-0.048 632 677 916 771 838 155 796 098 29 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.048 632 677 916 771 838 155 796 098 29(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
-0.048 632 677 916 771 838 155 796 098 29(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:

|-0.048 632 677 916 771 838 155 796 098 29| = 0.048 632 677 916 771 838 155 796 098 29


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: 0.048 632 677 916 771 838 155 796 098 29.

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) 0.048 632 677 916 771 838 155 796 098 29 × 2 = 0 + 0.097 265 355 833 543 676 311 592 196 58;
  • 2) 0.097 265 355 833 543 676 311 592 196 58 × 2 = 0 + 0.194 530 711 667 087 352 623 184 393 16;
  • 3) 0.194 530 711 667 087 352 623 184 393 16 × 2 = 0 + 0.389 061 423 334 174 705 246 368 786 32;
  • 4) 0.389 061 423 334 174 705 246 368 786 32 × 2 = 0 + 0.778 122 846 668 349 410 492 737 572 64;
  • 5) 0.778 122 846 668 349 410 492 737 572 64 × 2 = 1 + 0.556 245 693 336 698 820 985 475 145 28;
  • 6) 0.556 245 693 336 698 820 985 475 145 28 × 2 = 1 + 0.112 491 386 673 397 641 970 950 290 56;
  • 7) 0.112 491 386 673 397 641 970 950 290 56 × 2 = 0 + 0.224 982 773 346 795 283 941 900 581 12;
  • 8) 0.224 982 773 346 795 283 941 900 581 12 × 2 = 0 + 0.449 965 546 693 590 567 883 801 162 24;
  • 9) 0.449 965 546 693 590 567 883 801 162 24 × 2 = 0 + 0.899 931 093 387 181 135 767 602 324 48;
  • 10) 0.899 931 093 387 181 135 767 602 324 48 × 2 = 1 + 0.799 862 186 774 362 271 535 204 648 96;
  • 11) 0.799 862 186 774 362 271 535 204 648 96 × 2 = 1 + 0.599 724 373 548 724 543 070 409 297 92;
  • 12) 0.599 724 373 548 724 543 070 409 297 92 × 2 = 1 + 0.199 448 747 097 449 086 140 818 595 84;
  • 13) 0.199 448 747 097 449 086 140 818 595 84 × 2 = 0 + 0.398 897 494 194 898 172 281 637 191 68;
  • 14) 0.398 897 494 194 898 172 281 637 191 68 × 2 = 0 + 0.797 794 988 389 796 344 563 274 383 36;
  • 15) 0.797 794 988 389 796 344 563 274 383 36 × 2 = 1 + 0.595 589 976 779 592 689 126 548 766 72;
  • 16) 0.595 589 976 779 592 689 126 548 766 72 × 2 = 1 + 0.191 179 953 559 185 378 253 097 533 44;
  • 17) 0.191 179 953 559 185 378 253 097 533 44 × 2 = 0 + 0.382 359 907 118 370 756 506 195 066 88;
  • 18) 0.382 359 907 118 370 756 506 195 066 88 × 2 = 0 + 0.764 719 814 236 741 513 012 390 133 76;
  • 19) 0.764 719 814 236 741 513 012 390 133 76 × 2 = 1 + 0.529 439 628 473 483 026 024 780 267 52;
  • 20) 0.529 439 628 473 483 026 024 780 267 52 × 2 = 1 + 0.058 879 256 946 966 052 049 560 535 04;
  • 21) 0.058 879 256 946 966 052 049 560 535 04 × 2 = 0 + 0.117 758 513 893 932 104 099 121 070 08;
  • 22) 0.117 758 513 893 932 104 099 121 070 08 × 2 = 0 + 0.235 517 027 787 864 208 198 242 140 16;
  • 23) 0.235 517 027 787 864 208 198 242 140 16 × 2 = 0 + 0.471 034 055 575 728 416 396 484 280 32;
  • 24) 0.471 034 055 575 728 416 396 484 280 32 × 2 = 0 + 0.942 068 111 151 456 832 792 968 560 64;
  • 25) 0.942 068 111 151 456 832 792 968 560 64 × 2 = 1 + 0.884 136 222 302 913 665 585 937 121 28;
  • 26) 0.884 136 222 302 913 665 585 937 121 28 × 2 = 1 + 0.768 272 444 605 827 331 171 874 242 56;
  • 27) 0.768 272 444 605 827 331 171 874 242 56 × 2 = 1 + 0.536 544 889 211 654 662 343 748 485 12;
  • 28) 0.536 544 889 211 654 662 343 748 485 12 × 2 = 1 + 0.073 089 778 423 309 324 687 496 970 24;
  • 29) 0.073 089 778 423 309 324 687 496 970 24 × 2 = 0 + 0.146 179 556 846 618 649 374 993 940 48;
  • 30) 0.146 179 556 846 618 649 374 993 940 48 × 2 = 0 + 0.292 359 113 693 237 298 749 987 880 96;
  • 31) 0.292 359 113 693 237 298 749 987 880 96 × 2 = 0 + 0.584 718 227 386 474 597 499 975 761 92;
