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CompTIA A+ Certification All-in-One Exam Guide

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Because the 8088 had a 20-wire address bus, the most RAM it could

handle was 2 20 or 1,048,576 bytes. The 8088, therefore, had an address space

of 1,048,576 bytes. This is not to say that every computer with an 8088 CPU

had 1,048,576 bytes of RAM. Far from it! The original IBM PC only had a

measly 65,536 bytes—but that was considered plenty back in the Dark Ages

of Computing in the early 1980s.

Okay, so you know that the 8088 had 20 address wires and a total address

space of 1,048,576 bytes. Although this is accurate, no one uses such an exact

term to discuss the address space of the 8088. Instead you say that the 8088

had one megabyte (1 MB) of address space.

What’s a “mega”? Well, let’s get some terminology down. Dealing with

computers means constantly dealing with the number of patterns a set of

wires can handle. Certain powers of 2 have names used a lot in computing.

The following list explains.

1 kilo = 2 10 = 1024 (abbreviated as “K”)

1 kilobyte = 1024 bytes (abbreviated as “KB”)

1 mega = 2 20 = 1,048,576 (abbreviated as “M”)

1 megabyte = 1,048,576 bytes (abbreviated as “MB”)

1 giga = 2 30 = 1,073,741,824 (abbreviated as “G”)

1 gigabyte = 1,073,741,824 bytes (abbreviated as “GB”)

1 tera = 2 40 = 1,099,511,627,776 (abbreviated as “T”)

1 terabyte = 1,099,511,627,776 bytes (abbreviated as “TB”)

Metric System and Computer Memory

There’s a problem with that list you just read. If you asked a metric system

expert for explanation, she would say that a kilo is equal to exactly 1000, not

1024! Am I lying to you?

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