Convert the Number -1 302.123 456 789 012 345 67 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations
Number -1 302.123 456 789 012 345 67(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)
The first steps we'll go through to make the conversion:
Convert to binary (to base 2) the integer part of the number.
Convert to binary the fractional part of the number.
1. Start with the positive version of the number:
|-1 302.123 456 789 012 345 67| = 1 302.123 456 789 012 345 67
2. First, convert to binary (in base 2) the integer part: 1 302.
Divide the number repeatedly by 2.
Keep track of each remainder.
We stop when we get a quotient that is equal to zero.
- division = quotient + remainder;
- 1 302 ÷ 2 = 651 + 0;
- 651 ÷ 2 = 325 + 1;
- 325 ÷ 2 = 162 + 1;
- 162 ÷ 2 = 81 + 0;
- 81 ÷ 2 = 40 + 1;
- 40 ÷ 2 = 20 + 0;
- 20 ÷ 2 = 10 + 0;
- 10 ÷ 2 = 5 + 0;
- 5 ÷ 2 = 2 + 1;
- 2 ÷ 2 = 1 + 0;
- 1 ÷ 2 = 0 + 1;
3. Construct the base 2 representation of the integer part of the number.
Take all the remainders starting from the bottom of the list constructed above.
1 302(10) =
101 0001 0110(2)
4. Convert to binary (base 2) the fractional part: 0.123 456 789 012 345 67.
Multiply it repeatedly by 2.
Keep track of each integer part of the results.
Stop when we get a fractional part that is equal to zero.
- #) multiplying = integer + fractional part;
- 1) 0.123 456 789 012 345 67 × 2 = 0 + 0.246 913 578 024 691 34;
- 2) 0.246 913 578 024 691 34 × 2 = 0 + 0.493 827 156 049 382 68;
- 3) 0.493 827 156 049 382 68 × 2 = 0 + 0.987 654 312 098 765 36;
- 4) 0.987 654 312 098 765 36 × 2 = 1 + 0.975 308 624 197 530 72;
- 5) 0.975 308 624 197 530 72 × 2 = 1 + 0.950 617 248 395 061 44;
- 6) 0.950 617 248 395 061 44 × 2 = 1 + 0.901 234 496 790 122 88;
- 7) 0.901 234 496 790 122 88 × 2 = 1 + 0.802 468 993 580 245 76;
- 8) 0.802 468 993 580 245 76 × 2 = 1 + 0.604 937 987 160 491 52;
- 9) 0.604 937 987 160 491 52 × 2 = 1 + 0.209 875 974 320 983 04;
- 10) 0.209 875 974 320 983 04 × 2 = 0 + 0.419 751 948 641 966 08;
- 11) 0.419 751 948 641 966 08 × 2 = 0 + 0.839 503 897 283 932 16;
- 12) 0.839 503 897 283 932 16 × 2 = 1 + 0.679 007 794 567 864 32;
- 13) 0.679 007 794 567 864 32 × 2 = 1 + 0.358 015 589 135 728 64;
- 14) 0.358 015 589 135 728 64 × 2 = 0 + 0.716 031 178 271 457 28;
- 15) 0.716 031 178 271 457 28 × 2 = 1 + 0.432 062 356 542 914 56;
- 16) 0.432 062 356 542 914 56 × 2 = 0 + 0.864 124 713 085 829 12;
- 17) 0.864 124 713 085 829 12 × 2 = 1 + 0.728 249 426 171 658 24;
- 18) 0.728 249 426 171 658 24 × 2 = 1 + 0.456 498 852 343 316 48;
- 19) 0.456 498 852 343 316 48 × 2 = 0 + 0.912 997 704 686 632 96;
- 20) 0.912 997 704 686 632 96 × 2 = 1 + 0.825 995 409 373 265 92;
- 21) 0.825 995 409 373 265 92 × 2 = 1 + 0.651 990 818 746 531 84;
