Hexadecimal Subtraction and MultiplicationDate: 04/06/2000 at 11:53:40 From: Peggy Schnurr Subject: Hexadecimal multiplication I understand adding hexadecimal numbers, but I'm having trouble understanding multiplication and subtraction of hexadecimal numbers. I would appreciate any direction you can offer on the subject. Thank you. Peggy Schnurr Date: 04/06/2000 at 17:21:07 From: Doctor TWE Subject: Re: Hexadecimal multiplication Hi Peggy - thanks for writing to Dr. Math. Let's tackle subtraction first. Subtraction in hexadecimal works similar to subtraction in decimal. Let's do a pair of decimal problems and see if we can "parallel" the solution method in hex. Suppose we want to subtract 1586 - 243 (in decimal). We write the subtrahend (the number to be subtracted) underneath the minuend (the number to subtract from), lining up the place values at the decimal point - or at the right if there is no decimal point - like this: 1586 - 243 ---- Now, starting from the right, we subtract the digits. (Ones place) 6-3 = 3, (tens place) 8-4 = 4, (hundreds place) 5-2 = 3, (thousands place) 1-0 = 1. It looks like this: 1586 - 243 ---- 1343 We do the same thing in hexadecimal, except that we occasionally have a digit larger than 9. For example, suppose we want to subtract 5CD2 - 2A0 (hexadecimal). As with decimal, we write the subtrahend (the number to be subtracted) underneath the minuend (the number to subtract from), lining up the place values at the radix point - or at the right if there is no radix point - like this: 5CD2 h (The 'h' just indicates a hexadecimal value) - 2A0 h ------ Now, starting from the right, we subtract the digits. (Ones place) 2-0 = 2, (sixteens place) D-A = 13-10 = 3, (256's place) C-2 = 12-2 = 10 = A, (4096's place) 5-0 = 5. It looks like this: 5CD2 h - 2A0 h ------ 5A32 h That's fine - until you get a place value in which there's not enough to subtract from. What do we do in decimal? We borrow from the next higher place value. For example, suppose we want to subtract 3145 - 1976 (decimal). We'll align our numbers: 3 1 4 5 - 1 9 7 6 ---------- Now in the ones place, we can't subtract 6 from 5 (5 is less than 6) so we borrow 1 from the tens place. Since that has a place value (a "weight") of ten, we can exchange it for 10 ones. (This is kind of like changing a $10 bill into ten $1 bills.) This leaves us with 3 tens (4 minus the 1 we borrowed) and gives us 15 ones (5 plus the 10 we got from the borrow). We now subtract 15-6 = 9: 3 15 3 1 \ \ (There was a 4 and a 5 under the '\'s) - 1 9 7 6 ---------- 9 Similarly in the tens place, we can't subtract 7 from 3 so we borrow one from the hundreds place. Since that has a place value (a "weight") of one hundred, we can exchange it for 10 tens. (This is kind of like changing a $100 bill into ten $10 bills.) This leaves us with no hundreds (1 minus the 1 we borrowed) and gives us 13 ones (the 3 left over after the previous borrow plus the 10 we got from this borrow). We now subtract 13-7 = 6: 0 13 15 3 \ \ \ - 1 9 7 6 ---------- 6 9 We do the same for the hundreds, borrowing 1 from the thousands, then subtract the thousands. We end up like this: 2 10 13 15 \ \ \ \ - 1 9 7 6 ---------- 1 1 6 9 Why did we get 10 from each borrow? Because each time, the next place value was 10 times as large. This is because decimal is base-10. Now let's look at a hexadecimal problem that requires borrowing. For example, suppose we want to subtract A8D2 - 3EAC (hexadecimal). We'll align our numbers: A 8 D 2 h - 3 E A C h ------------- Now in the ones place, we can't subtract C (12) from 2 so we borrow 1 from the sixteens place. Since that has a place value (a "weight") of sixteen, we can exchange it for 16 ones. (This is kind of like changing a $16 bill into sixteen $1 bills.) This leaves us with 12 sixteens (D = 13 minus the 1 we borrowed) and gives us 18 ones (2 plus the 16 we got from the borrow). We now subtract 18-12 = 6: (Note that I use decimal here. Some people 12 18 write these as Ch and 12h.) A 8 \ \ h (There was a D and a 2 under the '\'s) - 3 E A C h ------------- 6 h In the sixteens place, we don't need to borrow because we can subtract 10 (A) from 12: 12 18 A 8 \ \ h - 3 E A C h ------------- 2 6 h In the 256's place, we again need to borrow. We'll borrow 1 from the 4096's place and exchange it for sixteen 256's (one 4096 equals sixteen 256's). This leaves us 9 in the 4096's place (A = 10 minus the 1 that we borrowed), and gives us 24 in the 256's place (8 plus the 16 from the borrow). We then can subtract 24-14 = 10 = A. So we have: 9 24 12 18 \ \ \ \ h - 3 E A C h ------------- A 2 6 h Finally, we subtract 9-3 = 6 in the 4096's place: 9 24 12 18 \ \ \ \ h - 3 E A C h ------------- 6 A 2 6 h Why did we get 16 from each borrow? Because each time, the next place value was 16 times as large. This is because hexadecimal is base-16. One final note: If the subtrahend (the bottom number) is larger than the minuend (the top number), flip the numbers around and make the final answer negative, just as you would in decimal. I'll send you information on multiplication tomorrow, as I have run out of time today. (I have a 5-year old I must pick up from kindergarten.) I hope this helps. If you have any more questions, write back. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/ Date: 04/06/2000 at 13:04:55 From: Doctor Peterson Subject: Re: Hexadecimal multiplication Hi, Peggy. All the operations work exactly the same in any base, as long as you use the right tables and remember to use the base when you carry or borrow. For example, here's 25 x 3A (37 x 58 base 10): 25 x 3A ---- 32 A x 5 (10 x 5 = 50) 14 A x 2 (10 x 2 = 20) F 3 x 5 (3 x 5 = 15) 6 3 x 2 (3 x 2 = 6) ---- 862 = 2146 (base 10) I wrote out each product of a pair of digits here, to save carrying, and to show explicitly the four numbers you would get from a multiplication table (though I just multiplied each pair in base 10 and converted to base 16); normally you'd write it as 25 x 3A ---- 172 = 32 + 140 6F = F + 60 ---- 862 Here, for example, 5 x A = 32, so I put down 2 and carried the 3, which I added to the result of A x 2 = 14 to get 17. When I added the partial products, I found that 7+F = 16, so I put down the 6 and carried the 1. If you haven't looked at our FAQ on number bases, please do so: http://mathforum.org/dr.math/faq/faq.bases.html If there's something you still have trouble with, show me your work on a sample problem so I can see what part to explain. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 04/07/2000 at 11:05:47 From: Doctor TWE Subject: Re: hexadecimal multiplication Hi again, Peggy. I see Dr. Peterson has already answered you regarding multiplication. But since I had already constructed most of this response yesterday, I thought I'd finish it and send it anyway. Let's look at a decimal multiplication problem; for example 325 * 18. Again, we align the numbers by the decimal point: 365 * 28 ----- We start with the ones digit of the multiplier (the bottom number). We multiply this by the ones digit of the multiplicand (the top number), 8*5 = 40. We record the 0 in the ones digit of the first partial product line, and keep the 4 as a carry: 4 365 * 28 ----- 0 Next, we multiply the ones digit of the multiplier by the tens digit of the multiplicand and add the carry; 8*6+4 = 48+4 = 52. We record the 2 in the tens digit of the first partial product line, and keep the 5 as a carry: 54 365 * 28 ----- 20 Now we multiply the ones digit of the multiplier by the hundreds digit of the multiplicand and add the carry; 8*3+5 = 24+5 = 29. We record the 2 in the tens digit of the first partial product line. Since this is the last digit of the multiplicand, we can bring down the carry and record a 29: 254 365 * 28 ----- 2920 Now we repeat the process with the tens digit of the multiplier, and start our answer in the tens digit in the second partial product line. I'll do this all at one (I hope you see how I got all this - I also "erased" my earlier carries): 11 365 * 28 ----- 2920 720 Finally, we add the partial products: 365 * 28 ----- 2920 730 ----- 10220 How did we know that 8*5 = 40, 2*6 = 12, etc.? We probably memorized these as a multiplication table in elementary school. We'll need such a multiplication table for hexadecimal. Here it is: X | 0 1 2 3 4 5 6 7 8 9 A B C D E F --+----------------------------------------------- 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 | 0 1 2 3 4 5 6 7 8 9 A B C D E F 2 | 0 2 4 6 8 A C E 10 12 14 16 18 1A 1C 1E 3 | 0 3 6 9 C F 12 15 18 1B 1E 21 24 27 2A 2D 4 | 0 4 8 C 10 14 18 1C 20 24 28 2C 30 34 38 3C 5 | 0 5 A F 14 19 1E 23 28 2D 32 37 3C 41 46 4B 6 | 0 6 C 12 18 1E 24 2A 30 36 3C 42 48 4E 54 5A 7 | 0 7 E 15 1C 23 2A 31 38 3F 46 4D 54 5B 62 69 8 | 0 8 10 18 20 28 30 38 40 48 50 58 60 68 70 78 9 | 0 9 12 1B 24 2D 36 3F 48 51 5A 63 6C 75 7E 87 A | 0 A 14 1E 28 32 3C 46 50 5A 64 6E 78 82 8C 96 B | 0 B 16 21 2C 37 42 4D 58 63 6E 79 84 8F 9A A5 C | 0 C 18 24 30 3C 48 54 60 6C 78 84 90 9C A8 B4 D | 0 D 1A 27 34 41 4E 5B 68 75 82 8F 9C A9 B6 C3 E | 0 E 1C 2A 38 46 54 62 70 7E 8C 9A A8 B6 C4 D2 F | 0 F 1E 2D 3C 4B 5A 69 78 87 96 A5 B4 C3 D2 E1 We do hexadecimal multiplication the same way we did the decimal, only we use the hex multiplication table above and hex addition. For example, 82E * B4. First, we align the numbers by the radix point: 82E * B4 ----- We start with the ones digit of the multiplier and multiply this by the ones digit of the multiplicand, 4*E = 38. We record the 8 in the ones digit of the first partial product line, and keep the 3 as a carry: 3 82E * B4 ----- 8 Next, we multiply the ones digit of the multiplier by the sixteens digit of the multiplicand and add the carry; 2*4+3 = 8+3 = B. We record the B in the tens digit of the first partial product line (there's no carry this time): 3 82E * B4 ----- B8 Now we multiply the ones digit of the multiplier by the 256's digit of the multiplicand (we'd add the carry if we had one); 4*8 = 20. We record the 0 in the sixteens digit of the first partial product line. Since this is the last digit of the multiplicand, we can bring down the carry and record a 20: 2 4 82E * B4 ----- 20B8 Now we repeat the process with the sixteens digit of the multiplier, and start our answer in the sixteens digit in the second partial product line. I'll do this all in one step (if you don't see how I got all this, write back): 519 82E * B4 ----- 20B8 59FA Finally, we add the partial products (in hexadecimal): 82E * B4 ----- 20B8 59FA ----- 5C058 I hope this helps. If you have any more questions, write back. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/ |
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