-0.000 000 000 001 637 250 83 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 001 637 250 83(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 001 637 250 83(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 001 637 250 83| = 0.000 000 000 001 637 250 83


2. 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;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 001 637 250 83.

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) 0.000 000 000 001 637 250 83 × 2 = 0 + 0.000 000 000 003 274 501 66;
  • 2) 0.000 000 000 003 274 501 66 × 2 = 0 + 0.000 000 000 006 549 003 32;
  • 3) 0.000 000 000 006 549 003 32 × 2 = 0 + 0.000 000 000 013 098 006 64;
  • 4) 0.000 000 000 013 098 006 64 × 2 = 0 + 0.000 000 000 026 196 013 28;
  • 5) 0.000 000 000 026 196 013 28 × 2 = 0 + 0.000 000 000 052 392 026 56;
  • 6) 0.000 000 000 052 392 026 56 × 2 = 0 + 0.000 000 000 104 784 053 12;
  • 7) 0.000 000 000 104 784 053 12 × 2 = 0 + 0.000 000 000 209 568 106 24;
  • 8) 0.000 000 000 209 568 106 24 × 2 = 0 + 0.000 000 000 419 136 212 48;
  • 9) 0.000 000 000 419 136 212 48 × 2 = 0 + 0.000 000 000 838 272 424 96;
  • 10) 0.000 000 000 838 272 424 96 × 2 = 0 + 0.000 000 001 676 544 849 92;
  • 11) 0.000 000 001 676 544 849 92 × 2 = 0 + 0.000 000 003 353 089 699 84;
  • 12) 0.000 000 003 353 089 699 84 × 2 = 0 + 0.000 000 006 706 179 399 68;
  • 13) 0.000 000 006 706 179 399 68 × 2 = 0 + 0.000 000 013 412 358 799 36;
  • 14) 0.000 000 013 412 358 799 36 × 2 = 0 + 0.000 000 026 824 717 598 72;
  • 15) 0.000 000 026 824 717 598 72 × 2 = 0 + 0.000 000 053 649 435 197 44;
  • 16) 0.000 000 053 649 435 197 44 × 2 = 0 + 0.000 000 107 298 870 394 88;
  • 17) 0.000 000 107 298 870 394 88 × 2 = 0 + 0.000 000 214 597 740 789 76;
  • 18) 0.000 000 214 597 740 789 76 × 2 = 0 + 0.000 000 429 195 481 579 52;
  • 19) 0.000 000 429 195 481 579 52 × 2 = 0 + 0.000 000 858 390 963 159 04;
  • 20) 0.000 000 858 390 963 159 04 × 2 = 0 + 0.000 001 716 781 926 318 08;
  • 21) 0.000 001 716 781 926 318 08 × 2 = 0 + 0.000 003 433 563 852 636 16;
  • 22) 0.000 003 433 563 852 636 16 × 2 = 0 + 0.000 006 867 127 705 272 32;
  • 23) 0.000 006 867 127 705 272 32 × 2 = 0 + 0.000 013 734 255 410 544 64;
  • 24) 0.000 013 734 255 410 544 64 × 2 = 0 + 0.000 027 468 510 821 089 28;
  • 25) 0.000 027 468 510 821 089 28 × 2 = 0 + 0.000 054 937 021 642 178 56;
  • 26) 0.000 054 937 021 642 178 56 × 2 = 0 + 0.000 109 874 043 284 357 12;
  • 27) 0.000 109 874 043 284 357 12 × 2 = 0 + 0.000 219 748 086 568 714 24;
  • 28) 0.000 219 748 086 568 714 24 × 2 = 0 + 0.000 439 496 173 137 428 48;
  • 29) 0.000 439 496 173 137 428 48 × 2 = 0 + 0.000 878 992 346 274 856 96;
  • 30) 0.000 878 992 346 274 856 96 × 2 = 0 + 0.001 757 984 692 549 713 92;
  • 31) 0.001 757 984 692 549 713 92 × 2 = 0 + 0.003 515 969 385 099 427 84;
  • 32) 0.003 515 969 385 099 427 84 × 2 = 0 + 0.007 031 938 770 198 855 68;
  • 33) 0.007 031 938 770 198 855 68 × 2 = 0 + 0.014 063 877 540 397 711 36;