- 22) 0.651 990 818 746 531 84 × 2 = 1 + 0.303 981 637 493 063 68;
- 23) 0.303 981 637 493 063 68 × 2 = 0 + 0.607 963 274 986 127 36;
- 24) 0.607 963 274 986 127 36 × 2 = 1 + 0.215 926 549 972 254 72;
- 25) 0.215 926 549 972 254 72 × 2 = 0 + 0.431 853 099 944 509 44;
- 26) 0.431 853 099 944 509 44 × 2 = 0 + 0.863 706 199 889 018 88;
- 27) 0.863 706 199 889 018 88 × 2 = 1 + 0.727 412 399 778 037 76;
- 28) 0.727 412 399 778 037 76 × 2 = 1 + 0.454 824 799 556 075 52;
- 29) 0.454 824 799 556 075 52 × 2 = 0 + 0.909 649 599 112 151 04;
- 30) 0.909 649 599 112 151 04 × 2 = 1 + 0.819 299 198 224 302 08;
- 31) 0.819 299 198 224 302 08 × 2 = 1 + 0.638 598 396 448 604 16;
- 32) 0.638 598 396 448 604 16 × 2 = 1 + 0.277 196 792 897 208 32;
- 33) 0.277 196 792 897 208 32 × 2 = 0 + 0.554 393 585 794 416 64;
- 34) 0.554 393 585 794 416 64 × 2 = 1 + 0.108 787 171 588 833 28;
- 35) 0.108 787 171 588 833 28 × 2 = 0 + 0.217 574 343 177 666 56;
- 36) 0.217 574 343 177 666 56 × 2 = 0 + 0.435 148 686 355 333 12;
- 37) 0.435 148 686 355 333 12 × 2 = 0 + 0.870 297 372 710 666 24;
- 38) 0.870 297 372 710 666 24 × 2 = 1 + 0.740 594 745 421 332 48;
- 39) 0.740 594 745 421 332 48 × 2 = 1 + 0.481 189 490 842 664 96;
- 40) 0.481 189 490 842 664 96 × 2 = 0 + 0.962 378 981 685 329 92;
- 41) 0.962 378 981 685 329 92 × 2 = 1 + 0.924 757 963 370 659 84;
- 42) 0.924 757 963 370 659 84 × 2 = 1 + 0.849 515 926 741 319 68;
- 43) 0.849 515 926 741 319 68 × 2 = 1 + 0.699 031 853 482 639 36;
- 44) 0.699 031 853 482 639 36 × 2 = 1 + 0.398 063 706 965 278 72;
- 45) 0.398 063 706 965 278 72 × 2 = 0 + 0.796 127 413 930 557 44;
- 46) 0.796 127 413 930 557 44 × 2 = 1 + 0.592 254 827 861 114 88;
- 47) 0.592 254 827 861 114 88 × 2 = 1 + 0.184 509 655 722 229 76;
- 48) 0.184 509 655 722 229 76 × 2 = 0 + 0.369 019 311 444 459 52;
- 49) 0.369 019 311 444 459 52 × 2 = 0 + 0.738 038 622 888 919 04;
- 50) 0.738 038 622 888 919 04 × 2 = 1 + 0.476 077 245 777 838 08;
- 51) 0.476 077 245 777 838 08 × 2 = 0 + 0.952 154 491 555 676 16;
- 52) 0.952 154 491 555 676 16 × 2 = 1 + 0.904 308 983 111 352 32;
- 53) 0.904 308 983 111 352 32 × 2 = 1 + 0.808 617 966 222 704 64;
We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)
5. Construct the base 2 representation of the fractional part of the number.
Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:
0.123 456 789 012 345 67(10) =
0.0001 1111 1001 1010 1101 1101 0011 0111 0100 0110 1111 0110 0101 1(2)
6. Positive number before normalization:
1 302.123 456 789 012 345 67(10) =
101 0001 0110.0001 1111 1001 1010 1101 1101 0011 0111 0100 0110 1111 0110 0101 1(2)
The last steps we'll go through to make the conversion:
Normalize the binary representation of the number.
Adjust the exponent.
Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.
Normalize the mantissa.
