-0.000 000 000 001 637 250 56 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 001 637 250 56(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 56(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 56| = 0.000 000 000 001 637 250 56


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 56.

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 56 × 2 = 0 + 0.000 000 000 003 274 501 12;
  • 2) 0.000 000 000 003 274 501 12 × 2 = 0 + 0.000 000 000 006 549 002 24;
  • 3) 0.000 000 000 006 549 002 24 × 2 = 0 + 0.000 000 000 013 098 004 48;
  • 4) 0.000 000 000 013 098 004 48 × 2 = 0 + 0.000 000 000 026 196 008 96;
  • 5) 0.000 000 000 026 196 008 96 × 2 = 0 + 0.000 000 000 052 392 017 92;
  • 6) 0.000 000 000 052 392 017 92 × 2 = 0 + 0.000 000 000 104 784 035 84;
  • 7) 0.000 000 000 104 784 035 84 × 2 = 0 + 0.000 000 000 209 568 071 68;
  • 8) 0.000 000 000 209 568 071 68 × 2 = 0 + 0.000 000 000 419 136 143 36;
  • 9) 0.000 000 000 419 136 143 36 × 2 = 0 + 0.000 000 000 838 272 286 72;
  • 10) 0.000 000 000 838 272 286 72 × 2 = 0 + 0.000 000 001 676 544 573 44;
  • 11) 0.000 000 001 676 544 573 44 × 2 = 0 + 0.000 000 003 353 089 146 88;
  • 12) 0.000 000 003 353 089 146 88 × 2 = 0 + 0.000 000 006 706 178 293 76;
  • 13) 0.000 000 006 706 178 293 76 × 2 = 0 + 0.000 000 013 412 356 587 52;
  • 14) 0.000 000 013 412 356 587 52 × 2 = 0 + 0.000 000 026 824 713 175 04;
  • 15) 0.000 000 026 824 713 175 04 × 2 = 0 + 0.000 000 053 649 426 350 08;
  • 16) 0.000 000 053 649 426 350 08 × 2 = 0 + 0.000 000 107 298 852 700 16;
  • 17) 0.000 000 107 298 852 700 16 × 2 = 0 + 0.000 000 214 597 705 400 32;
  • 18) 0.000 000 214 597 705 400 32 × 2 = 0 + 0.000 000 429 195 410 800 64;
  • 19) 0.000 000 429 195 410 800 64 × 2 = 0 + 0.000 000 858 390 821 601 28;
  • 20) 0.000 000 858 390 821 601 28 × 2 = 0 + 0.000 001 716 781 643 202 56;
  • 21) 0.000 001 716 781 643 202 56 × 2 = 0 + 0.000 003 433 563 286 405 12;
  • 22) 0.000 003 433 563 286 405 12 × 2 = 0 + 0.000 006 867 126 572 810 24;
  • 23) 0.000 006 867 126 572 810 24 × 2 = 0 + 0.000 013 734 253 145 620 48;
  • 24) 0.000 013 734 253 145 620 48 × 2 = 0 + 0.000 027 468 506 291 240 96;
  • 25) 0.000 027 468 506 291 240 96 × 2 = 0 + 0.000 054 937 012 582 481 92;
  • 26) 0.000 054 937 012 582 481 92 × 2 = 0 + 0.000 109 874 025 164 963 84;
  • 27) 0.000 109 874 025 164 963 84 × 2 = 0 + 0.000 219 748 050 329 927 68;
  • 28) 0.000 219 748 050 329 927 68 × 2 = 0 + 0.000 439 496 100 659 855 36;
  • 29) 0.000 439 496 100 659 855 36 × 2 = 0 + 0.000 878 992 201 319 710 72;
  • 30) 0.000 878 992 201 319 710 72 × 2 = 0 + 0.001 757 984 402 639 421 44;
  • 31) 0.001 757 984 402 639 421 44 × 2 = 0 + 0.003 515 968 805 278 842 88;
  • 32) 0.003 515 968 805 278 842 88 × 2 = 0 + 0.007 031 937 610 557 685 76;
  • 33) 0.007 031 937 610 557 685 76 × 2 = 0 + 0.014 063 875 221 115 371 52;
  • 34) 0.014 063 875 221 115 371 52 × 2 = 0 + 0.028 127 750 442 230 743 04;
