Convert the Number 0.000 000 127 591 192 722 320 556 68 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations
Number 0.000 000 127 591 192 722 320 556 68(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. First, convert to binary (in base 2) the integer part: 0.
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;
- 0 ÷ 2 = 0 + 0;
2. 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.
0(10) =
0(2)
3. Convert to binary (base 2) the fractional part: 0.000 000 127 591 192 722 320 556 68.
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) 0.000 000 127 591 192 722 320 556 68 × 2 = 0 + 0.000 000 255 182 385 444 641 113 36;
- 2) 0.000 000 255 182 385 444 641 113 36 × 2 = 0 + 0.000 000 510 364 770 889 282 226 72;
- 3) 0.000 000 510 364 770 889 282 226 72 × 2 = 0 + 0.000 001 020 729 541 778 564 453 44;
- 4) 0.000 001 020 729 541 778 564 453 44 × 2 = 0 + 0.000 002 041 459 083 557 128 906 88;
- 5) 0.000 002 041 459 083 557 128 906 88 × 2 = 0 + 0.000 004 082 918 167 114 257 813 76;
- 6) 0.000 004 082 918 167 114 257 813 76 × 2 = 0 + 0.000 008 165 836 334 228 515 627 52;
- 7) 0.000 008 165 836 334 228 515 627 52 × 2 = 0 + 0.000 016 331 672 668 457 031 255 04;
- 8) 0.000 016 331 672 668 457 031 255 04 × 2 = 0 + 0.000 032 663 345 336 914 062 510 08;
- 9) 0.000 032 663 345 336 914 062 510 08 × 2 = 0 + 0.000 065 326 690 673 828 125 020 16;
- 10) 0.000 065 326 690 673 828 125 020 16 × 2 = 0 + 0.000 130 653 381 347 656 250 040 32;
- 11) 0.000 130 653 381 347 656 250 040 32 × 2 = 0 + 0.000 261 306 762 695 312 500 080 64;
- 12) 0.000 261 306 762 695 312 500 080 64 × 2 = 0 + 0.000 522 613 525 390 625 000 161 28;
- 13) 0.000 522 613 525 390 625 000 161 28 × 2 = 0 + 0.001 045 227 050 781 250 000 322 56;
- 14) 0.001 045 227 050 781 250 000 322 56 × 2 = 0 + 0.002 090 454 101 562 500 000 645 12;
- 15) 0.002 090 454 101 562 500 000 645 12 × 2 = 0 + 0.004 180 908 203 125 000 001 290 24;
- 16) 0.004 180 908 203 125 000 001 290 24 × 2 = 0 + 0.008 361 816 406 250 000 002 580 48;
- 17) 0.008 361 816 406 250 000 002 580 48 × 2 = 0 + 0.016 723 632 812 500 000 005 160 96;
- 18) 0.016 723 632 812 500 000 005 160 96 × 2 = 0 + 0.033 447 265 625 000 000 010 321 92;
- 19) 0.033 447 265 625 000 000 010 321 92 × 2 = 0 + 0.066 894 531 250 000 000 020 643 84;
- 20) 0.066 894 531 250 000 000 020 643 84 × 2 = 0 + 0.133 789 062 500 000 000 041 287 68;
- 21) 0.133 789 062 500 000 000 041 287 68 × 2 = 0 + 0.267 578 125 000 000 000 082 575 36;
- 22) 0.267 578 125 000 000 000 082 575 36 × 2 = 0 + 0.535 156 250 000 000 000 165 150 72;
- 23) 0.535 156 250 000 000 000 165 150 72 × 2 = 1 + 0.070 312 500 000 000 000 330 301 44;
- 24) 0.070 312 500 000 000 000 330 301 44 × 2 = 0 + 0.140 625 000 000 000 000 660 602 88;
- 25) 0.140 625 000 000 000 000 660 602 88 × 2 = 0 + 0.281 250 000 000 000 001 321 205 76;
- 26) 0.281 250 000 000 000 001 321 205 76 × 2 = 0 + 0.562 500 000 000 000 002 642 411 52;
- 27) 0.562 500 000 000 000 002 642 411 52 × 2 = 1 + 0.125 000 000 000 000 005 284 823 04;
- 28) 0.125 000 000 000 000 005 284 823 04 × 2 = 0 + 0.250 000 000 000 000 010 569 646 08;
- 29) 0.250 000 000 000 000 010 569 646 08 × 2 = 0 + 0.500 000 000 000 000 021 139 292 16;
- 30) 0.500 000 000 000 000 021 139 292 16 × 2 = 1 + 0.000 000 000 000 000 042 278 584 32;
- 31) 0.000 000 000 000 000 042 278 584 32 × 2 = 0 + 0.000 000 000 000 000 084 557 168 64;
