-0.048 632 677 916 771 838 155 796 223 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.048 632 677 916 771 838 155 796 223(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
-0.048 632 677 916 771 838 155 796 223(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:

|-0.048 632 677 916 771 838 155 796 223| = 0.048 632 677 916 771 838 155 796 223


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: 0.048 632 677 916 771 838 155 796 223.

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) 0.048 632 677 916 771 838 155 796 223 × 2 = 0 + 0.097 265 355 833 543 676 311 592 446;
  • 2) 0.097 265 355 833 543 676 311 592 446 × 2 = 0 + 0.194 530 711 667 087 352 623 184 892;
  • 3) 0.194 530 711 667 087 352 623 184 892 × 2 = 0 + 0.389 061 423 334 174 705 246 369 784;
  • 4) 0.389 061 423 334 174 705 246 369 784 × 2 = 0 + 0.778 122 846 668 349 410 492 739 568;
  • 5) 0.778 122 846 668 349 410 492 739 568 × 2 = 1 + 0.556 245 693 336 698 820 985 479 136;
  • 6) 0.556 245 693 336 698 820 985 479 136 × 2 = 1 + 0.112 491 386 673 397 641 970 958 272;
  • 7) 0.112 491 386 673 397 641 970 958 272 × 2 = 0 + 0.224 982 773 346 795 283 941 916 544;
  • 8) 0.224 982 773 346 795 283 941 916 544 × 2 = 0 + 0.449 965 546 693 590 567 883 833 088;
  • 9) 0.449 965 546 693 590 567 883 833 088 × 2 = 0 + 0.899 931 093 387 181 135 767 666 176;
  • 10) 0.899 931 093 387 181 135 767 666 176 × 2 = 1 + 0.799 862 186 774 362 271 535 332 352;
  • 11) 0.799 862 186 774 362 271 535 332 352 × 2 = 1 + 0.599 724 373 548 724 543 070 664 704;
  • 12) 0.599 724 373 548 724 543 070 664 704 × 2 = 1 + 0.199 448 747 097 449 086 141 329 408;
  • 13) 0.199 448 747 097 449 086 141 329 408 × 2 = 0 + 0.398 897 494 194 898 172 282 658 816;
  • 14) 0.398 897 494 194 898 172 282 658 816 × 2 = 0 + 0.797 794 988 389 796 344 565 317 632;
  • 15) 0.797 794 988 389 796 344 565 317 632 × 2 = 1 + 0.595 589 976 779 592 689 130 635 264;
  • 16) 0.595 589 976 779 592 689 130 635 264 × 2 = 1 + 0.191 179 953 559 185 378 261 270 528;
  • 17) 0.191 179 953 559 185 378 261 270 528 × 2 = 0 + 0.382 359 907 118 370 756 522 541 056;
  • 18) 0.382 359 907 118 370 756 522 541 056 × 2 = 0 + 0.764 719 814 236 741 513 045 082 112;
  • 19) 0.764 719 814 236 741 513 045 082 112 × 2 = 1 + 0.529 439 628 473 483 026 090 164 224;
  • 20) 0.529 439 628 473 483 026 090 164 224 × 2 = 1 + 0.058 879 256 946 966 052 180 328 448;
  • 21) 0.058 879 256 946 966 052 180 328 448 × 2 = 0 + 0.117 758 513 893 932 104 360 656 896;
  • 22) 0.117 758 513 893 932 104 360 656 896 × 2 = 0 + 0.235 517 027 787 864 208 721 313 792;
  • 23) 0.235 517 027 787 864 208 721 313 792 × 2 = 0 + 0.471 034 055 575 728 417 442 627 584;
  • 24) 0.471 034 055 575 728 417 442 627 584 × 2 = 0 + 0.942 068 111 151 456 834 885 255 168;
  • 25) 0.942 068 111 151 456 834 885 255 168 × 2 = 1 + 0.884 136 222 302 913 669 770 510 336;
  • 26) 0.884 136 222 302 913 669 770 510 336 × 2 = 1 + 0.768 272 444 605 827 339 541 020 672;
  • 27) 0.768 272 444 605 827 339 541 020 672 × 2 = 1 + 0.536 544 889 211 654 679 082 041 344;
  • 28) 0.536 544 889 211 654 679 082 041 344 × 2 = 1 + 0.073 089 778 423 309 358 164 082 688;
