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A1 |
Before you can establish the numbers between 1 and 2, you must first establish the number 1 and, from that, then 2 ... 1 1/2 only makes sense as the number which is 1/2 of the unit distance between 1 and 2. So yes it is somewhat arbitrary what we accept as a unit (* standard *) of measurement - a foot, a meter, a hekat - but once a unit is established we can go on to construct the entire number line as outlined in one of the responses to your question. Though there are indeed infinitely many numbers between 1 and 2, they are actual and distinct numbers with their own characteristics ... and they are studied as if they possess objective existence. 3/2 = 1.5 is very different from Pi/2 = approximately 1.570796327 (accurate to 8 decimal places). This does not make the numbers 1 and 2 less real. The responses to your question (on the site) - though a bit excessive - also illustrate that one can come to the questions asked by you and MBM from multiple theoretical points of view. In my first post, I tried to provide an answer that was a bit more down to earth.
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"The Nobel Prize in mathematics was awarded to a California professor who has discovered a new number! The number is bleen, which he claims belongs between 6 and 7." --George Carlin
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A1 |
There is no Nobel Prize in mathematics ... |
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A1 |
Yeah, but I didn't want to alter the quote...
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A1 |
One famous example of a particular number between 0 and 1 is the Euler-Mascheroni constant.
If you know a little calculus, it can be defined by  Its numerical value to 50 decimal places is 0.57721 56649 01532 86060 65120 90082 40243 10421 59335 93992 ... It is unknown whether or not its actual value is fractional and this question is the subject of active research. |
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Phoenix Rising |
Thanks HB.... So when we utilize numbers we are actually creating a finite/false construct? (I'm not trying to use big words... I just don't know how else to express it) Meaning number one may have value.... but it is only the value that we give it...
Ummm... HULLO???  I wasn't playing when I said my brain shut down after I read his response... I was just staring at the screen.. like  Usually they give great information... they even recommended me to another site where I could get good critique and skill development for my poetry....
Peace, Khalliqa "The Goddess emerges as the evanescence of the inferior dissipates.... " |
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A1 |
It - the number 1 - has value. I'm not sure I would call it "false" though. In some sense, it is a formal construct. But it's a construct which seems to derive its basic meaning from fundamental human intuitions. So it's not entirely artificial. Otherwise, all human language games are "false". After all, what does any word really mean? The integers and the rational numbers (integer fractions) can be considered finite constructs. But the number line as a whole is not. It requires in-finite processes both to construct and to understand it. Pi is an example of one such number.
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D4 |
Salaam (Peace) to all.
Numbers are symbols that represent concepts. We can use these symbols to represent both concrete and abstract concepts. However, the realm of abstract (mental) concepts is far less bound by the laws of physical nature than is the realm of concrete (material) phenomena. So then, we can consider applications of these symbols and their associations in the "abstract" world, which simply are not operable in the "concrete" world (as far as we know). One can "imagine" the number "1" as a symbol of "infinity", but try to demonstrate that in the physical realm....not so easy. This is not to say that the "concrete" realm is "more real" than the "abstract" realm. Quite the contrary, for the "concrete" emanates from the "abstract" and not vice versa. However, we cannot expect to successfully impose upon the "concrete" realm the possibilities that exist in the "abstract" realm. The natural order is that we first get acquainted with how these concepts interplay in the "concrete" realm, and then we eventually, through our study and contemplation, conceive of higher applications of these concepts. Therefore, If I hold in my hand "1" baseball, it cannot be demonstrated in the physical realm that the "1" baseball is actually an "infinitude" of baseballs. Nor can it be demonstrated, because of the limitations of physical material, that the "1" baseball can be divided an "infinitude" of times; although these applications can and do work in the "abstract" realm. That being said... How many "moments" are there in a Lifetime? And is Life really "too short"? (smile) RM
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Well said.  |
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A1 |
The mother is 21 years older than her child. 6 years from now, the mother will be 5 times as old as the child.
Question: Where is the child's father right now?
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A1 |
Question: How wealthy would I be if I had a penny for every @$$hole with an ax to grind who couldn't correctly interpret a statistic to save his life? |
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A1 |
Time for a Polish joke: Theorem: The commutative law for + implies the associative law for +. proof: Express the laws in Polish notation. Commutative Law: +xy = +yx for all x and y. Associative Law: ++abc = +a+bc for all a, b and c. Now, assume the Commutative Law. Let x = + and y = a. Then we have ++abc = +xybc = +yxbc = +a+bc. QED
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A1 |
I was teaching my college algebra class the other day. We were talking about functions, graphs, and transformations. I told them that a "horizontal shrink" is a psychiatrist that lays down on his own couch. PS: For what it's worth, I've taken you off "ignore". I forgive you. PPS: That binary joke was funny.