- 32) 0.000 000 000 000 000 084 557 168 64 × 2 = 0 + 0.000 000 000 000 000 169 114 337 28;
- 33) 0.000 000 000 000 000 169 114 337 28 × 2 = 0 + 0.000 000 000 000 000 338 228 674 56;
- 34) 0.000 000 000 000 000 338 228 674 56 × 2 = 0 + 0.000 000 000 000 000 676 457 349 12;
- 35) 0.000 000 000 000 000 676 457 349 12 × 2 = 0 + 0.000 000 000 000 001 352 914 698 24;
- 36) 0.000 000 000 000 001 352 914 698 24 × 2 = 0 + 0.000 000 000 000 002 705 829 396 48;
- 37) 0.000 000 000 000 002 705 829 396 48 × 2 = 0 + 0.000 000 000 000 005 411 658 792 96;
- 38) 0.000 000 000 000 005 411 658 792 96 × 2 = 0 + 0.000 000 000 000 010 823 317 585 92;
- 39) 0.000 000 000 000 010 823 317 585 92 × 2 = 0 + 0.000 000 000 000 021 646 635 171 84;
- 40) 0.000 000 000 000 021 646 635 171 84 × 2 = 0 + 0.000 000 000 000 043 293 270 343 68;
- 41) 0.000 000 000 000 043 293 270 343 68 × 2 = 0 + 0.000 000 000 000 086 586 540 687 36;
- 42) 0.000 000 000 000 086 586 540 687 36 × 2 = 0 + 0.000 000 000 000 173 173 081 374 72;
- 43) 0.000 000 000 000 173 173 081 374 72 × 2 = 0 + 0.000 000 000 000 346 346 162 749 44;
- 44) 0.000 000 000 000 346 346 162 749 44 × 2 = 0 + 0.000 000 000 000 692 692 325 498 88;
- 45) 0.000 000 000 000 692 692 325 498 88 × 2 = 0 + 0.000 000 000 001 385 384 650 997 76;
- 46) 0.000 000 000 001 385 384 650 997 76 × 2 = 0 + 0.000 000 000 002 770 769 301 995 52;
- 47) 0.000 000 000 002 770 769 301 995 52 × 2 = 0 + 0.000 000 000 005 541 538 603 991 04;
- 48) 0.000 000 000 005 541 538 603 991 04 × 2 = 0 + 0.000 000 000 011 083 077 207 982 08;
- 49) 0.000 000 000 011 083 077 207 982 08 × 2 = 0 + 0.000 000 000 022 166 154 415 964 16;
- 50) 0.000 000 000 022 166 154 415 964 16 × 2 = 0 + 0.000 000 000 044 332 308 831 928 32;
- 51) 0.000 000 000 044 332 308 831 928 32 × 2 = 0 + 0.000 000 000 088 664 617 663 856 64;
- 52) 0.000 000 000 088 664 617 663 856 64 × 2 = 0 + 0.000 000 000 177 329 235 327 713 28;
- 53) 0.000 000 000 177 329 235 327 713 28 × 2 = 0 + 0.000 000 000 354 658 470 655 426 56;
- 54) 0.000 000 000 354 658 470 655 426 56 × 2 = 0 + 0.000 000 000 709 316 941 310 853 12;
- 55) 0.000 000 000 709 316 941 310 853 12 × 2 = 0 + 0.000 000 001 418 633 882 621 706 24;
- 56) 0.000 000 001 418 633 882 621 706 24 × 2 = 0 + 0.000 000 002 837 267 765 243 412 48;
- 57) 0.000 000 002 837 267 765 243 412 48 × 2 = 0 + 0.000 000 005 674 535 530 486 824 96;
- 58) 0.000 000 005 674 535 530 486 824 96 × 2 = 0 + 0.000 000 011 349 071 060 973 649 92;
- 59) 0.000 000 011 349 071 060 973 649 92 × 2 = 0 + 0.000 000 022 698 142 121 947 299 84;
- 60) 0.000 000 022 698 142 121 947 299 84 × 2 = 0 + 0.000 000 045 396 284 243 894 599 68;
- 61) 0.000 000 045 396 284 243 894 599 68 × 2 = 0 + 0.000 000 090 792 568 487 789 199 36;
- 62) 0.000 000 090 792 568 487 789 199 36 × 2 = 0 + 0.000 000 181 585 136 975 578 398 72;
- 63) 0.000 000 181 585 136 975 578 398 72 × 2 = 0 + 0.000 000 363 170 273 951 156 797 44;
- 64) 0.000 000 363 170 273 951 156 797 44 × 2 = 0 + 0.000 000 726 340 547 902 313 594 88;
- 65) 0.000 000 726 340 547 902 313 594 88 × 2 = 0 + 0.000 001 452 681 095 804 627 189 76;
- 66) 0.000 001 452 681 095 804 627 189 76 × 2 = 0 + 0.000 002 905 362 191 609 254 379 52;
- 67) 0.000 002 905 362 191 609 254 379 52 × 2 = 0 + 0.000 005 810 724 383 218 508 759 04;
- 68) 0.000 005 810 724 383 218 508 759 04 × 2 = 0 + 0.000 011 621 448 766 437 017 518 08;