  • 29) 0.073 089 778 423 309 358 164 082 688 × 2 = 0 + 0.146 179 556 846 618 716 328 165 376;
  • 30) 0.146 179 556 846 618 716 328 165 376 × 2 = 0 + 0.292 359 113 693 237 432 656 330 752;
  • 31) 0.292 359 113 693 237 432 656 330 752 × 2 = 0 + 0.584 718 227 386 474 865 312 661 504;
  • 32) 0.584 718 227 386 474 865 312 661 504 × 2 = 1 + 0.169 436 454 772 949 730 625 323 008;
  • 33) 0.169 436 454 772 949 730 625 323 008 × 2 = 0 + 0.338 872 909 545 899 461 250 646 016;
  • 34) 0.338 872 909 545 899 461 250 646 016 × 2 = 0 + 0.677 745 819 091 798 922 501 292 032;
  • 35) 0.677 745 819 091 798 922 501 292 032 × 2 = 1 + 0.355 491 638 183 597 845 002 584 064;
  • 36) 0.355 491 638 183 597 845 002 584 064 × 2 = 0 + 0.710 983 276 367 195 690 005 168 128;
  • 37) 0.710 983 276 367 195 690 005 168 128 × 2 = 1 + 0.421 966 552 734 391 380 010 336 256;
  • 38) 0.421 966 552 734 391 380 010 336 256 × 2 = 0 + 0.843 933 105 468 782 760 020 672 512;
  • 39) 0.843 933 105 468 782 760 020 672 512 × 2 = 1 + 0.687 866 210 937 565 520 041 345 024;
  • 40) 0.687 866 210 937 565 520 041 345 024 × 2 = 1 + 0.375 732 421 875 131 040 082 690 048;
  • 41) 0.375 732 421 875 131 040 082 690 048 × 2 = 0 + 0.751 464 843 750 262 080 165 380 096;
  • 42) 0.751 464 843 750 262 080 165 380 096 × 2 = 1 + 0.502 929 687 500 524 160 330 760 192;
  • 43) 0.502 929 687 500 524 160 330 760 192 × 2 = 1 + 0.005 859 375 001 048 320 661 520 384;
  • 44) 0.005 859 375 001 048 320 661 520 384 × 2 = 0 + 0.011 718 750 002 096 641 323 040 768;
  • 45) 0.011 718 750 002 096 641 323 040 768 × 2 = 0 + 0.023 437 500 004 193 282 646 081 536;
  • 46) 0.023 437 500 004 193 282 646 081 536 × 2 = 0 + 0.046 875 000 008 386 565 292 163 072;
  • 47) 0.046 875 000 008 386 565 292 163 072 × 2 = 0 + 0.093 750 000 016 773 130 584 326 144;
  • 48) 0.093 750 000 016 773 130 584 326 144 × 2 = 0 + 0.187 500 000 033 546 261 168 652 288;
  • 49) 0.187 500 000 033 546 261 168 652 288 × 2 = 0 + 0.375 000 000 067 092 522 337 304 576;
  • 50) 0.375 000 000 067 092 522 337 304 576 × 2 = 0 + 0.750 000 000 134 185 044 674 609 152;
  • 51) 0.750 000 000 134 185 044 674 609 152 × 2 = 1 + 0.500 000 000 268 370 089 349 218 304;
  • 52) 0.500 000 000 268 370 089 349 218 304 × 2 = 1 + 0.000 000 000 536 740 178 698 436 608;
  • 53) 0.000 000 000 536 740 178 698 436 608 × 2 = 0 + 0.000 000 001 073 480 357 396 873 216;
  • 54) 0.000 000 001 073 480 357 396 873 216 × 2 = 0 + 0.000 000 002 146 960 714 793 746 432;
  • 55) 0.000 000 002 146 960 714 793 746 432 × 2 = 0 + 0.000 000 004 293 921 429 587 492 864;
  • 56) 0.000 000 004 293 921 429 587 492 864 × 2 = 0 + 0.000 000 008 587 842 859 174 985 728;
  • 57) 0.000 000 008 587 842 859 174 985 728 × 2 = 0 + 0.000 000 017 175 685 718 349 971 456;

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:


0.048 632 677 916 771 838 155 796 223(10) =

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

6. Positive number before normalization:

0.048 632 677 916 771 838 155 796 223(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:


0.048 632 677 916 771 838 155 796 223(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 -0.048 632 677 916 771 838 155 796 223 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