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A2 |
 Okay Brotha...Honest Brother...please explain this word problem. I have not a freaking CLUE! I need help. Here goes. And please whateva you do don't laugh I know this is soooo elementary for you. "Dan Wheldon won the Indy Japan 300 with a time of 1:49:482611 for the 300 mi race. At one point, Wheldon was 80 mi closer to the finish than the start. How far had Wheldon traveled at that point? How do I begin And the following problem is to make each pair of inequaliies equivalent BUT...my underline button doesn't work for this next problem...let's pretend the underline is in fact under the < and >. Here goes: -5x <30; x > -6 Please explain. Is it a trick to this? And while you're at it: # consecutive numbers in a roll.  I remember the professor saying 1st: x; 2nd: x + 1; and 3rd: x + 2. The method is to read the entire problem; Represent the unknown; Write the Guideline; substitute into guideline; solve the equation. Okay I got that but when I try to incorporate it in a problem...I get ALL twisted. Look at this: Sharon invested money in a saving account at a rate of 6% simple interest. After 1 year, she has $6996 in the account. How much did Sharon orginally invest?  Any help you can provide...appreciate it  BTW: this is the ONLY subject that totally INTIMATES me...where I have NOTHING absolutely nothing to say...not a thing. |
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A2 |
Okay HonestBrother....here's another word problem that is driving me MAD! The houses on the west side of Lincoln Avenue are consecutive ODD numbers. Sam and Colleen are next door neighbors and the sum of their house number is 572. Find their house number.[Do I use the method: 1x + 1x+ 1 = 572? Or do I use the medthod: 1x +1x + 2 = 572?]....Help! |
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A1 |
Koco, I'll be with you in a minute (more like an hour). I just got out of class.
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A1 |
Answer: 285 and 287 SOLUTION: Here are two consecutive ODD numbers: 3 and 5. Notice that 5 = 3 + 2 This would be the case for ANY two odd consecutive numbers. So if the first number is x then the second number is x + 2: x x + 2 If their sum is 572 this means that when I add these two numbers together I should get 572: x + x + 2 = 572 If you simplify you get 2x + 2 = 572 Subtract 2 from both sides: 2x + 2 - 2 = 572 - 2 So that 2x = 570 Then divide both sides by 2: x = 570/2 or x = 285 So one house number is 285. The other one is 285+2=287. 285 and 287 are two consecutive odd numbers. Notice that 285+287= 572.
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A2 |
Got it my brotha  .......but PLEASE look at the post prior to the last one I entered. I have about 3 questions. These are much much harder for me to comprehend. My brain...it's hurting! However, you can get back to me whenever you're able. I have an exam MONDAY So I'll be coming back and forth all weekend...awaiting your "brilliant" explanations. You don't KNOW how much I appreciate this extra help. Thankyou my brotha...much love to ya |
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A1 |
Answer: $6600 SOLUTION: The formula for simple interest is: I = P*r*t I= the Interest earned over the relevant period of time P= the Principle (which just means the amount of your original investment) r= the interest rate t= the time period of your investment. In this example, r= 6% which is 0.06 in decimal form t= 1 year We have a bit more information. I, which is the interest accrued on the principle after 1 year, is the amount after one year OVER and above the principle P that Sharon started with. Since Sharon has $6996 in her account after 1 year that means that I = 6996 - P Go back to the formula I = P*r*t and plug in 6996 - P for I: 6996 - P = P*r*t This equation must be solved for P, the principle or the amount originally invested. P is the quantity you have to find. 6996 - P = P*r*t Add P to both sides 6996 - P + P = P + P*r*t which simplifies to 6996 = P + P*r*t Factor the right hand side 6996 = P*(1 + rt) Then divide both sides by 1+rt to find that P = 6996/(1 + rt) Then plug in the values r=0.06 and t=1 to get P = 6996/1.06 = $6600
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A1 |
Answer: He had traveled 190 miles. SOLUTION If I'm interpreting this question correctly, there's some irrelevant information in the problem. For example, it doesn't matter that his time was 1:49:482611 It doesn't even matter if the track was circular or a straight line. So I'll begin by representing the track with a graphic: Start (0 miles) __________________________X________Finish (300 miles) Where the X represents his location (distance wise) between Start and Finish .... and the total distance separating Start and Finish is 300 miles. Notice that the distance from Start is X and the distance from Finish is 300 - X. You are asked for the value of X when Dan is 80 miles closer to Finish than Start ... Or when the distance from Finish is (=) 80 miles less than the distance from Start. Translating this into an equation: You're asked for the value of X when Distance from Finish = Distance from Start - 80 or 300 - X = X - 80 Solve this for X: Add 80 to both sides: 380 - X = X Then add X to both sides: 380 = 2X then divide both sides by 2: X = 190 miles Notice that he is 190 miles from the Start but 110 miles from the Finish. There is an 80 mile difference.
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