  • 35) 0.028 127 750 442 230 743 04 × 2 = 0 + 0.056 255 500 884 461 486 08;
  • 36) 0.056 255 500 884 461 486 08 × 2 = 0 + 0.112 511 001 768 922 972 16;
  • 37) 0.112 511 001 768 922 972 16 × 2 = 0 + 0.225 022 003 537 845 944 32;
  • 38) 0.225 022 003 537 845 944 32 × 2 = 0 + 0.450 044 007 075 691 888 64;
  • 39) 0.450 044 007 075 691 888 64 × 2 = 0 + 0.900 088 014 151 383 777 28;
  • 40) 0.900 088 014 151 383 777 28 × 2 = 1 + 0.800 176 028 302 767 554 56;
  • 41) 0.800 176 028 302 767 554 56 × 2 = 1 + 0.600 352 056 605 535 109 12;
  • 42) 0.600 352 056 605 535 109 12 × 2 = 1 + 0.200 704 113 211 070 218 24;
  • 43) 0.200 704 113 211 070 218 24 × 2 = 0 + 0.401 408 226 422 140 436 48;
  • 44) 0.401 408 226 422 140 436 48 × 2 = 0 + 0.802 816 452 844 280 872 96;
  • 45) 0.802 816 452 844 280 872 96 × 2 = 1 + 0.605 632 905 688 561 745 92;
  • 46) 0.605 632 905 688 561 745 92 × 2 = 1 + 0.211 265 811 377 123 491 84;
  • 47) 0.211 265 811 377 123 491 84 × 2 = 0 + 0.422 531 622 754 246 983 68;
  • 48) 0.422 531 622 754 246 983 68 × 2 = 0 + 0.845 063 245 508 493 967 36;
  • 49) 0.845 063 245 508 493 967 36 × 2 = 1 + 0.690 126 491 016 987 934 72;
  • 50) 0.690 126 491 016 987 934 72 × 2 = 1 + 0.380 252 982 033 975 869 44;
  • 51) 0.380 252 982 033 975 869 44 × 2 = 0 + 0.760 505 964 067 951 738 88;
  • 52) 0.760 505 964 067 951 738 88 × 2 = 1 + 0.521 011 928 135 903 477 76;
  • 53) 0.521 011 928 135 903 477 76 × 2 = 1 + 0.042 023 856 271 806 955 52;
  • 54) 0.042 023 856 271 806 955 52 × 2 = 0 + 0.084 047 712 543 613 911 04;
  • 55) 0.084 047 712 543 613 911 04 × 2 = 0 + 0.168 095 425 087 227 822 08;
  • 56) 0.168 095 425 087 227 822 08 × 2 = 0 + 0.336 190 850 174 455 644 16;
  • 57) 0.336 190 850 174 455 644 16 × 2 = 0 + 0.672 381 700 348 911 288 32;
  • 58) 0.672 381 700 348 911 288 32 × 2 = 1 + 0.344 763 400 697 822 576 64;
  • 59) 0.344 763 400 697 822 576 64 × 2 = 0 + 0.689 526 801 395 645 153 28;
  • 60) 0.689 526 801 395 645 153 28 × 2 = 1 + 0.379 053 602 791 290 306 56;
  • 61) 0.379 053 602 791 290 306 56 × 2 = 0 + 0.758 107 205 582 580 613 12;
  • 62) 0.758 107 205 582 580 613 12 × 2 = 1 + 0.516 214 411 165 161 226 24;
  • 63) 0.516 214 411 165 161 226 24 × 2 = 1 + 0.032 428 822 330 322 452 48;
  • 64) 0.032 428 822 330 322 452 48 × 2 = 0 + 0.064 857 644 660 644 904 96;
  • 65) 0.064 857 644 660 644 904 96 × 2 = 0 + 0.129 715 289 321 289 809 92;
  • 66) 0.129 715 289 321 289 809 92 × 2 = 0 + 0.259 430 578 642 579 619 84;
  • 67) 0.259 430 578 642 579 619 84 × 2 = 0 + 0.518 861 157 285 159 239 68;
  • 68) 0.518 861 157 285 159 239 68 × 2 = 1 + 0.037 722 314 570 318 479 36;
  • 69) 0.037 722 314 570 318 479 36 × 2 = 0 + 0.075 444 629 140 636 958 72;
  • 70) 0.075 444 629 140 636 958 72 × 2 = 0 + 0.150 889 258 281 273 917 44;
  • 71) 0.150 889 258 281 273 917 44 × 2 = 0 + 0.301 778 516 562 547 834 88;
  • 72) 0.301 778 516 562 547 834 88 × 2 = 0 + 0.603 557 033 125 095 669 76;
  • 73) 0.603 557 033 125 095 669 76 × 2 = 1 + 0.207 114 066 250 191 339 52;