  • 32) 0.584 718 227 386 474 597 499 975 761 92 × 2 = 1 + 0.169 436 454 772 949 194 999 951 523 84;
  • 33) 0.169 436 454 772 949 194 999 951 523 84 × 2 = 0 + 0.338 872 909 545 898 389 999 903 047 68;
  • 34) 0.338 872 909 545 898 389 999 903 047 68 × 2 = 0 + 0.677 745 819 091 796 779 999 806 095 36;
  • 35) 0.677 745 819 091 796 779 999 806 095 36 × 2 = 1 + 0.355 491 638 183 593 559 999 612 190 72;
  • 36) 0.355 491 638 183 593 559 999 612 190 72 × 2 = 0 + 0.710 983 276 367 187 119 999 224 381 44;
  • 37) 0.710 983 276 367 187 119 999 224 381 44 × 2 = 1 + 0.421 966 552 734 374 239 998 448 762 88;
  • 38) 0.421 966 552 734 374 239 998 448 762 88 × 2 = 0 + 0.843 933 105 468 748 479 996 897 525 76;
  • 39) 0.843 933 105 468 748 479 996 897 525 76 × 2 = 1 + 0.687 866 210 937 496 959 993 795 051 52;
  • 40) 0.687 866 210 937 496 959 993 795 051 52 × 2 = 1 + 0.375 732 421 874 993 919 987 590 103 04;
  • 41) 0.375 732 421 874 993 919 987 590 103 04 × 2 = 0 + 0.751 464 843 749 987 839 975 180 206 08;
  • 42) 0.751 464 843 749 987 839 975 180 206 08 × 2 = 1 + 0.502 929 687 499 975 679 950 360 412 16;
  • 43) 0.502 929 687 499 975 679 950 360 412 16 × 2 = 1 + 0.005 859 374 999 951 359 900 720 824 32;
  • 44) 0.005 859 374 999 951 359 900 720 824 32 × 2 = 0 + 0.011 718 749 999 902 719 801 441 648 64;
  • 45) 0.011 718 749 999 902 719 801 441 648 64 × 2 = 0 + 0.023 437 499 999 805 439 602 883 297 28;
  • 46) 0.023 437 499 999 805 439 602 883 297 28 × 2 = 0 + 0.046 874 999 999 610 879 205 766 594 56;
  • 47) 0.046 874 999 999 610 879 205 766 594 56 × 2 = 0 + 0.093 749 999 999 221 758 411 533 189 12;
  • 48) 0.093 749 999 999 221 758 411 533 189 12 × 2 = 0 + 0.187 499 999 998 443 516 823 066 378 24;
  • 49) 0.187 499 999 998 443 516 823 066 378 24 × 2 = 0 + 0.374 999 999 996 887 033 646 132 756 48;
  • 50) 0.374 999 999 996 887 033 646 132 756 48 × 2 = 0 + 0.749 999 999 993 774 067 292 265 512 96;
  • 51) 0.749 999 999 993 774 067 292 265 512 96 × 2 = 1 + 0.499 999 999 987 548 134 584 531 025 92;
  • 52) 0.499 999 999 987 548 134 584 531 025 92 × 2 = 0 + 0.999 999 999 975 096 269 169 062 051 84;
  • 53) 0.999 999 999 975 096 269 169 062 051 84 × 2 = 1 + 0.999 999 999 950 192 538 338 124 103 68;
  • 54) 0.999 999 999 950 192 538 338 124 103 68 × 2 = 1 + 0.999 999 999 900 385 076 676 248 207 36;
  • 55) 0.999 999 999 900 385 076 676 248 207 36 × 2 = 1 + 0.999 999 999 800 770 153 352 496 414 72;
  • 56) 0.999 999 999 800 770 153 352 496 414 72 × 2 = 1 + 0.999 999 999 601 540 306 704 992 829 44;
  • 57) 0.999 999 999 601 540 306 704 992 829 44 × 2 = 1 + 0.999 999 999 203 080 613 409 985 658 88;

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:


0.048 632 677 916 771 838 155 796 098 29(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0010 1111 1(2)

6. Positive number before normalization:

0.048 632 677 916 771 838 155 796 098 29(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0010 1111 1(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:


0.048 632 677 916 771 838 155 796 098 29(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0010 1111 1(2) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0010 1111 1(2) × 20 =

1.1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0101 1111(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 0101 1111


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 0101 1111 =

1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0101 1111


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 0101 1111


Decimal number -0.048 632 677 916 771 838 155 796 098 29 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 0101 1111

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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