-0.048 632 677 916 771 838 155 796 087 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.048 632 677 916 771 838 155 796 087 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
-0.048 632 677 916 771 838 155 796 087 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:

|-0.048 632 677 916 771 838 155 796 087 9| = 0.048 632 677 916 771 838 155 796 087 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: 0.048 632 677 916 771 838 155 796 087 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) 0.048 632 677 916 771 838 155 796 087 9 × 2 = 0 + 0.097 265 355 833 543 676 311 592 175 8;
  • 2) 0.097 265 355 833 543 676 311 592 175 8 × 2 = 0 + 0.194 530 711 667 087 352 623 184 351 6;
  • 3) 0.194 530 711 667 087 352 623 184 351 6 × 2 = 0 + 0.389 061 423 334 174 705 246 368 703 2;
  • 4) 0.389 061 423 334 174 705 246 368 703 2 × 2 = 0 + 0.778 122 846 668 349 410 492 737 406 4;
  • 5) 0.778 122 846 668 349 410 492 737 406 4 × 2 = 1 + 0.556 245 693 336 698 820 985 474 812 8;
  • 6) 0.556 245 693 336 698 820 985 474 812 8 × 2 = 1 + 0.112 491 386 673 397 641 970 949 625 6;
  • 7) 0.112 491 386 673 397 641 970 949 625 6 × 2 = 0 + 0.224 982 773 346 795 283 941 899 251 2;
  • 8) 0.224 982 773 346 795 283 941 899 251 2 × 2 = 0 + 0.449 965 546 693 590 567 883 798 502 4;
  • 9) 0.449 965 546 693 590 567 883 798 502 4 × 2 = 0 + 0.899 931 093 387 181 135 767 597 004 8;
  • 10) 0.899 931 093 387 181 135 767 597 004 8 × 2 = 1 + 0.799 862 186 774 362 271 535 194 009 6;
  • 11) 0.799 862 186 774 362 271 535 194 009 6 × 2 = 1 + 0.599 724 373 548 724 543 070 388 019 2;
  • 12) 0.599 724 373 548 724 543 070 388 019 2 × 2 = 1 + 0.199 448 747 097 449 086 140 776 038 4;
  • 13) 0.199 448 747 097 449 086 140 776 038 4 × 2 = 0 + 0.398 897 494 194 898 172 281 552 076 8;
  • 14) 0.398 897 494 194 898 172 281 552 076 8 × 2 = 0 + 0.797 794 988 389 796 344 563 104 153 6;
  • 15) 0.797 794 988 389 796 344 563 104 153 6 × 2 = 1 + 0.595 589 976 779 592 689 126 208 307 2;
  • 16) 0.595 589 976 779 592 689 126 208 307 2 × 2 = 1 + 0.191 179 953 559 185 378 252 416 614 4;
  • 17) 0.191 179 953 559 185 378 252 416 614 4 × 2 = 0 + 0.382 359 907 118 370 756 504 833 228 8;
  • 18) 0.382 359 907 118 370 756 504 833 228 8 × 2 = 0 + 0.764 719 814 236 741 513 009 666 457 6;
  • 19) 0.764 719 814 236 741 513 009 666 457 6 × 2 = 1 + 0.529 439 628 473 483 026 019 332 915 2;
  • 20) 0.529 439 628 473 483 026 019 332 915 2 × 2 = 1 + 0.058 879 256 946 966 052 038 665 830 4;
  • 21) 0.058 879 256 946 966 052 038 665 830 4 × 2 = 0 + 0.117 758 513 893 932 104 077 331 660 8;
  • 22) 0.117 758 513 893 932 104 077 331 660 8 × 2 = 0 + 0.235 517 027 787 864 208 154 663 321 6;
  • 23) 0.235 517 027 787 864 208 154 663 321 6 × 2 = 0 + 0.471 034 055 575 728 416 309 326 643 2;
  • 24) 0.471 034 055 575 728 416 309 326 643 2 × 2 = 0 + 0.942 068 111 151 456 832 618 653 286 4;
  • 25) 0.942 068 111 151 456 832 618 653 286 4 × 2 = 1 + 0.884 136 222 302 913 665 237 306 572 8;
  • 26) 0.884 136 222 302 913 665 237 306 572 8 × 2 = 1 + 0.768 272 444 605 827 330 474 613 145 6;
  • 27) 0.768 272 444 605 827 330 474 613 145 6 × 2 = 1 + 0.536 544 889 211 654 660 949 226 291 2;
  • 28) 0.536 544 889 211 654 660 949 226 291 2 × 2 = 1 + 0.073 089 778 423 309 321 898 452 582 4;
  • 29) 0.073 089 778 423 309 321 898 452 582 4 × 2 = 0 + 0.146 179 556 846 618 643 796 905 164 8;
