text: Hi:\n\nthe truth table for binary addition is :\n\n0+0 =0 \n0+1 =1\n1+0 =1\n1+1 = 10 { the one is carried over to the next digit}\n\nfor example 7 + 3 = 10 decimal \n\n111 is 7 decimal in binary  because 4(1)+2(1) +1(1) = 4+2+1=7\n011 is 3 decimal in binay   because 4(0) +2(1) +1(1) = 2+1= 3 \n\nokay let add \n\n{ ignore the # symbol it for postioning thing here it add nothing to the problem okay.\n\n## 111\n#+ 011\n----------\n\n\nfrom our truth table of adding number 1+1 = 10 \nso :\n\n####1 - ( carry section here)\n## 111\n#+ 011\n----------\n#####0\n\nand 1+1= 10 however we have a carry of 1here so get added on \nso 1+1+1= 11 So we do the following:\n\n## 11 - ( carry section here)\n## 111\n#+ 011\n----------\n### 10\n\nNow we add the next column. Like so: 1+0= 1 but we have a carry of 1 so: 1+1= 10 and we do like so :\n\n# 111 -( carry section here) \n## 111\n#+ 011\n----------\n##010 \n\nNow in the next column we have a 1 in the carry section but nothing in that column so a zero is put here and added. So 1+0 = 1. And we do the following:\n\n\n#111 -( carry section here)\n+0111\n##011\n----------\n#1010 \n\nand 1010 is the answer because   8(1)+(0)4+(1)2+(0)1= 8+2 = 10 decimal \n\nSo 3 decimal which is 011 binary and 7 decimal in binary is 111 when add together is 10 decimal; which in binary is 1010 \n\nand that is how you add numbers in Binary\n\nMulitiply in is the same as multiplication of 1 and 0 \n\nfor example let do 5 times 5 in binary \n\n\nSo 5= 101 binary \n\n\n##101\n#x101\n----------\nStarting with the right column Multiply 101 by 1 which equals 101\n\nSo :\n\n\n##101\n#x101\n----------\n### 101\n\nnow with the next colum from the right multiply the 101 by 0 it equals 0 \n\nSo \n\n\n##101\n#x101\n----------\n### 101\n## 000\n\nnow for the final column multiply 101 by 1 and it equal 101 \n\nSo \n\n\n##101\n#x101\n----------\n### 101\n## 000\n# 101\n-------------\n\n\nAdd the partial additions here to get :\n\n##101\n#x101\n----------\n### 101\n## 000\n# 101\n-------------\n11001 \n\nLet see if this is the right answer \n\n16(1) +8(1) +4(0) +2(0)+1(1) = 16 + 8+1= 25\n\nSo 5 x 5 = 25 \nor 101(5 binary) time 101( 5 binary) = 11001(25 binary ) \nthis check out \n\n\nDivision is a bit more involved but lets do it:\n\n5 divided by 2 \n\nSo 5 =101 and 2= 10 \n\nsetup the problem like so:\n\n\n##___________________\n10/101.00000000000000\n\n\nSo let start by how many 10 will go into 10 the answer 1 \n\nSo \n\n##_1_________________\n10/101.00000000000000\n-10\n----------- \n### 1\n\nhow many 10 will go in to 1 answer 0 So \n\n##_10._________________\n10/101.00000000000000\n-10\n----------- \n####1\n### -0 \n------------\n####10 \n\nPut our decimal point here{ because we reached it } and \nHow many 10 will go into 10 answer 1 \n\nSo :\n\n##_10.1_________________\n10/101.00000000000000\n-10\n----------- \n####1\n### -0 \n------------\n####10 \n### -10\n--------------- \n######0 \n\nNo need to go any farther all the rest are zeros from here\n\nSo the answer is 10.1 binary which is \n\n2(1)+ 1(0) + (1)(1/2) = 2 1/2 decimal which is the correct answer \n\nuse this  to help you :\n\n\n\n\n 4096, 2048, 1024, 512 ,  256 , 128 ,  64 , 32 , 16 , 8, 4, 2  1\n##0####0####0###0####0####1###1###1###1# 1#1#0#1 \n\nput in the 1's and zero's in the 2 place holder like I got above here.\n\nmultiply the 1 and the zeros where they appear in the 2's place holder { I ignore the the leading zeros here} like so:\n\n 128(1)+64(1)+32(1)+ 16(1)+8(1) +4(1) +2(0)+1(1) \n\n128+64+32+16+8+4+1 =  253  \n\nso: 11111101 binary is 253 decimal \n\nthe best way to do this is to set your computer's calulator to Binary by clicking the bin circle enter your binary number and select the dec circle to tell you what the decimal number is \n\n                                                                                                                                                                                        \n\n\nHope this helps