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      "body": "Congratulations @yey! You received a personal award!\n\n<table><tr><td>https://steemitimages.com/70x70/http://steemitboard.com/@yey/birthday2.png</td><td>Happy Birthday! - You are on the Steem blockchain for 2 years!</td></tr></table>\n\n<sub>_You can view [your badges on your Steem Board](https://steemitboard.com/@yey) and compare to others on the [Steem Ranking](https://steemitboard.com/ranking/index.php?name=yey)_</sub>\n\n\n**Do not miss the last post from @steemitboard:**\n<table><tr><td><a href=\"https://steemit.com/steemtoolbar/@steemitboard/steemtoolbar-update-display-bug-fixed\"><img src=\"https://steemitimages.com/64x128/http://i.cubeupload.com/7CiQEO.png\"></a></td><td><a href=\"https://steemit.com/steemtoolbar/@steemitboard/steemtoolbar-update-display-bug-fixed\">SteemitBoard - Witness Update</a></td></tr><tr><td><a href=\"https://steemit.com/steem/@steemitboard/do-not-miss-the-coming-rocky-mountain-steem-meetup-and-get-a-new-community-badge\"><img src=\"https://steemitimages.com/64x128/https://cdn.steemitimages.com/DQmUphCGZFWgt6bJ1XTtunV7esnwy6bxnGqcLcHAV3NEqnQ/meetup-rocky-mountain.png\"></a></td><td><a href=\"https://steemit.com/steem/@steemitboard/do-not-miss-the-coming-rocky-mountain-steem-meetup-and-get-a-new-community-badge\">Do not miss the coming Rocky Mountain Steem Meetup and get a new community badge!</a></td></tr></table>\n\n###### [Vote for @Steemitboard as a witness](https://v2.steemconnect.com/sign/account-witness-vote?witness=steemitboard&approve=1) to get one more award and increased upvotes!",
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      "body": "I did not take that into account. It would seem that having rational coefficients is required, but I think it would work for real or complex coefficients as well, just as long as the zero is rational.",
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      "permlink": "re-yey-re-voice-of-apollo-math-challenge-for-the-brave-20171011t074956889z",
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      "body": "I think the easiest way to prove that is the following:\nLet f(x)=x^{n}+a_{n-1} x^{n-1} + ... a_{0} be a monic polynomial with rational coefficients and let p/q be a root of this polynomial. (Here, q is nonzero, p and q are coprime.)\nThen we have\n(p/q)^{n} + a_{n-1} (p/q)^{n-1} + ... + a_{0} = 0.\nBy multiplying q^{n} on the both sides, we get\np^{n} + a_{n-1} p^{n-1}q + ... a_{0} q^{n} = 0,\ni.e.,  p^{n}=-(   a_{n-1} p^{n-1}q + ... a_{0} q^{n}   ) .\nSince ( a_{n-1} p^{n-1}q + ... a_{0} q^{n} )is a multiple of q, q must divide p^{n}. Also, since p and q are coprime, q must divide p as well. Now we can conclude that q=1 and p/q is an integer .",
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      "body": "Do you mean a monic polynomial with 'rational' coefficients?\n(By the way, I received 3SBD. Thank you:)  )",
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      "body": "극좌표에서 넓이는 (r^2)/2 를 theta에 대해 적분해 주면 되는 거 아시죠?  곡선이 r^2=2cos(2theta)니까 cos(2theta)를 theta에 대해 적분해주시면 되실거구.. 문제에서 8자 모양 전체 넓이 구하라고 한거면 theta 범위 -pi/4부터 pi/4 까지 해서 적분해주시면 8자 모양 중 절반 부분인데 이거 넓이가 1이 나오니까 8자 전체 넓이는 2가 되겠네요!",
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      "body": "I was just going over proofs on StackExchange and I think you're correct. I would be very impressed if you found this organically by yourself, but still since you answered it right I owe you 3 SD. I'll send that to you when I have that amount. \nHere are some examples of the proof I found:\n![proof1.png](https://steemitimages.com/DQmRci6gSpRGsRr2DEDmYRUSMo6ZPmDo45BkUsAw3ogjeWU/proof1.png)\n![proof2.png](https://steemitimages.com/DQme1ZYg7QdDaJG8H73EF15N9U8tNYiGQcPhTiExuRofrsU/proof2.png)\n![proof3.png](https://steemitimages.com/DQmYfX6kif7jNntAxTDh3Y4nznKAs7c8FQRPx1jjS4sk4DY/proof3.png)",
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      "body": "Will do, thanks. I'm checking the proof right now. Thanks for the response.",
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      "permlink": "re-danbain-re-yey-re-danbain-5mpypw-3-20170902t054657086z",
      "title": "",
      "body": "아 맞다 외국에 계셨죠 ㅋㅋ",
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      "body": "@@ -36,32 +36,53 @@\n n acyclic graph \n+with finite vertices \n must have a leaf\n@@ -220,16 +220,37 @@\n  graph)%22\n+ with finite vertices\n  has a l\n@@ -333,16 +333,37 @@\n  graph)%22\n+ with finite vertices\n  must ha\n",
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      "author": "danbain",
      "permlink": "re-yey-re-danbain-5mpypw-3-20170902t050914906z",
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      "body": "그림은 무료이미지에서 퍼왔습니다 ㅋㅋ 날씨가 풀리는게 보이긴하네요 여기도",
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      "permlink": "re-beoped-re-yey-re-yey-untitled-20170902t044316756z",
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      "body": "옙, 감사합니다 @beoped 님. 날씨가 많이 바뀌어서 주변에도 감기 걸리시는 분들 좀 있던데 @beoped님도 감기 조심하세요! 좋은 하루 되세요~",
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      "body": "그렇죠 ㅋㅋㅋ 내일 새 글을 기대해 볼게요 ㅎㅎ  저도 요새 감기나 여러 일들이 있어서 글을 쓰기가 참 힘들더라구요 ㅎㅎ",
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      "body": "For the proof of the theorem that an acyclic graph must have a leaf, see the article \"Acyclic graph must have a leaf\" on math.stackexchange.com\nThere is a reply that shows a \"tree(connected, acyclic graph)\" has a leaf. One can easily generalize this theorem and prove that a \"forest(acyclic graph)\" must have a leaf.",
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      "body": "For the proof of the theorem that an acyclic graph must have a leaf, see the article \"Acyclic graph must have a leaf\" on math.stackexchange.com\nThere is a reply that shows a \"tree(connected, acyclic graph)\" has a leaf. One can easily generalize this theorem and prove that a \"forest(acyclic graph)\" must have a leaf.",
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      "body": "I think you can use induction to the number of vertices. Let G be a graph with only 1 vertex. To satisfy the condition \"e is not less than v\", it must have a loop.  So, G has a cycle.\nNow suppose that the statement is true for all graphs with number of vertices at most k. Consider a graph G with (k+1) vertices and suppose that |E(G)| is not less than |V(G)|. If G has no cycles, than it must have a \"leaf\", v_0. (A leaf means a vertex of a graph of degree 1.)\nNow consider the graph H we get by removing the vertex v_0 from G. Then H is a graph with k=|V(G)|-1 vertices and |E(G)|-1 edges. So, the number of edges of H is not less than the number of vertices of H. By the induction hypothesis, H must contain a cycle. However, since H is a subgraph of an acyclc graph G, H cannot contain a cycle. This is a contradiction. \nNow we see that the statement holds for all graphs with finite number of vertices.",
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      "body": "스팀잇도 꾸준히 하기 힘드네요;;",
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      "body": "스팀잇도 꾸준히 하기 힘드네요.",
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      "title": "제목 없음. Untitled.",
      "body": "<html>\n<p>&nbsp;요 며칠 사이 바빠서 스팀잇에 거의 들어오질 못했다. 내일 중으로 새 글을 올려야겠다.</p>\n<p>&nbsp;I've been so busy that I couldn't log into my Steemit account for a while. Planning to upload a new post tomorrow.\n<center><img src=\"https://cdn.pixabay.com/photo/2014/09/09/19/10/pier-440339_640.jpg\" width=\"640\" height=\"426\"/></center>\n</html>",
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