  • 30) 0.146 179 556 846 618 643 796 905 164 8 × 2 = 0 + 0.292 359 113 693 237 287 593 810 329 6;
  • 31) 0.292 359 113 693 237 287 593 810 329 6 × 2 = 0 + 0.584 718 227 386 474 575 187 620 659 2;
  • 32) 0.584 718 227 386 474 575 187 620 659 2 × 2 = 1 + 0.169 436 454 772 949 150 375 241 318 4;
  • 33) 0.169 436 454 772 949 150 375 241 318 4 × 2 = 0 + 0.338 872 909 545 898 300 750 482 636 8;
  • 34) 0.338 872 909 545 898 300 750 482 636 8 × 2 = 0 + 0.677 745 819 091 796 601 500 965 273 6;
  • 35) 0.677 745 819 091 796 601 500 965 273 6 × 2 = 1 + 0.355 491 638 183 593 203 001 930 547 2;
  • 36) 0.355 491 638 183 593 203 001 930 547 2 × 2 = 0 + 0.710 983 276 367 186 406 003 861 094 4;
  • 37) 0.710 983 276 367 186 406 003 861 094 4 × 2 = 1 + 0.421 966 552 734 372 812 007 722 188 8;
  • 38) 0.421 966 552 734 372 812 007 722 188 8 × 2 = 0 + 0.843 933 105 468 745 624 015 444 377 6;
  • 39) 0.843 933 105 468 745 624 015 444 377 6 × 2 = 1 + 0.687 866 210 937 491 248 030 888 755 2;
  • 40) 0.687 866 210 937 491 248 030 888 755 2 × 2 = 1 + 0.375 732 421 874 982 496 061 777 510 4;
  • 41) 0.375 732 421 874 982 496 061 777 510 4 × 2 = 0 + 0.751 464 843 749 964 992 123 555 020 8;
  • 42) 0.751 464 843 749 964 992 123 555 020 8 × 2 = 1 + 0.502 929 687 499 929 984 247 110 041 6;
  • 43) 0.502 929 687 499 929 984 247 110 041 6 × 2 = 1 + 0.005 859 374 999 859 968 494 220 083 2;
  • 44) 0.005 859 374 999 859 968 494 220 083 2 × 2 = 0 + 0.011 718 749 999 719 936 988 440 166 4;
  • 45) 0.011 718 749 999 719 936 988 440 166 4 × 2 = 0 + 0.023 437 499 999 439 873 976 880 332 8;
  • 46) 0.023 437 499 999 439 873 976 880 332 8 × 2 = 0 + 0.046 874 999 998 879 747 953 760 665 6;
  • 47) 0.046 874 999 998 879 747 953 760 665 6 × 2 = 0 + 0.093 749 999 997 759 495 907 521 331 2;
  • 48) 0.093 749 999 997 759 495 907 521 331 2 × 2 = 0 + 0.187 499 999 995 518 991 815 042 662 4;
  • 49) 0.187 499 999 995 518 991 815 042 662 4 × 2 = 0 + 0.374 999 999 991 037 983 630 085 324 8;
  • 50) 0.374 999 999 991 037 983 630 085 324 8 × 2 = 0 + 0.749 999 999 982 075 967 260 170 649 6;
  • 51) 0.749 999 999 982 075 967 260 170 649 6 × 2 = 1 + 0.499 999 999 964 151 934 520 341 299 2;
  • 52) 0.499 999 999 964 151 934 520 341 299 2 × 2 = 0 + 0.999 999 999 928 303 869 040 682 598 4;
  • 53) 0.999 999 999 928 303 869 040 682 598 4 × 2 = 1 + 0.999 999 999 856 607 738 081 365 196 8;
  • 54) 0.999 999 999 856 607 738 081 365 196 8 × 2 = 1 + 0.999 999 999 713 215 476 162 730 393 6;
  • 55) 0.999 999 999 713 215 476 162 730 393 6 × 2 = 1 + 0.999 999 999 426 430 952 325 460 787 2;
  • 56) 0.999 999 999 426 430 952 325 460 787 2 × 2 = 1 + 0.999 999 998 852 861 904 650 921 574 4;
  • 57) 0.999 999 998 852 861 904 650 921 574 4 × 2 = 1 + 0.999 999 997 705 723 809 301 843 148 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:


0.048 632 677 916 771 838 155 796 087 9(10) =

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

6. Positive number before normalization:

0.048 632 677 916 771 838 155 796 087 9(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:


0.048 632 677 916 771 838 155 796 087 9(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 -0.048 632 677 916 771 838 155 796 087 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 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