  • 34) 0.014 063 877 540 397 711 36 × 2 = 0 + 0.028 127 755 080 795 422 72;
  • 35) 0.028 127 755 080 795 422 72 × 2 = 0 + 0.056 255 510 161 590 845 44;
  • 36) 0.056 255 510 161 590 845 44 × 2 = 0 + 0.112 511 020 323 181 690 88;
  • 37) 0.112 511 020 323 181 690 88 × 2 = 0 + 0.225 022 040 646 363 381 76;
  • 38) 0.225 022 040 646 363 381 76 × 2 = 0 + 0.450 044 081 292 726 763 52;
  • 39) 0.450 044 081 292 726 763 52 × 2 = 0 + 0.900 088 162 585 453 527 04;
  • 40) 0.900 088 162 585 453 527 04 × 2 = 1 + 0.800 176 325 170 907 054 08;
  • 41) 0.800 176 325 170 907 054 08 × 2 = 1 + 0.600 352 650 341 814 108 16;
  • 42) 0.600 352 650 341 814 108 16 × 2 = 1 + 0.200 705 300 683 628 216 32;
  • 43) 0.200 705 300 683 628 216 32 × 2 = 0 + 0.401 410 601 367 256 432 64;
  • 44) 0.401 410 601 367 256 432 64 × 2 = 0 + 0.802 821 202 734 512 865 28;
  • 45) 0.802 821 202 734 512 865 28 × 2 = 1 + 0.605 642 405 469 025 730 56;
  • 46) 0.605 642 405 469 025 730 56 × 2 = 1 + 0.211 284 810 938 051 461 12;
  • 47) 0.211 284 810 938 051 461 12 × 2 = 0 + 0.422 569 621 876 102 922 24;
  • 48) 0.422 569 621 876 102 922 24 × 2 = 0 + 0.845 139 243 752 205 844 48;
  • 49) 0.845 139 243 752 205 844 48 × 2 = 1 + 0.690 278 487 504 411 688 96;
  • 50) 0.690 278 487 504 411 688 96 × 2 = 1 + 0.380 556 975 008 823 377 92;
  • 51) 0.380 556 975 008 823 377 92 × 2 = 0 + 0.761 113 950 017 646 755 84;
  • 52) 0.761 113 950 017 646 755 84 × 2 = 1 + 0.522 227 900 035 293 511 68;
  • 53) 0.522 227 900 035 293 511 68 × 2 = 1 + 0.044 455 800 070 587 023 36;
  • 54) 0.044 455 800 070 587 023 36 × 2 = 0 + 0.088 911 600 141 174 046 72;
  • 55) 0.088 911 600 141 174 046 72 × 2 = 0 + 0.177 823 200 282 348 093 44;
  • 56) 0.177 823 200 282 348 093 44 × 2 = 0 + 0.355 646 400 564 696 186 88;
  • 57) 0.355 646 400 564 696 186 88 × 2 = 0 + 0.711 292 801 129 392 373 76;
  • 58) 0.711 292 801 129 392 373 76 × 2 = 1 + 0.422 585 602 258 784 747 52;
  • 59) 0.422 585 602 258 784 747 52 × 2 = 0 + 0.845 171 204 517 569 495 04;
  • 60) 0.845 171 204 517 569 495 04 × 2 = 1 + 0.690 342 409 035 138 990 08;
  • 61) 0.690 342 409 035 138 990 08 × 2 = 1 + 0.380 684 818 070 277 980 16;
  • 62) 0.380 684 818 070 277 980 16 × 2 = 0 + 0.761 369 636 140 555 960 32;
  • 63) 0.761 369 636 140 555 960 32 × 2 = 1 + 0.522 739 272 281 111 920 64;
  • 64) 0.522 739 272 281 111 920 64 × 2 = 1 + 0.045 478 544 562 223 841 28;
  • 65) 0.045 478 544 562 223 841 28 × 2 = 0 + 0.090 957 089 124 447 682 56;
  • 66) 0.090 957 089 124 447 682 56 × 2 = 0 + 0.181 914 178 248 895 365 12;
  • 67) 0.181 914 178 248 895 365 12 × 2 = 0 + 0.363 828 356 497 790 730 24;
  • 68) 0.363 828 356 497 790 730 24 × 2 = 0 + 0.727 656 712 995 581 460 48;
  • 69) 0.727 656 712 995 581 460 48 × 2 = 1 + 0.455 313 425 991 162 920 96;
  • 70) 0.455 313 425 991 162 920 96 × 2 = 0 + 0.910 626 851 982 325 841 92;
  • 71) 0.910 626 851 982 325 841 92 × 2 = 1 + 0.821 253 703 964 651 683 84;
