-0.048 632 677 916 771 838 155 796 088 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.048 632 677 916 771 838 155 796 088 8(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
-0.048 632 677 916 771 838 155 796 088 8(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:

|-0.048 632 677 916 771 838 155 796 088 8| = 0.048 632 677 916 771 838 155 796 088 8


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: 0.048 632 677 916 771 838 155 796 088 8.

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) 0.048 632 677 916 771 838 155 796 088 8 × 2 = 0 + 0.097 265 355 833 543 676 311 592 177 6;
  • 2) 0.097 265 355 833 543 676 311 592 177 6 × 2 = 0 + 0.194 530 711 667 087 352 623 184 355 2;
  • 3) 0.194 530 711 667 087 352 623 184 355 2 × 2 = 0 + 0.389 061 423 334 174 705 246 368 710 4;
  • 4) 0.389 061 423 334 174 705 246 368 710 4 × 2 = 0 + 0.778 122 846 668 349 410 492 737 420 8;
  • 5) 0.778 122 846 668 349 410 492 737 420 8 × 2 = 1 + 0.556 245 693 336 698 820 985 474 841 6;
  • 6) 0.556 245 693 336 698 820 985 474 841 6 × 2 = 1 + 0.112 491 386 673 397 641 970 949 683 2;
  • 7) 0.112 491 386 673 397 641 970 949 683 2 × 2 = 0 + 0.224 982 773 346 795 283 941 899 366 4;
  • 8) 0.224 982 773 346 795 283 941 899 366 4 × 2 = 0 + 0.449 965 546 693 590 567 883 798 732 8;
  • 9) 0.449 965 546 693 590 567 883 798 732 8 × 2 = 0 + 0.899 931 093 387 181 135 767 597 465 6;
  • 10) 0.899 931 093 387 181 135 767 597 465 6 × 2 = 1 + 0.799 862 186 774 362 271 535 194 931 2;
  • 11) 0.799 862 186 774 362 271 535 194 931 2 × 2 = 1 + 0.599 724 373 548 724 543 070 389 862 4;
  • 12) 0.599 724 373 548 724 543 070 389 862 4 × 2 = 1 + 0.199 448 747 097 449 086 140 779 724 8;
  • 13) 0.199 448 747 097 449 086 140 779 724 8 × 2 = 0 + 0.398 897 494 194 898 172 281 559 449 6;
  • 14) 0.398 897 494 194 898 172 281 559 449 6 × 2 = 0 + 0.797 794 988 389 796 344 563 118 899 2;
  • 15) 0.797 794 988 389 796 344 563 118 899 2 × 2 = 1 + 0.595 589 976 779 592 689 126 237 798 4;
  • 16) 0.595 589 976 779 592 689 126 237 798 4 × 2 = 1 + 0.191 179 953 559 185 378 252 475 596 8;
  • 17) 0.191 179 953 559 185 378 252 475 596 8 × 2 = 0 + 0.382 359 907 118 370 756 504 951 193 6;
  • 18) 0.382 359 907 118 370 756 504 951 193 6 × 2 = 0 + 0.764 719 814 236 741 513 009 902 387 2;
  • 19) 0.764 719 814 236 741 513 009 902 387 2 × 2 = 1 + 0.529 439 628 473 483 026 019 804 774 4;
  • 20) 0.529 439 628 473 483 026 019 804 774 4 × 2 = 1 + 0.058 879 256 946 966 052 039 609 548 8;
  • 21) 0.058 879 256 946 966 052 039 609 548 8 × 2 = 0 + 0.117 758 513 893 932 104 079 219 097 6;
  • 22) 0.117 758 513 893 932 104 079 219 097 6 × 2 = 0 + 0.235 517 027 787 864 208 158 438 195 2;
  • 23) 0.235 517 027 787 864 208 158 438 195 2 × 2 = 0 + 0.471 034 055 575 728 416 316 876 390 4;
  • 24) 0.471 034 055 575 728 416 316 876 390 4 × 2 = 0 + 0.942 068 111 151 456 832 633 752 780 8;
  • 25) 0.942 068 111 151 456 832 633 752 780 8 × 2 = 1 + 0.884 136 222 302 913 665 267 505 561 6;
  • 26) 0.884 136 222 302 913 665 267 505 561 6 × 2 = 1 + 0.768 272 444 605 827 330 535 011 123 2;
  • 27) 0.768 272 444 605 827 330 535 011 123 2 × 2 = 1 + 0.536 544 889 211 654 661 070 022 246 4;
  • 28) 0.536 544 889 211 654 661 070 022 246 4 × 2 = 1 + 0.073 089 778 423 309 322 140 044 492 8;
