![]() At that time there were only about three stamp dealers in Washington, the most widely known being the late H. Hibbs & Co., stock brokers and bankers where I was at that time employed, I became interested in the collection of postage stamps, especially those of the United States. C.Ībout two years prior to that date, through the efforts of a fellow worker in the office of W. Nearly twenty years have passed since that memorable Tuesday of May 14, 1918, when the sheet of one hundred of the twenty-four cent Airmail stamps with the center inverted were passed over the window to me here in Washington, D. I have decided to acquaint the philatelic world with the details. I have often thought that I would like to write all of these details in order that many of the present and all future generations of philatelists may know these facts. During all of these years that have passed since I purchased these stamps I do not recall having seen an article that has given all of the details connected with the purchase and sale of the sheet. The last price seen by the writer was $3,900.00 for a single copy. At this time these stamps are considered one of the rarest items of all stamps that have been issued by the United States as well as foreign countries. We have used all the basis steps, and now we get $P(16)\implies P(20)$, $P(17)\implies P(21)$ and so on.For years the newspapers and the philatelic press of this and foreign countries have told many stories of the discovery of the sheet of one hundred twenty-four cent Air Mail stamps with the center inverted. You have already proved $P(12)-P(15)$ by hand. Take postage for $k+1$ and $k+1=(k-3)+4$, that is you can get $k+1$ adding a 4-cent stamp to a $k-3$ postage: if you have $P(k-3)$, you get also $P(k+1)$. You want to prove that $P(k)\implies P(k+1)$, so assume $P(n),\forall n\le k$ and let's see if we can deduce $P(k+1)$. ![]() Now, your problem talks about postage $\ge 12$, so it would have no sense trying to prove that property for - let's say - $5$, so they have shown as basis step $P(12), P(13), P(14), P(15)$. But if you could also prove $P(1)$, then you would also get $P(1)\implies P(3)\implies P(5)\implies.$ and so $P$ would hold for $odd$ numbers too. Can you conclude that $P$ holds for all the natural numbers? No, you can just deduce that $P(0)\implies P(2)\implies P(4)\implies.$ that is your property holds for all the $even$ numbers. This is the problem because now we cannot argue that $P(k-3)$ is true all the time because it is not true when $k-312 P(j)$Īssume a property is true for $0$, and we write $P(0)$ assume also that $\forall n\in\mathbb,P(n)\implies P(n+2)$ - where $2$ is just an example - and $P(0)$. Inductive Step: Now, since we have only done the basic step for $P(12)$, we will have to let $12\le k$ instead of $15\le k$ as it was done with the $4$ base cases. I will, here, try to explain the problem that will arise if we don't do the 4 base cases.īasic Step: $P(12) = 4\times 3$ or four stamps of 3 cents The main question here is, why $4$ cases in the basic step. Like can someone please explain this to me? Inductive hypothesis is true, then P (k + 1) is also true. Stamp to the stamps we used to form postage of k − 3 cents. To form postage of k + 1 cents, we need only add another 4-cent That P (k − 3) is true because k − 3 ≥ 12, that is, we can form postage of k − 3 cents using Using the inductive hypothesis, we can assume Is true, we can also form postage of k + 1 cents. We need to show that under the assumption that P (k + 1) To complete the inductive step, we assume that we can form INDUCTIVE STEP: The inductive hypothesis is the statement that P (j )is true for 12 ≤ j ≤ k, So can someone please help me?īASIS STEP: We can form postage of 12, 13, 14, and 15 cents using three 4-cent stamps, twoĤ-cent stamps and one 5-cent stamp, one 4-cent stamp and two 5-cent stamps, and three 5-cent ![]() I know this question has been posted before, there is a solution in my text-book for this question and also this is posted on so many websites with full solution but I still don't understand.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |