{"id":396,"date":"2023-09-25T14:10:48","date_gmt":"2023-09-25T14:10:48","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=396"},"modified":"2023-09-25T14:13:59","modified_gmt":"2023-09-25T14:13:59","slug":"week-5-jss-1-second-term-mathematics-lesson-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-5-jss-1-second-term-mathematics-lesson-notes\/","title":{"rendered":"Week 5 – Jss 1 Second Term Mathematics Lesson Notes"},"content":{"rendered":"

\u00a0
\n\u00a0WEEK 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 NUMBER SYSTEM<\/p>\n

\u00a05.1\u00a0\u00a0\u00a0\u00a0 Converting Numbers Base From One Baseto Other Bases
\n<\/strong>This is a 2-stage conversion process. The steps to do this are simple. First convert the given base to base ten and then convert the result to the required base.
\nExample5.1:<\/strong> Convert 21356 <\/sub>to a number in base two
\nSolution:
\n<\/strong>Step one:<\/em> Converting to base ten
\n\u00a0\u00a0\u00a0\u00a023<\/sup>12<\/sup>31<\/sup>50 <\/sup>= (2X63<\/sup>) + (1×62<\/sup>) + (3×61<\/sup>) + (5×60<\/sup>)
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= (2x 216) + (1x 36) + (3x 6) + (5 x 1)
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 432 + 36 + 18 + 5 = 491
\nStep two:<\/em> Convert the answer in base ten to a number in base two
\n\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0491
\n\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0245 R 1
\n\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0122 R 1
\n\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a061 R 0
\n\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a030 R 1
\n\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a015 R 0
\n\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a07 R 1
\n\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a03 R 1
\n\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a01 R 1
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00 R 1
\nTherefore 21356 <\/sub>= 1111010112<\/sub><\/p>\n

\u00a0
\n\u00a0
\n\u00a0
\n\u00a0Examples 5.2<\/strong>: Express 101112<\/sub> to a number in base three
\nSolution:<\/strong>
\n\t\t1. 14<\/sup>03<\/sup>12<\/sup>11<\/sup>10<\/sup>= (1 X 24<\/sup>) + (0 X 23<\/sup>) + (1 X 22<\/sup>) + (1 X 21<\/sup>) + (1 X 20<\/sup>)
\n= 1 X 16 + 0 X 8 + 1 X 4 + 1 X 2 + 1 X 1
\n= 16 + 0 + 4 + 2 + 1
\n = 2310<\/sub>
\n\t\t\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a023
\n\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a07 R 2
\n\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a02 R 1
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00 R 2
\nTherefore 101112 <\/sub>= 2123<\/sub><\/p>\n

\u00a0
\n\u00a05.2\u00a0\u00a0\u00a0\u00a0Addition and Subtraction of Number Base
\n<\/strong>
\n\u00a0Note the following STEPS in adding and subtracting.
\n<\/em><\/strong>To add in base two
\n<\/em><\/strong>0+0 = 0
\n<\/em><\/strong>1+ 0 = 1
\n<\/em><\/strong>1 + 1 = 10
\n<\/em><\/strong>1 + 1+ 1 = 11
\n<\/em><\/strong>
\n\u00a0To subtract in base two
\n<\/em><\/strong>0 \u2013 0 =0
\n<\/em><\/strong>1 \u2013 0 = 1
\n<\/em><\/strong>10 \u2013 1 = 1
\n<\/em><\/strong>11 \u2013 1 = 10\u00a0\u00a0\u00a0\u00a0
\n<\/em><\/strong>
\n\u00a0Example 5.3:<\/strong> Add the following
\n1.\u00a0\u00a0\u00a0\u00a011012<\/sub> and 11112<\/sub>
\n\t2.\u00a0\u00a0\u00a0\u00a0110112<\/sub>, 101012<\/sub>, and 10012<\/sub>
\n\tSolution:
\n<\/strong>1.\u00a0\u00a0\u00a0\u00a01101 + 1111 = 1 1 0 1 [using the addition steps 1 + 1 = 10, write 0 down and move the 1 to the next number]
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 + 1 1 1 1 [also take the 1 to next, and move them till you get to last one].
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 1 1 0 02<\/sub><\/p>\n

\u00a02.\u00a0\u00a0\u00a0\u00a011011 + 10101 + 1001 = 1 1 0 1 12<\/sub>
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 0 1 0 12<\/sub>
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+ 1 0 0 12<\/sub>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 1 1 0 0 12<\/sub><\/p>\n

\u00a0Example 5.4:<\/strong> Subtract (a) 10112<\/sub> from 101012<\/sub> (b) 110112<\/sub> from 1110102
\n<\/sub>Solution:
\n<\/strong>\u00a0\u00a0\u00a0\u00a01.\u00a0\u00a0\u00a0\u00a01 0 1 0 1[using the subtraction steps 1- 1 = 0, write 0 down and move to the next number]
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-1 0 1 12<\/sub>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 0 1 02<\/sub>
\n\t\t\u00a0\u00a0\u00a0\u00a02.\u00a0\u00a0\u00a0\u00a01 1 1 0 1 02<\/sub>
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-1 1 0 1 12<\/sub>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 1 1 1 12<\/sub><\/p>\n

\u00a05.3\u00a0\u00a0\u00a0\u00a0Multiplication of Binary
\n<\/strong>Example 5.5:<\/strong> Find the product of 10112<\/sub> and 1012<\/sub>
\n\tSolution:
\n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 0 1 12<\/sub>
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x 1 0 12<\/sub>
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 0 1 1
\n\u00a0\u00a0\u00a0\u00a0 0 0 0 0
\n\"\"\/\u00a0\u00a0\u00a0\u00a01 0 1 1
\n1 1 0 1 1 12<\/sub>
\n\t\t\u00a0\u00a0\u00a0\u00a0
\nExample 5.6:<\/strong> multiply 1111 by 110
\nSolution:
\n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 1 1 12<\/sub>
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 1 02<\/sub>
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00 0 0 0
\n\u00a0\u00a0\u00a0\u00a0 1 1 1 1
\n\"\"\/\u00a0\u00a0\u00a0\u00a0 1 1 1 1
\n 1 0 1 1 0 1 02<\/sub><\/p>\n

\u00a0
\n\u00a0EVALUATION:<\/strong> DO THESE
\n1.\u00a0\u00a0\u00a0\u00a0Add the following (a) 10012<\/sub> + 11112<\/sub>
\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub>(b) 1011102<\/sub>+100102
\n<\/sub>2.\u00a0\u00a0\u00a0\u00a0Subtract 10112<\/sub> from 11011012<\/sub>
\n\t3.\u00a0\u00a0\u00a0\u00a0Multiply 1101012<\/sub> by 10112
\n<\/sub>4. find the value of the following binary numbers 1011+10-111
\n<\/strong>
\n\u00a0ASSIGNMENT:
\n<\/strong>EXERCISE 9.5 page 107 NO 1(B,C,D), NO 2 A and B, NO 3 (E,F,G,H)
\nEXERCISE 9.6 NO 2,5,8 and 11. PAGE 108<\/p>\n","protected":false},"excerpt":{"rendered":"

\u00a0 \u00a0WEEK 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 NUMBER SYSTEM \u00a05.1\u00a0\u00a0\u00a0\u00a0 Converting Numbers Base From One Baseto Other Bases This…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,66],"tags":[],"class_list":["post-396","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-jss1-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/396","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=396"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/396\/revisions"}],"predecessor-version":[{"id":397,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/396\/revisions\/397"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=396"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=396"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=396"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}