-0.048 632 677 916 771 838 155 868 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.048 632 677 916 771 838 155 868 9(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
-0.048 632 677 916 771 838 155 868 9(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:

|-0.048 632 677 916 771 838 155 868 9| = 0.048 632 677 916 771 838 155 868 9


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: 0.048 632 677 916 771 838 155 868 9.

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) 0.048 632 677 916 771 838 155 868 9 × 2 = 0 + 0.097 265 355 833 543 676 311 737 8;
  • 2) 0.097 265 355 833 543 676 311 737 8 × 2 = 0 + 0.194 530 711 667 087 352 623 475 6;
  • 3) 0.194 530 711 667 087 352 623 475 6 × 2 = 0 + 0.389 061 423 334 174 705 246 951 2;
  • 4) 0.389 061 423 334 174 705 246 951 2 × 2 = 0 + 0.778 122 846 668 349 410 493 902 4;
  • 5) 0.778 122 846 668 349 410 493 902 4 × 2 = 1 + 0.556 245 693 336 698 820 987 804 8;
  • 6) 0.556 245 693 336 698 820 987 804 8 × 2 = 1 + 0.112 491 386 673 397 641 975 609 6;
  • 7) 0.112 491 386 673 397 641 975 609 6 × 2 = 0 + 0.224 982 773 346 795 283 951 219 2;
  • 8) 0.224 982 773 346 795 283 951 219 2 × 2 = 0 + 0.449 965 546 693 590 567 902 438 4;
  • 9) 0.449 965 546 693 590 567 902 438 4 × 2 = 0 + 0.899 931 093 387 181 135 804 876 8;
  • 10) 0.899 931 093 387 181 135 804 876 8 × 2 = 1 + 0.799 862 186 774 362 271 609 753 6;
  • 11) 0.799 862 186 774 362 271 609 753 6 × 2 = 1 + 0.599 724 373 548 724 543 219 507 2;
  • 12) 0.599 724 373 548 724 543 219 507 2 × 2 = 1 + 0.199 448 747 097 449 086 439 014 4;
  • 13) 0.199 448 747 097 449 086 439 014 4 × 2 = 0 + 0.398 897 494 194 898 172 878 028 8;
  • 14) 0.398 897 494 194 898 172 878 028 8 × 2 = 0 + 0.797 794 988 389 796 345 756 057 6;
  • 15) 0.797 794 988 389 796 345 756 057 6 × 2 = 1 + 0.595 589 976 779 592 691 512 115 2;
  • 16) 0.595 589 976 779 592 691 512 115 2 × 2 = 1 + 0.191 179 953 559 185 383 024 230 4;
  • 17) 0.191 179 953 559 185 383 024 230 4 × 2 = 0 + 0.382 359 907 118 370 766 048 460 8;
  • 18) 0.382 359 907 118 370 766 048 460 8 × 2 = 0 + 0.764 719 814 236 741 532 096 921 6;
  • 19) 0.764 719 814 236 741 532 096 921 6 × 2 = 1 + 0.529 439 628 473 483 064 193 843 2;
  • 20) 0.529 439 628 473 483 064 193 843 2 × 2 = 1 + 0.058 879 256 946 966 128 387 686 4;
  • 21) 0.058 879 256 946 966 128 387 686 4 × 2 = 0 + 0.117 758 513 893 932 256 775 372 8;
  • 22) 0.117 758 513 893 932 256 775 372 8 × 2 = 0 + 0.235 517 027 787 864 513 550 745 6;
  • 23) 0.235 517 027 787 864 513 550 745 6 × 2 = 0 + 0.471 034 055 575 729 027 101 491 2;
  • 24) 0.471 034 055 575 729 027 101 491 2 × 2 = 0 + 0.942 068 111 151 458 054 202 982 4;
  • 25) 0.942 068 111 151 458 054 202 982 4 × 2 = 1 + 0.884 136 222 302 916 108 405 964 8;
  • 26) 0.884 136 222 302 916 108 405 964 8 × 2 = 1 + 0.768 272 444 605 832 216 811 929 6;
  • 27) 0.768 272 444 605 832 216 811 929 6 × 2 = 1 + 0.536 544 889 211 664 433 623 859 2;
  • 28) 0.536 544 889 211 664 433 623 859 2 × 2 = 1 + 0.073 089 778 423 328 867 247 718 4;
  • 29) 0.073 089 778 423 328 867 247 718 4 × 2 = 0 + 0.146 179 556 846 657 734 495 436 8;
  • 30) 0.146 179 556 846 657 734 495 436 8 × 2 = 0 + 0.292 359 113 693 315 468 990 873 6;
  • 31) 0.292 359 113 693 315 468 990 873 6 × 2 = 0 + 0.584 718 227 386 630 937 981 747 2;
  • 32) 0.584 718 227 386 630 937 981 747 2 × 2 = 1 + 0.169 436 454 773 261 875 963 494 4;
  • 33) 0.169 436 454 773 261 875 963 494 4 × 2 = 0 + 0.338 872 909 546 523 751 926 988 8;
  • 34) 0.338 872 909 546 523 751 926 988 8 × 2 = 0 + 0.677 745 819 093 047 503 853 977 6;
  • 35) 0.677 745 819 093 047 503 853 977 6 × 2 = 1 + 0.355 491 638 186 095 007 707 955 2;
  • 36) 0.355 491 638 186 095 007 707 955 2 × 2 = 0 + 0.710 983 276 372 190 015 415 910 4;
  • 37) 0.710 983 276 372 190 015 415 910 4 × 2 = 1 + 0.421 966 552 744 380 030 831 820 8;
  • 38) 0.421 966 552 744 380 030 831 820 8 × 2 = 0 + 0.843 933 105 488 760 061 663 641 6;
  • 39) 0.843 933 105 488 760 061 663 641 6 × 2 = 1 + 0.687 866 210 977 520 123 327 283 2;
  • 40) 0.687 866 210 977 520 123 327 283 2 × 2 = 1 + 0.375 732 421 955 040 246 654 566 4;
  • 41) 0.375 732 421 955 040 246 654 566 4 × 2 = 0 + 0.751 464 843 910 080 493 309 132 8;
  • 42) 0.751 464 843 910 080 493 309 132 8 × 2 = 1 + 0.502 929 687 820 160 986 618 265 6;
  • 43) 0.502 929 687 820 160 986 618 265 6 × 2 = 1 + 0.005 859 375 640 321 973 236 531 2;
  • 44) 0.005 859 375 640 321 973 236 531 2 × 2 = 0 + 0.011 718 751 280 643 946 473 062 4;
  • 45) 0.011 718 751 280 643 946 473 062 4 × 2 = 0 + 0.023 437 502 561 287 892 946 124 8;
  • 46) 0.023 437 502 561 287 892 946 124 8 × 2 = 0 + 0.046 875 005 122 575 785 892 249 6;
  • 47) 0.046 875 005 122 575 785 892 249 6 × 2 = 0 + 0.093 750 010 245 151 571 784 499 2;
  • 48) 0.093 750 010 245 151 571 784 499 2 × 2 = 0 + 0.187 500 020 490 303 143 568 998 4;
  • 49) 0.187 500 020 490 303 143 568 998 4 × 2 = 0 + 0.375 000 040 980 606 287 137 996 8;
  • 50) 0.375 000 040 980 606 287 137 996 8 × 2 = 0 + 0.750 000 081 961 212 574 275 993 6;
  • 51) 0.750 000 081 961 212 574 275 993 6 × 2 = 1 + 0.500 000 163 922 425 148 551 987 2;
  • 52) 0.500 000 163 922 425 148 551 987 2 × 2 = 1 + 0.000 000 327 844 850 297 103 974 4;
  • 53) 0.000 000 327 844 850 297 103 974 4 × 2 = 0 + 0.000 000 655 689 700 594 207 948 8;
  • 54) 0.000 000 655 689 700 594 207 948 8 × 2 = 0 + 0.000 001 311 379 401 188 415 897 6;
  • 55) 0.000 001 311 379 401 188 415 897 6 × 2 = 0 + 0.000 002 622 758 802 376 831 795 2;
  • 56) 0.000 002 622 758 802 376 831 795 2 × 2 = 0 + 0.000 005 245 517 604 753 663 590 4;
  • 57) 0.000 005 245 517 604 753 663 590 4 × 2 = 0 + 0.000 010 491 035 209 507 327 180 8;

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:


0.048 632 677 916 771 838 155 868 9(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0011 0000 0(2)

6. Positive number before normalization:

0.048 632 677 916 771 838 155 868 9(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0011 0000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.048 632 677 916 771 838 155 868 9(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0011 0000 0(2) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0011 0000 0(2) × 20 =

1.1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0110 0000(2) × 2-5


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): -5

Mantissa (not normalized):
1.1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0110 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0110 0000 =

1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0110 0000


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 1111 1010

Mantissa (52 bits) =
1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0110 0000


Decimal number -0.048 632 677 916 771 838 155 868 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0110 0000

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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