  • 72) 0.821 253 703 964 651 683 84 × 2 = 1 + 0.642 507 407 929 303 367 68;
  • 73) 0.642 507 407 929 303 367 68 × 2 = 1 + 0.285 014 815 858 606 735 36;
  • 74) 0.285 014 815 858 606 735 36 × 2 = 0 + 0.570 029 631 717 213 470 72;
  • 75) 0.570 029 631 717 213 470 72 × 2 = 1 + 0.140 059 263 434 426 941 44;
  • 76) 0.140 059 263 434 426 941 44 × 2 = 0 + 0.280 118 526 868 853 882 88;
  • 77) 0.280 118 526 868 853 882 88 × 2 = 0 + 0.560 237 053 737 707 765 76;
  • 78) 0.560 237 053 737 707 765 76 × 2 = 1 + 0.120 474 107 475 415 531 52;
  • 79) 0.120 474 107 475 415 531 52 × 2 = 0 + 0.240 948 214 950 831 063 04;
  • 80) 0.240 948 214 950 831 063 04 × 2 = 0 + 0.481 896 429 901 662 126 08;
  • 81) 0.481 896 429 901 662 126 08 × 2 = 0 + 0.963 792 859 803 324 252 16;
  • 82) 0.963 792 859 803 324 252 16 × 2 = 1 + 0.927 585 719 606 648 504 32;
  • 83) 0.927 585 719 606 648 504 32 × 2 = 1 + 0.855 171 439 213 297 008 64;
  • 84) 0.855 171 439 213 297 008 64 × 2 = 1 + 0.710 342 878 426 594 017 28;
  • 85) 0.710 342 878 426 594 017 28 × 2 = 1 + 0.420 685 756 853 188 034 56;
  • 86) 0.420 685 756 853 188 034 56 × 2 = 0 + 0.841 371 513 706 376 069 12;
  • 87) 0.841 371 513 706 376 069 12 × 2 = 1 + 0.682 743 027 412 752 138 24;
  • 88) 0.682 743 027 412 752 138 24 × 2 = 1 + 0.365 486 054 825 504 276 48;
  • 89) 0.365 486 054 825 504 276 48 × 2 = 0 + 0.730 972 109 651 008 552 96;
  • 90) 0.730 972 109 651 008 552 96 × 2 = 1 + 0.461 944 219 302 017 105 92;
  • 91) 0.461 944 219 302 017 105 92 × 2 = 0 + 0.923 888 438 604 034 211 84;
  • 92) 0.923 888 438 604 034 211 84 × 2 = 1 + 0.847 776 877 208 068 423 68;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 000 001 637 250 83(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101(2)

6. Positive number before normalization:

0.000 000 000 001 637 250 83(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 40 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 001 637 250 83(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101(2) × 20 =

1.1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101(2) × 2-40


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): -40

Mantissa (not normalized):
1.1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-40 + 2(11-1) - 1 =

(-40 + 1 023)(10) =

983(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;
  • 983 ÷ 2 = 491 + 1;
  • 491 ÷ 2 = 245 + 1;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

983(10) =

011 1101 0111(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101 =

1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101


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) =
011 1101 0111

Mantissa (52 bits) =
1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101


Decimal number -0.000 000 000 001 637 250 83 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0111 - 1100 1100 1101 1000 0101 1011 0000 1011 1010 0100 0111 1011 0101

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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