- 69) 0.000 011 621 448 766 437 017 518 08 × 2 = 0 + 0.000 023 242 897 532 874 035 036 16;
- 70) 0.000 023 242 897 532 874 035 036 16 × 2 = 0 + 0.000 046 485 795 065 748 070 072 32;
- 71) 0.000 046 485 795 065 748 070 072 32 × 2 = 0 + 0.000 092 971 590 131 496 140 144 64;
- 72) 0.000 092 971 590 131 496 140 144 64 × 2 = 0 + 0.000 185 943 180 262 992 280 289 28;
- 73) 0.000 185 943 180 262 992 280 289 28 × 2 = 0 + 0.000 371 886 360 525 984 560 578 56;
- 74) 0.000 371 886 360 525 984 560 578 56 × 2 = 0 + 0.000 743 772 721 051 969 121 157 12;
- 75) 0.000 743 772 721 051 969 121 157 12 × 2 = 0 + 0.001 487 545 442 103 938 242 314 24;
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...)
4. 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:
0.000 000 127 591 192 722 320 556 68(10) =
0.0000 0000 0000 0000 0000 0010 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 000(2)
5. Positive number before normalization:
0.000 000 127 591 192 722 320 556 68(10) =
0.0000 0000 0000 0000 0000 0010 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 000(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.
6. Normalize the binary representation of the number.
Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:
0.000 000 127 591 192 722 320 556 68(10) =
0.0000 0000 0000 0000 0000 0010 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 000(2) =
0.0000 0000 0000 0000 0000 0010 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 000(2) × 20 =
1.0001 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000(2) × 2-23
7. 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 0 (a positive number)
Exponent (unadjusted): -23
Mantissa (not normalized):
1.0001 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
8. Adjust the exponent.
Use the 11 bit excess/bias notation:
Exponent (adjusted) =
Exponent (unadjusted) + 2(11-1) - 1 =
-23 + 2(11-1) - 1 =
(-23 + 1 023)(10) =
1 000(10)
9. 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;
- 1 000 ÷ 2 = 500 + 0;
- 500 ÷ 2 = 250 + 0;
- 250 ÷ 2 = 125 + 0;
- 125 ÷ 2 = 62 + 1;
- 62 ÷ 2 = 31 + 0;
- 31 ÷ 2 = 15 + 1;
- 15 ÷ 2 = 7 + 1;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 1 ÷ 2 = 0 + 1;
10. Construct the base 2 representation of the adjusted exponent.
Take all the remainders starting from the bottom of the list constructed above.
Exponent (adjusted) =
1000(10) =
011 1110 1000(2)
11. 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, only if necessary (not the case here).
Mantissa (normalized) =
1. 0001 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 =
0001 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:
Sign (1 bit) =
0 (a positive number)
Exponent (11 bits) =
011 1110 1000
Mantissa (52 bits) =
0001 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
The base ten decimal number 0.000 000 127 591 192 722 320 556 68 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1110 1000 - 0001 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
Sign (1 bit):
Exponent (11 bits):
0
621
611
601
591
581
570
561
550
540
530
52
Mantissa (52 bits):
0
510
500
491
480
470
461
450
440
430
420
410
400
390
380
370
360
350
340
330
320
310
300
290
280
270
260
250
240
230
220
210
200
190
180
170
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
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