  • 29) 0.073 089 778 423 309 322 140 044 492 8 × 2 = 0 + 0.146 179 556 846 618 644 280 088 985 6;
  • 30) 0.146 179 556 846 618 644 280 088 985 6 × 2 = 0 + 0.292 359 113 693 237 288 560 177 971 2;
  • 31) 0.292 359 113 693 237 288 560 177 971 2 × 2 = 0 + 0.584 718 227 386 474 577 120 355 942 4;
  • 32) 0.584 718 227 386 474 577 120 355 942 4 × 2 = 1 + 0.169 436 454 772 949 154 240 711 884 8;
  • 33) 0.169 436 454 772 949 154 240 711 884 8 × 2 = 0 + 0.338 872 909 545 898 308 481 423 769 6;
  • 34) 0.338 872 909 545 898 308 481 423 769 6 × 2 = 0 + 0.677 745 819 091 796 616 962 847 539 2;
  • 35) 0.677 745 819 091 796 616 962 847 539 2 × 2 = 1 + 0.355 491 638 183 593 233 925 695 078 4;
  • 36) 0.355 491 638 183 593 233 925 695 078 4 × 2 = 0 + 0.710 983 276 367 186 467 851 390 156 8;
  • 37) 0.710 983 276 367 186 467 851 390 156 8 × 2 = 1 + 0.421 966 552 734 372 935 702 780 313 6;
  • 38) 0.421 966 552 734 372 935 702 780 313 6 × 2 = 0 + 0.843 933 105 468 745 871 405 560 627 2;
  • 39) 0.843 933 105 468 745 871 405 560 627 2 × 2 = 1 + 0.687 866 210 937 491 742 811 121 254 4;
  • 40) 0.687 866 210 937 491 742 811 121 254 4 × 2 = 1 + 0.375 732 421 874 983 485 622 242 508 8;
  • 41) 0.375 732 421 874 983 485 622 242 508 8 × 2 = 0 + 0.751 464 843 749 966 971 244 485 017 6;
  • 42) 0.751 464 843 749 966 971 244 485 017 6 × 2 = 1 + 0.502 929 687 499 933 942 488 970 035 2;
  • 43) 0.502 929 687 499 933 942 488 970 035 2 × 2 = 1 + 0.005 859 374 999 867 884 977 940 070 4;
  • 44) 0.005 859 374 999 867 884 977 940 070 4 × 2 = 0 + 0.011 718 749 999 735 769 955 880 140 8;
  • 45) 0.011 718 749 999 735 769 955 880 140 8 × 2 = 0 + 0.023 437 499 999 471 539 911 760 281 6;
  • 46) 0.023 437 499 999 471 539 911 760 281 6 × 2 = 0 + 0.046 874 999 998 943 079 823 520 563 2;
  • 47) 0.046 874 999 998 943 079 823 520 563 2 × 2 = 0 + 0.093 749 999 997 886 159 647 041 126 4;
  • 48) 0.093 749 999 997 886 159 647 041 126 4 × 2 = 0 + 0.187 499 999 995 772 319 294 082 252 8;
  • 49) 0.187 499 999 995 772 319 294 082 252 8 × 2 = 0 + 0.374 999 999 991 544 638 588 164 505 6;
  • 50) 0.374 999 999 991 544 638 588 164 505 6 × 2 = 0 + 0.749 999 999 983 089 277 176 329 011 2;
  • 51) 0.749 999 999 983 089 277 176 329 011 2 × 2 = 1 + 0.499 999 999 966 178 554 352 658 022 4;
  • 52) 0.499 999 999 966 178 554 352 658 022 4 × 2 = 0 + 0.999 999 999 932 357 108 705 316 044 8;
  • 53) 0.999 999 999 932 357 108 705 316 044 8 × 2 = 1 + 0.999 999 999 864 714 217 410 632 089 6;
  • 54) 0.999 999 999 864 714 217 410 632 089 6 × 2 = 1 + 0.999 999 999 729 428 434 821 264 179 2;
  • 55) 0.999 999 999 729 428 434 821 264 179 2 × 2 = 1 + 0.999 999 999 458 856 869 642 528 358 4;
  • 56) 0.999 999 999 458 856 869 642 528 358 4 × 2 = 1 + 0.999 999 998 917 713 739 285 056 716 8;
  • 57) 0.999 999 998 917 713 739 285 056 716 8 × 2 = 1 + 0.999 999 997 835 427 478 570 113 433 6;

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:


0.048 632 677 916 771 838 155 796 088 8(10) =

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

6. Positive number before normalization:

0.048 632 677 916 771 838 155 796 088 8(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:


0.048 632 677 916 771 838 155 796 088 8(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 -0.048 632 677 916 771 838 155 796 088 8 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