7. Normalize the binary representation of the number.
Shift the decimal mark 10 positions to the left, so that only one non zero digit remains to the left of it:
1 302.123 456 789 012 345 67(10) =
101 0001 0110.0001 1111 1001 1010 1101 1101 0011 0111 0100 0110 1111 0110 0101 1(2) =
101 0001 0110.0001 1111 1001 1010 1101 1101 0011 0111 0100 0110 1111 0110 0101 1(2) × 20 =
1.0100 0101 1000 0111 1110 0110 1011 0111 0100 1101 1101 0001 1011 1101 1001 011(2) × 210
8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:
Sign 1 (a negative number)
Exponent (unadjusted): 10
Mantissa (not normalized):
1.0100 0101 1000 0111 1110 0110 1011 0111 0100 1101 1101 0001 1011 1101 1001 011
9. Adjust the exponent.
Use the 11 bit excess/bias notation:
Exponent (adjusted) =
Exponent (unadjusted) + 2(11-1) - 1 =
10 + 2(11-1) - 1 =
(10 + 1 023)(10) =
1 033(10)
10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.
Use the same technique of repeatedly dividing by 2:
- division = quotient + remainder;
- 1 033 ÷ 2 = 516 + 1;
- 516 ÷ 2 = 258 + 0;
- 258 ÷ 2 = 129 + 0;
- 129 ÷ 2 = 64 + 1;
- 64 ÷ 2 = 32 + 0;
- 32 ÷ 2 = 16 + 0;
- 16 ÷ 2 = 8 + 0;
- 8 ÷ 2 = 4 + 0;
- 4 ÷ 2 = 2 + 0;
- 2 ÷ 2 = 1 + 0;
- 1 ÷ 2 = 0 + 1;
11. Construct the base 2 representation of the adjusted exponent.
Take all the remainders starting from the bottom of the list constructed above.
Exponent (adjusted) =
1033(10) =
100 0000 1001(2)
12. Normalize the mantissa.
a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.
b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).
Mantissa (normalized) =
1. 0100 0101 1000 0111 1110 0110 1011 0111 0100 1101 1101 0001 1011 110 1100 1011 =
0100 0101 1000 0111 1110 0110 1011 0111 0100 1101 1101 0001 1011
13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:
Sign (1 bit) =
1 (a negative number)
Exponent (11 bits) =
100 0000 1001
Mantissa (52 bits) =
0100 0101 1000 0111 1110 0110 1011 0111 0100 1101 1101 0001 1011
The base ten decimal number -1 302.123 456 789 012 345 67 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 1001 - 0100 0101 1000 0111 1110 0110 1011 0111 0100 1101 1101 0001 1011
Sign (1 bit):
Exponent (11 bits):
1
620
610
600
590
580
570
561
550
540
531
52
Mantissa (52 bits):
0
511
500
490
480
471
460
451
441
430
420
410
400
391
381
371
361
351
341
330
320
311
301
290
281
270
261
251
240
231
221
211
200
191
180
170
161
151
140
131
121
111
100
91
80
70
60
51
41
30
21
11
0
Convert to 64 bit double precision IEEE 754 binary floating point representation standard
A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)
The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation
How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard
Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:
- 1. If the number to be converted is negative, start with its the positive version.
- 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
- 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
- 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
- 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
- 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
- 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 - 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
- 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.
Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:
- 1. Start with the positive version of the number:
|-31.640 215| = 31.640 215
- 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
- division = quotient + remainder;
- 31 ÷ 2 = 15 + 1;
- 15 ÷ 2 = 7 + 1;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 1 ÷ 2 = 0 + 1;
- We have encountered a quotient that is ZERO => FULL STOP
- 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:
31(10) = 1 1111(2)
- 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
- #) multiplying = integer + fractional part;
- 1) 0.640 215 × 2 = 1 + 0.280 43;
- 2) 0.280 43 × 2 = 0 + 0.560 86;
- 3) 0.560 86 × 2 = 1 + 0.121 72;
- 4) 0.121 72 × 2 = 0 + 0.243 44;
- 5) 0.243 44 × 2 = 0 + 0.486 88;
- 6) 0.486 88 × 2 = 0 + 0.973 76;
- 7) 0.973 76 × 2 = 1 + 0.947 52;
- 8) 0.947 52 × 2 = 1 + 0.895 04;
- 9) 0.895 04 × 2 = 1 + 0.790 08;
- 10) 0.790 08 × 2 = 1 + 0.580 16;
- 11) 0.580 16 × 2 = 1 + 0.160 32;
- 12) 0.160 32 × 2 = 0 + 0.320 64;
- 13) 0.320 64 × 2 = 0 + 0.641 28;
- 14) 0.641 28 × 2 = 1 + 0.282 56;
- 15) 0.282 56 × 2 = 0 + 0.565 12;
- 16) 0.565 12 × 2 = 1 + 0.130 24;
- 17) 0.130 24 × 2 = 0 + 0.260 48;
- 18) 0.260 48 × 2 = 0 + 0.520 96;
- 19) 0.520 96 × 2 = 1 + 0.041 92;
- 20) 0.041 92 × 2 = 0 + 0.083 84;
- 21) 0.083 84 × 2 = 0 + 0.167 68;
- 22) 0.167 68 × 2 = 0 + 0.335 36;
- 23) 0.335 36 × 2 = 0 + 0.670 72;
- 24) 0.670 72 × 2 = 1 + 0.341 44;
- 25) 0.341 44 × 2 = 0 + 0.682 88;
- 26) 0.682 88 × 2 = 1 + 0.365 76;
- 27) 0.365 76 × 2 = 0 + 0.731 52;
- 28) 0.731 52 × 2 = 1 + 0.463 04;
- 29) 0.463 04 × 2 = 0 + 0.926 08;
- 30) 0.926 08 × 2 = 1 + 0.852 16;
- 31) 0.852 16 × 2 = 1 + 0.704 32;
- 32) 0.704 32 × 2 = 1 + 0.408 64;
- 33) 0.408 64 × 2 = 0 + 0.817 28;
- 34) 0.817 28 × 2 = 1 + 0.634 56;
- 35) 0.634 56 × 2 = 1 + 0.269 12;
- 36) 0.269 12 × 2 = 0 + 0.538 24;
- 37) 0.538 24 × 2 = 1 + 0.076 48;
- 38) 0.076 48 × 2 = 0 + 0.152 96;
- 39) 0.152 96 × 2 = 0 + 0.305 92;
- 40) 0.305 92 × 2 = 0 + 0.611 84;
- 41) 0.611 84 × 2 = 1 + 0.223 68;
- 42) 0.223 68 × 2 = 0 + 0.447 36;
- 43) 0.447 36 × 2 = 0 + 0.894 72;
- 44) 0.894 72 × 2 = 1 + 0.789 44;
- 45) 0.789 44 × 2 = 1 + 0.578 88;
- 46) 0.578 88 × 2 = 1 + 0.157 76;
- 47) 0.157 76 × 2 = 0 + 0.315 52;
- 48) 0.315 52 × 2 = 0 + 0.631 04;
- 49) 0.631 04 × 2 = 1 + 0.262 08;
- 50) 0.262 08 × 2 = 0 + 0.524 16;
- 51) 0.524 16 × 2 = 1 + 0.048 32;
- 52) 0.048 32 × 2 = 0 + 0.096 64;
- 53) 0.096 64 × 2 = 0 + 0.193 28;
- We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
- 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:
0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)
- 6. Summarizing - the positive number before normalization:
31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)
- 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:
31.640 215(10) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24
- 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:
Sign: 1 (a negative number)
Exponent (unadjusted): 4
Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0
- 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
100 0000 0011(2)
- 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):
Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0
Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
- Conclusion:
Sign (1 bit) = 1 (a negative number)
Exponent (8 bits) = 100 0000 0011
Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
Available Base Conversions Between Decimal and Binary Systems
Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):
1. Integer -> Binary
2. Decimal -> Binary
3. Binary -> Integer
4. Binary -> Decimal