  • 74) 0.207 114 066 250 191 339 52 × 2 = 0 + 0.414 228 132 500 382 679 04;
  • 75) 0.414 228 132 500 382 679 04 × 2 = 0 + 0.828 456 265 000 765 358 08;
  • 76) 0.828 456 265 000 765 358 08 × 2 = 1 + 0.656 912 530 001 530 716 16;
  • 77) 0.656 912 530 001 530 716 16 × 2 = 1 + 0.313 825 060 003 061 432 32;
  • 78) 0.313 825 060 003 061 432 32 × 2 = 0 + 0.627 650 120 006 122 864 64;
  • 79) 0.627 650 120 006 122 864 64 × 2 = 1 + 0.255 300 240 012 245 729 28;
  • 80) 0.255 300 240 012 245 729 28 × 2 = 0 + 0.510 600 480 024 491 458 56;
  • 81) 0.510 600 480 024 491 458 56 × 2 = 1 + 0.021 200 960 048 982 917 12;
  • 82) 0.021 200 960 048 982 917 12 × 2 = 0 + 0.042 401 920 097 965 834 24;
  • 83) 0.042 401 920 097 965 834 24 × 2 = 0 + 0.084 803 840 195 931 668 48;
  • 84) 0.084 803 840 195 931 668 48 × 2 = 0 + 0.169 607 680 391 863 336 96;
  • 85) 0.169 607 680 391 863 336 96 × 2 = 0 + 0.339 215 360 783 726 673 92;
  • 86) 0.339 215 360 783 726 673 92 × 2 = 0 + 0.678 430 721 567 453 347 84;
  • 87) 0.678 430 721 567 453 347 84 × 2 = 1 + 0.356 861 443 134 906 695 68;
  • 88) 0.356 861 443 134 906 695 68 × 2 = 0 + 0.713 722 886 269 813 391 36;
  • 89) 0.713 722 886 269 813 391 36 × 2 = 1 + 0.427 445 772 539 626 782 72;
  • 90) 0.427 445 772 539 626 782 72 × 2 = 0 + 0.854 891 545 079 253 565 44;
  • 91) 0.854 891 545 079 253 565 44 × 2 = 1 + 0.709 783 090 158 507 130 88;
  • 92) 0.709 783 090 158 507 130 88 × 2 = 1 + 0.419 566 180 317 014 261 76;

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 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 1100 1101 1000 0101 0110 0001 0000 1001 1010 1000 0010 1011(2)

6. Positive number before normalization:

0.000 000 000 001 637 250 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 1100 1101 1000 0101 0110 0001 0000 1001 1010 1000 0010 1011(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 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 1100 1101 1000 0101 0110 0001 0000 1001 1010 1000 0010 1011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 1100 1101 1000 0101 0110 0001 0000 1001 1010 1000 0010 1011(2) × 20 =

1.1100 1100 1101 1000 0101 0110 0001 0000 1001 1010 1000 0010 1011(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 0110 0001 0000 1001 1010 1000 0010 1011


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 0110 0001 0000 1001 1010 1000 0010 1011 =

1100 1100 1101 1000 0101 0110 0001 0000 1001 1010 1000 0010 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) =
011 1101 0111

Mantissa (52 bits) =
1100 1100 1101 1000 0101 0110 0001 0000 1001 1010 1000 0010 1011


Decimal number -0.000 000 000 001 637 250 56 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0111 - 1100 1100 1101 1000 0101 0110 0001 0000 1001 1010 1000 0010 1011

Share

» Convert decimal -0.000 000 000 001 637 250 83 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity. Gemini Notebook YouTube Video of 6.59 minutes

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