,5) and (6,-1).

I have a "nonstandard" method for finding slope.

Put the points one over another - it doesn't matter which point goes over the other - like this:

(2 , 5)

(6 , -1)

Now, just "subtract" the second from the first. This gives:

(-4 , 6)

Now, put the first thing "under" the second.........6 /-4  = -3 / 2 and there's the slope !!!

Uh, OK.....I'll take your word for it.....Is there a question there??

Basically, it's just a shorthand notation. It comes in handy for expressing extremely "large" or "small" numbers.

For instance, suppose we had 9,100,000,000,000

Instead of having to write this "long" number out, we can write it as......  9.1 x 10¹²

The rule is to move the decimal point to the right of the first non-zero digit and count how many places we moved it - in this case, 12 - and that's the exponent that goes on the "10." Moving the decimal point to the left results in a positive exponent. Moving it to the right results in a negative exponent, as in the following example:

.0000000000002345 = 2.345 x 10^-13

It also comes in handy for multiplying/dividing large or small numbers, in certain situations.

This article explains it more thoroughly than I can: