{"id":52,"date":"2023-01-25T16:33:57","date_gmt":"2023-01-25T16:33:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/positional-systems-and-bases-dig-deeper-page-1\/"},"modified":"2025-08-23T00:47:51","modified_gmt":"2025-08-23T00:47:51","slug":"positional-systems-and-bases-fresh-take","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/positional-systems-and-bases-fresh-take\/","title":{"raw":"Positional Systems and Bases: Fresh Take","rendered":"Positional Systems and Bases: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Understand different number systems<\/li>\r\n\t<li>Convert different number systems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox recall\">Recall that in our base-[latex]10[\/latex] number system, each place value in a number represents a power of ten. Our numbers take the form thousands, hundreds, tens, ones or [latex]10^3[\/latex], [latex]10^2[\/latex], [latex]10^1[\/latex], [latex]10^0[\/latex]. Remember [latex]10^{0}=1[\/latex], any number raised to the [latex]\\text{zero}^{th}[\/latex] power equals one.We have an intuitive understanding that [latex]10^{1}=10[\/latex] because we are using [latex]10[\/latex] as a factor [latex]1[\/latex] time. And certainly [latex]10^{2} = 10\\times 10 = 100, 10^{3}=1000,[\/latex] and so on. This pattern works with bases other than [latex]10[\/latex] as well. As you\u2019ll see below, using [latex]5[\/latex] as a base yields the following:<center>[latex]5^{0}=1[\/latex]<\/center><center>[latex]5^{1}=5[\/latex]<\/center><center>[latex]5^{2}=5\\times 5 = 25[\/latex]<\/center><center>[latex]5^{3}=5\\times 5\\times 5 = 125,[\/latex]<\/center>and so on.<\/section>\r\n<p>In our base-[latex]10[\/latex] system, a number like [latex]5,783,216[\/latex] has meaning to us because we are familiar with the system and its places. As we know, there are six ones, since there is a [latex]6[\/latex] in the ones place. Likewise, there are seven \u201chundred thousands,\u201d since the [latex]7[\/latex] resides in that place. Each digit has a value that is explicitly determined by its position within the number. We make a distinction between digit, which is just a symbol such as [latex]5[\/latex], and a number, which is made up of one or more digits. We can take this number and assign each of its digits a value. One way to do this is with a table, which follows:<\/p>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: center;\">[latex]5,000,000[\/latex]<\/td>\r\n<td colspan=\"1\">[latex]= 5 \u00d7 1,000,000[\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 5 \u00d7 10^6[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\text{Five million}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]+700,000[\/latex]<\/td>\r\n<td colspan=\"1\">[latex]=7 \u00d7 100,000[\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"2\">[latex]=7 \u00d7 10^5[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\begin{align}\\text{Seven hundred} \\text{ thousand}\\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]+80,000[\/latex]<\/td>\r\n<td colspan=\"1\">[latex]= 8 \u00d7 10,000[\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 8 \u00d7 10^4[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\text{Eighty thousand}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]+3,000[\/latex]<\/td>\r\n<td colspan=\"1\">[latex]= 3 \u00d7 1000[\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 3 \u00d7 10^3[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\text{Three thousand}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]+200[\/latex]<\/td>\r\n<td colspan=\"1\">[latex]= 2 \u00d7 100[\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 2 \u00d7 10^2[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\text{Two hundred}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]+10[\/latex]<\/td>\r\n<td colspan=\"1\">[latex]= 1 \u00d7 10[\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 1 \u00d7 10^1[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\text{Ten}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]+6[\/latex]<\/td>\r\n<td colspan=\"1\">[latex]= 6 \u00d7 1[\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 6 \u00d7 10^0[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\text{Six}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]5,783,216[\/latex]<\/td>\r\n<td colspan=\"1\">[latex][\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"2\">[latex][\/latex]<\/td>\r\n<td style=\"text-align: center;\" colspan=\"4\">[latex]\\begin{align} \\hspace{0.8cm} \\text{Five million, seven hundred} \\\\ \\hspace{0.8cm} \\text{eighty-three thousand, two hundred sixteen}\\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>&nbsp;<\/p>\r\n<p>From the third column in the table, we can see that each place is simply a multiple of ten. Of course, this makes sense given that our base is ten. The digits that are multiplying each place simply tell us how many of that place we have. We are restricted to having at most [latex]9[\/latex] in any one place before we have to \u201ccarry\u201d over to the next place. We cannot, for example, have [latex]11[\/latex] in the hundreds place. Instead, we would carry [latex]1[\/latex] to the thousands place and retain [latex]1[\/latex] in the hundreds place. This comes as no surprise to us since we readily see that [latex]11[\/latex] hundreds is the same as one thousand, one hundred. Carrying is a pretty typical occurrence in a base system.<\/p>\r\n<p>Watch this video to see more examples of converting numbers in bases other than [latex]10[\/latex] into a base-[latex]10[\/latex] number.<span style=\"font-size: 0.9em; font-style: italic; text-align: initial; background-color: initial;\">\u00a0<\/span><\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/TjvexIVV_gI\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Convert+Numbers+in+Base+Ten+to+Different+Bases_+Remainder+Method.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cConvert Numbers in Base Ten to Different Bases: Remainder Method\u201d here (opens in new window).<\/a><\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">Convert [latex]41065_{7}[\/latex] to a base-[latex]10[\/latex] number.<br \/>\r\n[reveal-answer q=\"896067\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"896067\"][latex]41065_{7} = 9994_{10}[\/latex][\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]2331[\/ohm2_question]<\/section>\r\n<h2>Another Method For Converting From Base-[latex]10[\/latex] to Other Bases<\/h2>\r\n<section class=\"textbox recall\">This section presents a method of converting bases that uses a calculator to do the heavy lifting for you. You'll often find, after learning a method to compute a mathematical result by hand, that there is an easier or faster way to do it with a calculator. But it is still beneficial to learn the manual method because the underlying process can contain a logic hidden by the calculator. That logic is often transportable to other situations.<\/section>\r\n<p>As budding mathematicians, you should always be asking questions like \u201cHow could I simplify this process?\u201d In general, that is one of the main things that mathematicians do: they look for ways to take complicated situations and make them easier or more familiar. In this section we will attempt to do that. To do so, we will start by looking at our own decimal system. What we do may seem obvious and maybe even intuitive but that\u2019s the point. We want to find a process that we readily recognize works and makes sense to us in a familiar system and then use it to extend our results to a different, unfamiliar system.<\/p>\r\n<p>Let\u2019s start with the decimal number, [latex]4863_{10}[\/latex]. We will convert this number to base-[latex]10[\/latex]. Yeah, I know it\u2019s already in base-[latex]10[\/latex], but if you carefully follow what we\u2019re doing, you\u2019ll see it makes things work out very nicely with other bases later on. We first note that the highest power of [latex]10[\/latex] that will divide into [latex]4863[\/latex] at least once is [latex]10^3 = 1000[\/latex]. <em>In general, this is the first step in our new process; we find the highest power of a given base that will divide at least once into our given number.<\/em><\/p>\r\n<p style=\"text-align: center;\">We now divide [latex]1000[\/latex] into [latex]4863[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]4863 \u00f7 1000 = 4.863[\/latex]<\/p>\r\n<p>This says that there are four thousands in [latex]4863[\/latex] (obviously). However, it also says that there are [latex]0.863[\/latex] thousands in [latex]4863[\/latex]. This fractional part is our remainder and will be converted to lower powers of our base ([latex]10[\/latex]). If we take that decimal and multiply by [latex]10[\/latex] (since that\u2019s the base we\u2019re in) we get the following:<\/p>\r\n<p style=\"text-align: center;\">[latex]0.863 \u00d7 10 = 8.63[\/latex]<\/p>\r\n<p>Why multiply by [latex]10[\/latex] at this point? We need to recognize here that [latex]0.863[\/latex] thousands is the same as [latex]8.63[\/latex] hundreds. Think about that until it sinks in.<\/p>\r\n<p style=\"text-align: center;\">[latex](0.863)(1000) = 863[\/latex]<br \/>\r\n[latex](8.63)(100) = 863[\/latex]<\/p>\r\n<p>These two statements are equivalent. So, what we are really doing here by multiplying by [latex]10[\/latex] is rephrasing or converting from one place (thousands) to the next place down (hundreds).<\/p>\r\n<p style=\"text-align: center;\">[latex]0.863 \u00d7 10 \\Rightarrow 8.63[\/latex]<br \/>\r\n[latex](\\text{Parts of Thousands}) \u00d7 10 \\Rightarrow \\text{Hundreds}[\/latex]<\/p>\r\n<p>What we have now is [latex]8[\/latex] hundreds and a remainder of [latex]0.63[\/latex] hundreds, which is the same as [latex]6.3[\/latex] tens. We can do this again with the [latex]0.63[\/latex] that remains after this first step.<\/p>\r\n<p style=\"text-align: center;\">[latex]0.63 \u00d7 10 \\Rightarrow 6.3[\/latex]<br \/>\r\n[latex]\\text{Hundreds} \u00d7 10 \\Rightarrow \\text{Tens}[\/latex]<\/p>\r\n<p>So we have six tens and [latex]0.3[\/latex] tens, which is the same as [latex]3[\/latex] ones, our last place value.<\/p>\r\n<p>Now here\u2019s the punch line. Let\u2019s put all of the together in one place:<\/p>\r\n<center>\r\n[caption id=\"attachment_300\" align=\"aligncenter\" width=\"237\"]<img class=\"wp-image-300 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/282\/2016\/01\/20155212\/Fig5_1_20.png\" alt=\"4863 divided by 1000 = 4.863, 0.863x 10 = 8.63, 0.63x10 = 6.3, 0.3x10 = 3.0\" width=\"237\" height=\"200\" \/> Figure 1. Converting bases[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>Note that in each step, the remainder is carried down to the next step and multiplied by [latex]10[\/latex], the base. Also, at each step, the whole number part, which is circled, gives the digit that belongs in that particular place. What is amazing is that this works for any base! So, to convert from a base-[latex]10[\/latex] number to some other base, [latex]b[\/latex], we have the following steps we can follow:<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How to: Convert from Base-[latex]10[\/latex] to Base-[latex]b[\/latex]: Another method<\/strong><\/p>\r\n<ol>\r\n\t<li>Find the highest power of the base-[latex]b[\/latex] that will divide into the given number at least once and then divide.<\/li>\r\n\t<li>Keep the whole number part, and multiply the fractional part by the base-[latex]b[\/latex].<\/li>\r\n\t<li>Repeat step two, keeping the whole number part (including [latex]0[\/latex]), carrying the fractional part to the next step until only a whole number result is obtained.<\/li>\r\n\t<li>Collect all your whole number parts to get your number in base-[latex]b[\/latex] notation.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<p>We will illustrate this procedure with some examples.<\/p>\r\n<section class=\"textbox seeExample\">Convert the base-[latex]10[\/latex] number, [latex]348_{10}[\/latex], to base-[latex]5[\/latex].<br \/>\r\n[reveal-answer q=\"881622\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"881622\"]This is actually a conversion that we have done in a previous example. The powers of five are:<br \/>\r\n[latex]5^{0} = 1[\/latex]<br \/>\r\n[latex]5^{1} = 5[\/latex]<br \/>\r\n[latex]5^{2} = 25[\/latex]<br \/>\r\n[latex]5^{3} = 125[\/latex]<br \/>\r\n[latex]5^{4} = 625[\/latex]<br \/>\r\nEtc\u2026The highest power of five that will go into [latex]348[\/latex] at least once is [latex]5^{3}[\/latex]. We divide by [latex]125[\/latex] and then proceed.<center><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5207\/2020\/04\/09213605\/Screen-Shot-2020-11-09-at-1.35.48-PM.png\"><img class=\"alignnone wp-image-5343 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5207\/2020\/04\/09213605\/Screen-Shot-2020-11-09-at-1.35.48-PM-211x300.png\" alt=\"348\/125=2.784, .784(5)=3.92, .92(5)=4.6, .6(5)=3.0\" width=\"211\" height=\"300\" \/><\/a><\/center>\r\n<p>&nbsp;<\/p>\r\n\r\nBy keeping all the whole number parts, from top bottom, gives [latex]2343[\/latex] as our base-[latex]5[\/latex] number. Thus, [latex]2343_{5}\u00a0= 348_{10}.[\/latex][\/hidden-answer]<\/section>\r\n<p>We can compare our result with what we saw earlier, or simply check with our calculator, and find that these two numbers really are equivalent to each other.<\/p>\r\n<section class=\"textbox seeExample\">Convert the base-[latex]10[\/latex] number, [latex]3007_{10}[\/latex], to base-[latex]5[\/latex].<br \/>\r\n[reveal-answer q=\"462788\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"462788\"]The highest power of [latex]5[\/latex] that divides at least once into [latex]3007[\/latex] is [latex]5^{4} = 625[\/latex]. Thus, we have:<center>[latex]3007 \u00f7 625 = 4.8112[\/latex][latex]0.8112 \u00d7 5 = 4.056[\/latex]<\/center><center>[latex]0.056 \u00d7 5 = 0.28[\/latex]<\/center><center>[latex]0.28 \u00d7 5 = 1.4[\/latex]<\/center><center>[latex]0.4 \u00d7 5 = 2.0[\/latex]<\/center>This gives us that [latex]3007_{10} = 44012_{5}[\/latex]. <br \/>\r\nNotice that in the third line that multiplying by [latex]5[\/latex] gave us [latex]0[\/latex] for our whole number part. We don\u2019t discard that! The zero tells us that a zero is in that place. That is, there are no [latex]5^{2}[\/latex]s in this number.[\/hidden-answer]<\/section>\r\n<p>This last example shows the importance of using a calculator in certain situations and taking care to avoid clearing the calculator\u2019s memory or display until you get to the very end of the process.<\/p>\r\n<section class=\"textbox seeExample\">Convert the base-[latex]10[\/latex] number, [latex]63201_{10}[\/latex], to base-[latex]7[\/latex].<br \/>\r\n[reveal-answer q=\"186862\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"186862\"]The powers of [latex]7[\/latex] are:<center>[latex]7^{0} = 1[\/latex][latex]7^{1} = 7[\/latex]<\/center><center>[latex]7^{2} = 49[\/latex]<\/center><center>[latex]7^{3} = 343[\/latex]<\/center><center>[latex]7^{4} = 2401[\/latex]<\/center><center>[latex]7^{5} = 16807[\/latex]<\/center>\r\n<p>etc\u2026<\/p>\r\n\r\nThe highest power of [latex]7[\/latex] that will divide at least once into [latex]63201[\/latex] is [latex]7^{5}[\/latex]. When we do the initial division on a calculator, we get the following:<center>[latex]63201 \u00f7 7^{5} = 3.760397453[\/latex]<br \/>\r\n<br \/>\r\n<\/center>The decimal part actually fills up the calculators display and we don\u2019t know if it terminates at some point or perhaps even repeats down the road. So if we clear our calculator at this point, we will introduce error that is likely to keep this process from ever ending. To avoid this problem, we leave the result in the calculator and simply subtract [latex]3[\/latex] from this to get the fractional part all by itself. <strong>Do not round off!<\/strong> Subtraction and then multiplication by seven gives:<br \/>\r\n<br \/>\r\n<center>[latex]63201 \u00f7 7^{5} = \u2462.760397453[\/latex]<\/center><center>[latex]0.760397453 \u00d7 7 = \u2464.322782174[\/latex]<\/center><center>[latex]0.322782174 \u00d7 7 = \u2461.259475219[\/latex]<\/center><center>[latex]0.259475219 \u00d7 7 = \u2460.816326531[\/latex]<\/center><center>[latex]0.816326531 \u00d7 7 = \u2464.714285714[\/latex]&lt;<\/center><center>[latex]0.714285714 \u00d7 7 = \u2464.000000000[\/latex]<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>Yes, believe it or not, that last product is exactly [latex]5[\/latex], <em>as long as you don\u2019t clear anything out on your calculator<\/em>. This gives us our final result: [latex]63201_{10} = 352155_{7}[\/latex].<\/p>\r\n\r\n[\/hidden-answer]<\/section>\r\n<p>If we round, even to two decimal places in each step, clearing our calculator out at each step along the way, we will get a series of numbers that do not terminate but begin repeating themselves endlessly. (Try it!) We end up with something that doesn\u2019t make any sense, at least not in this context. So be careful to use your calculator cautiously on these conversion problems.<\/p>\r\n<p>Also, remember that if your first division is by [latex]7^{5}[\/latex], then you expect to have [latex]6[\/latex] digits in the final answer, corresponding to the places for [latex]7^{5}[\/latex], [latex]7^{4}[\/latex], and so on down to [latex]7^{0}[\/latex]. If you find yourself with more than [latex]6[\/latex] digits due to rounding errors, you know something went wrong.<\/p>\r\n<p>The following video shows how to use a calculator to convert numbers in base-[latex]10[\/latex] into other bases.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/YNPTYelCeIs\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Convert+Numbers+in+Base+Ten+to+Different+Bases_+Calculator+Method.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cConvert Numbers in Base Ten to Different Bases: Calculator Method\u201d here (opens in new window).<\/a><\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand different number systems<\/li>\n<li>Convert different number systems<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox recall\">Recall that in our base-[latex]10[\/latex] number system, each place value in a number represents a power of ten. Our numbers take the form thousands, hundreds, tens, ones or [latex]10^3[\/latex], [latex]10^2[\/latex], [latex]10^1[\/latex], [latex]10^0[\/latex]. Remember [latex]10^{0}=1[\/latex], any number raised to the [latex]\\text{zero}^{th}[\/latex] power equals one.We have an intuitive understanding that [latex]10^{1}=10[\/latex] because we are using [latex]10[\/latex] as a factor [latex]1[\/latex] time. And certainly [latex]10^{2} = 10\\times 10 = 100, 10^{3}=1000,[\/latex] and so on. This pattern works with bases other than [latex]10[\/latex] as well. As you\u2019ll see below, using [latex]5[\/latex] as a base yields the following:<\/p>\n<div style=\"text-align: center;\">[latex]5^{0}=1[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]5^{1}=5[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]5^{2}=5\\times 5 = 25[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]5^{3}=5\\times 5\\times 5 = 125,[\/latex]<\/div>\n<p>and so on.<\/section>\n<p>In our base-[latex]10[\/latex] system, a number like [latex]5,783,216[\/latex] has meaning to us because we are familiar with the system and its places. As we know, there are six ones, since there is a [latex]6[\/latex] in the ones place. Likewise, there are seven \u201chundred thousands,\u201d since the [latex]7[\/latex] resides in that place. Each digit has a value that is explicitly determined by its position within the number. We make a distinction between digit, which is just a symbol such as [latex]5[\/latex], and a number, which is made up of one or more digits. We can take this number and assign each of its digits a value. One way to do this is with a table, which follows:<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"text-align: center;\">[latex]5,000,000[\/latex]<\/td>\n<td colspan=\"1\">[latex]= 5 \u00d7 1,000,000[\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 5 \u00d7 10^6[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\text{Five million}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]+700,000[\/latex]<\/td>\n<td colspan=\"1\">[latex]=7 \u00d7 100,000[\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"2\">[latex]=7 \u00d7 10^5[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\begin{align}\\text{Seven hundred} \\text{ thousand}\\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]+80,000[\/latex]<\/td>\n<td colspan=\"1\">[latex]= 8 \u00d7 10,000[\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 8 \u00d7 10^4[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\text{Eighty thousand}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]+3,000[\/latex]<\/td>\n<td colspan=\"1\">[latex]= 3 \u00d7 1000[\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 3 \u00d7 10^3[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\text{Three thousand}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]+200[\/latex]<\/td>\n<td colspan=\"1\">[latex]= 2 \u00d7 100[\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 2 \u00d7 10^2[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\text{Two hundred}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]+10[\/latex]<\/td>\n<td colspan=\"1\">[latex]= 1 \u00d7 10[\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 1 \u00d7 10^1[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\text{Ten}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]+6[\/latex]<\/td>\n<td colspan=\"1\">[latex]= 6 \u00d7 1[\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"2\">[latex]= 6 \u00d7 10^0[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\text{Six}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]5,783,216[\/latex]<\/td>\n<td colspan=\"1\">[latex][\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"2\">[latex][\/latex]<\/td>\n<td style=\"text-align: center;\" colspan=\"4\">[latex]\\begin{align} \\hspace{0.8cm} \\text{Five million, seven hundred} \\\\ \\hspace{0.8cm} \\text{eighty-three thousand, two hundred sixteen}\\end{align}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>From the third column in the table, we can see that each place is simply a multiple of ten. Of course, this makes sense given that our base is ten. The digits that are multiplying each place simply tell us how many of that place we have. We are restricted to having at most [latex]9[\/latex] in any one place before we have to \u201ccarry\u201d over to the next place. We cannot, for example, have [latex]11[\/latex] in the hundreds place. Instead, we would carry [latex]1[\/latex] to the thousands place and retain [latex]1[\/latex] in the hundreds place. This comes as no surprise to us since we readily see that [latex]11[\/latex] hundreds is the same as one thousand, one hundred. Carrying is a pretty typical occurrence in a base system.<\/p>\n<p>Watch this video to see more examples of converting numbers in bases other than [latex]10[\/latex] into a base-[latex]10[\/latex] number.<span style=\"font-size: 0.9em; font-style: italic; text-align: initial; background-color: initial;\">\u00a0<\/span><\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/TjvexIVV_gI\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Convert+Numbers+in+Base+Ten+to+Different+Bases_+Remainder+Method.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cConvert Numbers in Base Ten to Different Bases: Remainder Method\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<section class=\"textbox tryIt\">Convert [latex]41065_{7}[\/latex] to a base-[latex]10[\/latex] number.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q896067\">Show Solution<\/button><\/p>\n<div id=\"q896067\" class=\"hidden-answer\" style=\"display: none\">[latex]41065_{7} = 9994_{10}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm2331\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=2331&theme=lumen&iframe_resize_id=ohm2331&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h2>Another Method For Converting From Base-[latex]10[\/latex] to Other Bases<\/h2>\n<section class=\"textbox recall\">This section presents a method of converting bases that uses a calculator to do the heavy lifting for you. You&#8217;ll often find, after learning a method to compute a mathematical result by hand, that there is an easier or faster way to do it with a calculator. But it is still beneficial to learn the manual method because the underlying process can contain a logic hidden by the calculator. That logic is often transportable to other situations.<\/section>\n<p>As budding mathematicians, you should always be asking questions like \u201cHow could I simplify this process?\u201d In general, that is one of the main things that mathematicians do: they look for ways to take complicated situations and make them easier or more familiar. In this section we will attempt to do that. To do so, we will start by looking at our own decimal system. What we do may seem obvious and maybe even intuitive but that\u2019s the point. We want to find a process that we readily recognize works and makes sense to us in a familiar system and then use it to extend our results to a different, unfamiliar system.<\/p>\n<p>Let\u2019s start with the decimal number, [latex]4863_{10}[\/latex]. We will convert this number to base-[latex]10[\/latex]. Yeah, I know it\u2019s already in base-[latex]10[\/latex], but if you carefully follow what we\u2019re doing, you\u2019ll see it makes things work out very nicely with other bases later on. We first note that the highest power of [latex]10[\/latex] that will divide into [latex]4863[\/latex] at least once is [latex]10^3 = 1000[\/latex]. <em>In general, this is the first step in our new process; we find the highest power of a given base that will divide at least once into our given number.<\/em><\/p>\n<p style=\"text-align: center;\">We now divide [latex]1000[\/latex] into [latex]4863[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]4863 \u00f7 1000 = 4.863[\/latex]<\/p>\n<p>This says that there are four thousands in [latex]4863[\/latex] (obviously). However, it also says that there are [latex]0.863[\/latex] thousands in [latex]4863[\/latex]. This fractional part is our remainder and will be converted to lower powers of our base ([latex]10[\/latex]). If we take that decimal and multiply by [latex]10[\/latex] (since that\u2019s the base we\u2019re in) we get the following:<\/p>\n<p style=\"text-align: center;\">[latex]0.863 \u00d7 10 = 8.63[\/latex]<\/p>\n<p>Why multiply by [latex]10[\/latex] at this point? We need to recognize here that [latex]0.863[\/latex] thousands is the same as [latex]8.63[\/latex] hundreds. Think about that until it sinks in.<\/p>\n<p style=\"text-align: center;\">[latex](0.863)(1000) = 863[\/latex]<br \/>\n[latex](8.63)(100) = 863[\/latex]<\/p>\n<p>These two statements are equivalent. So, what we are really doing here by multiplying by [latex]10[\/latex] is rephrasing or converting from one place (thousands) to the next place down (hundreds).<\/p>\n<p style=\"text-align: center;\">[latex]0.863 \u00d7 10 \\Rightarrow 8.63[\/latex]<br \/>\n[latex](\\text{Parts of Thousands}) \u00d7 10 \\Rightarrow \\text{Hundreds}[\/latex]<\/p>\n<p>What we have now is [latex]8[\/latex] hundreds and a remainder of [latex]0.63[\/latex] hundreds, which is the same as [latex]6.3[\/latex] tens. We can do this again with the [latex]0.63[\/latex] that remains after this first step.<\/p>\n<p style=\"text-align: center;\">[latex]0.63 \u00d7 10 \\Rightarrow 6.3[\/latex]<br \/>\n[latex]\\text{Hundreds} \u00d7 10 \\Rightarrow \\text{Tens}[\/latex]<\/p>\n<p>So we have six tens and [latex]0.3[\/latex] tens, which is the same as [latex]3[\/latex] ones, our last place value.<\/p>\n<p>Now here\u2019s the punch line. Let\u2019s put all of the together in one place:<\/p>\n<div style=\"text-align: center;\">\n<figure id=\"attachment_300\" aria-describedby=\"caption-attachment-300\" style=\"width: 237px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-300 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/282\/2016\/01\/20155212\/Fig5_1_20.png\" alt=\"4863 divided by 1000 = 4.863, 0.863x 10 = 8.63, 0.63x10 = 6.3, 0.3x10 = 3.0\" width=\"237\" height=\"200\" \/><figcaption id=\"caption-attachment-300\" class=\"wp-caption-text\">Figure 1. Converting bases<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Note that in each step, the remainder is carried down to the next step and multiplied by [latex]10[\/latex], the base. Also, at each step, the whole number part, which is circled, gives the digit that belongs in that particular place. What is amazing is that this works for any base! So, to convert from a base-[latex]10[\/latex] number to some other base, [latex]b[\/latex], we have the following steps we can follow:<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>How to: Convert from Base-[latex]10[\/latex] to Base-[latex]b[\/latex]: Another method<\/strong><\/p>\n<ol>\n<li>Find the highest power of the base-[latex]b[\/latex] that will divide into the given number at least once and then divide.<\/li>\n<li>Keep the whole number part, and multiply the fractional part by the base-[latex]b[\/latex].<\/li>\n<li>Repeat step two, keeping the whole number part (including [latex]0[\/latex]), carrying the fractional part to the next step until only a whole number result is obtained.<\/li>\n<li>Collect all your whole number parts to get your number in base-[latex]b[\/latex] notation.<\/li>\n<\/ol>\n<\/section>\n<p>We will illustrate this procedure with some examples.<\/p>\n<section class=\"textbox seeExample\">Convert the base-[latex]10[\/latex] number, [latex]348_{10}[\/latex], to base-[latex]5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q881622\">Show Solution<\/button><\/p>\n<div id=\"q881622\" class=\"hidden-answer\" style=\"display: none\">This is actually a conversion that we have done in a previous example. The powers of five are:<br \/>\n[latex]5^{0} = 1[\/latex]<br \/>\n[latex]5^{1} = 5[\/latex]<br \/>\n[latex]5^{2} = 25[\/latex]<br \/>\n[latex]5^{3} = 125[\/latex]<br \/>\n[latex]5^{4} = 625[\/latex]<br \/>\nEtc\u2026The highest power of five that will go into [latex]348[\/latex] at least once is [latex]5^{3}[\/latex]. We divide by [latex]125[\/latex] and then proceed.<\/p>\n<div style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5207\/2020\/04\/09213605\/Screen-Shot-2020-11-09-at-1.35.48-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-5343 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5207\/2020\/04\/09213605\/Screen-Shot-2020-11-09-at-1.35.48-PM-211x300.png\" alt=\"348\/125=2.784, .784(5)=3.92, .92(5)=4.6, .6(5)=3.0\" width=\"211\" height=\"300\" \/><\/a><\/div>\n<p>&nbsp;<\/p>\n<p>By keeping all the whole number parts, from top bottom, gives [latex]2343[\/latex] as our base-[latex]5[\/latex] number. Thus, [latex]2343_{5}\u00a0= 348_{10}.[\/latex]<\/p><\/div>\n<\/div>\n<\/section>\n<p>We can compare our result with what we saw earlier, or simply check with our calculator, and find that these two numbers really are equivalent to each other.<\/p>\n<section class=\"textbox seeExample\">Convert the base-[latex]10[\/latex] number, [latex]3007_{10}[\/latex], to base-[latex]5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q462788\">Show Solution<\/button><\/p>\n<div id=\"q462788\" class=\"hidden-answer\" style=\"display: none\">The highest power of [latex]5[\/latex] that divides at least once into [latex]3007[\/latex] is [latex]5^{4} = 625[\/latex]. Thus, we have:<\/p>\n<div style=\"text-align: center;\">[latex]3007 \u00f7 625 = 4.8112[\/latex][latex]0.8112 \u00d7 5 = 4.056[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]0.056 \u00d7 5 = 0.28[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]0.28 \u00d7 5 = 1.4[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]0.4 \u00d7 5 = 2.0[\/latex]<\/div>\n<p>This gives us that [latex]3007_{10} = 44012_{5}[\/latex]. <br \/>\nNotice that in the third line that multiplying by [latex]5[\/latex] gave us [latex]0[\/latex] for our whole number part. We don\u2019t discard that! The zero tells us that a zero is in that place. That is, there are no [latex]5^{2}[\/latex]s in this number.<\/div>\n<\/div>\n<\/section>\n<p>This last example shows the importance of using a calculator in certain situations and taking care to avoid clearing the calculator\u2019s memory or display until you get to the very end of the process.<\/p>\n<section class=\"textbox seeExample\">Convert the base-[latex]10[\/latex] number, [latex]63201_{10}[\/latex], to base-[latex]7[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q186862\">Show Solution<\/button><\/p>\n<div id=\"q186862\" class=\"hidden-answer\" style=\"display: none\">The powers of [latex]7[\/latex] are:<\/p>\n<div style=\"text-align: center;\">[latex]7^{0} = 1[\/latex][latex]7^{1} = 7[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]7^{2} = 49[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]7^{3} = 343[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]7^{4} = 2401[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]7^{5} = 16807[\/latex]<\/div>\n<p>etc\u2026<\/p>\n<p>The highest power of [latex]7[\/latex] that will divide at least once into [latex]63201[\/latex] is [latex]7^{5}[\/latex]. When we do the initial division on a calculator, we get the following:<\/p>\n<div style=\"text-align: center;\">[latex]63201 \u00f7 7^{5} = 3.760397453[\/latex]<\/p>\n<\/div>\n<p>The decimal part actually fills up the calculators display and we don\u2019t know if it terminates at some point or perhaps even repeats down the road. So if we clear our calculator at this point, we will introduce error that is likely to keep this process from ever ending. To avoid this problem, we leave the result in the calculator and simply subtract [latex]3[\/latex] from this to get the fractional part all by itself. <strong>Do not round off!<\/strong> Subtraction and then multiplication by seven gives:<\/p>\n<div style=\"text-align: center;\">[latex]63201 \u00f7 7^{5} = \u2462.760397453[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]0.760397453 \u00d7 7 = \u2464.322782174[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]0.322782174 \u00d7 7 = \u2461.259475219[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]0.259475219 \u00d7 7 = \u2460.816326531[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]0.816326531 \u00d7 7 = \u2464.714285714[\/latex]&lt;<\/div>\n<div style=\"text-align: center;\">[latex]0.714285714 \u00d7 7 = \u2464.000000000[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Yes, believe it or not, that last product is exactly [latex]5[\/latex], <em>as long as you don\u2019t clear anything out on your calculator<\/em>. This gives us our final result: [latex]63201_{10} = 352155_{7}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>If we round, even to two decimal places in each step, clearing our calculator out at each step along the way, we will get a series of numbers that do not terminate but begin repeating themselves endlessly. (Try it!) We end up with something that doesn\u2019t make any sense, at least not in this context. So be careful to use your calculator cautiously on these conversion problems.<\/p>\n<p>Also, remember that if your first division is by [latex]7^{5}[\/latex], then you expect to have [latex]6[\/latex] digits in the final answer, corresponding to the places for [latex]7^{5}[\/latex], [latex]7^{4}[\/latex], and so on down to [latex]7^{0}[\/latex]. If you find yourself with more than [latex]6[\/latex] digits due to rounding errors, you know something went wrong.<\/p>\n<p>The following video shows how to use a calculator to convert numbers in base-[latex]10[\/latex] into other bases.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/YNPTYelCeIs\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Convert+Numbers+in+Base+Ten+to+Different+Bases_+Calculator+Method.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cConvert Numbers in Base Ten to Different Bases: Calculator Method\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":40,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/52"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":70,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/52\/revisions"}],"predecessor-version":[{"id":15567,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/52\/revisions\/15567"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/40"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/52\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=52"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=52"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=52"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=52"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}