-0.048 632 677 916 771 838 155 796 097 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.048 632 677 916 771 838 155 796 097 59(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
-0.048 632 677 916 771 838 155 796 097 59(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:

|-0.048 632 677 916 771 838 155 796 097 59| = 0.048 632 677 916 771 838 155 796 097 59


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: 0.048 632 677 916 771 838 155 796 097 59.

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) 0.048 632 677 916 771 838 155 796 097 59 × 2 = 0 + 0.097 265 355 833 543 676 311 592 195 18;
  • 2) 0.097 265 355 833 543 676 311 592 195 18 × 2 = 0 + 0.194 530 711 667 087 352 623 184 390 36;
  • 3) 0.194 530 711 667 087 352 623 184 390 36 × 2 = 0 + 0.389 061 423 334 174 705 246 368 780 72;
  • 4) 0.389 061 423 334 174 705 246 368 780 72 × 2 = 0 + 0.778 122 846 668 349 410 492 737 561 44;
  • 5) 0.778 122 846 668 349 410 492 737 561 44 × 2 = 1 + 0.556 245 693 336 698 820 985 475 122 88;
  • 6) 0.556 245 693 336 698 820 985 475 122 88 × 2 = 1 + 0.112 491 386 673 397 641 970 950 245 76;
  • 7) 0.112 491 386 673 397 641 970 950 245 76 × 2 = 0 + 0.224 982 773 346 795 283 941 900 491 52;
  • 8) 0.224 982 773 346 795 283 941 900 491 52 × 2 = 0 + 0.449 965 546 693 590 567 883 800 983 04;
  • 9) 0.449 965 546 693 590 567 883 800 983 04 × 2 = 0 + 0.899 931 093 387 181 135 767 601 966 08;
  • 10) 0.899 931 093 387 181 135 767 601 966 08 × 2 = 1 + 0.799 862 186 774 362 271 535 203 932 16;
  • 11) 0.799 862 186 774 362 271 535 203 932 16 × 2 = 1 + 0.599 724 373 548 724 543 070 407 864 32;
  • 12) 0.599 724 373 548 724 543 070 407 864 32 × 2 = 1 + 0.199 448 747 097 449 086 140 815 728 64;
  • 13) 0.199 448 747 097 449 086 140 815 728 64 × 2 = 0 + 0.398 897 494 194 898 172 281 631 457 28;
  • 14) 0.398 897 494 194 898 172 281 631 457 28 × 2 = 0 + 0.797 794 988 389 796 344 563 262 914 56;
  • 15) 0.797 794 988 389 796 344 563 262 914 56 × 2 = 1 + 0.595 589 976 779 592 689 126 525 829 12;
  • 16) 0.595 589 976 779 592 689 126 525 829 12 × 2 = 1 + 0.191 179 953 559 185 378 253 051 658 24;
  • 17) 0.191 179 953 559 185 378 253 051 658 24 × 2 = 0 + 0.382 359 907 118 370 756 506 103 316 48;
  • 18) 0.382 359 907 118 370 756 506 103 316 48 × 2 = 0 + 0.764 719 814 236 741 513 012 206 632 96;
  • 19) 0.764 719 814 236 741 513 012 206 632 96 × 2 = 1 + 0.529 439 628 473 483 026 024 413 265 92;
  • 20) 0.529 439 628 473 483 026 024 413 265 92 × 2 = 1 + 0.058 879 256 946 966 052 048 826 531 84;
  • 21) 0.058 879 256 946 966 052 048 826 531 84 × 2 = 0 + 0.117 758 513 893 932 104 097 653 063 68;
  • 22) 0.117 758 513 893 932 104 097 653 063 68 × 2 = 0 + 0.235 517 027 787 864 208 195 306 127 36;
  • 23) 0.235 517 027 787 864 208 195 306 127 36 × 2 = 0 + 0.471 034 055 575 728 416 390 612 254 72;
  • 24) 0.471 034 055 575 728 416 390 612 254 72 × 2 = 0 + 0.942 068 111 151 456 832 781 224 509 44;
  • 25) 0.942 068 111 151 456 832 781 224 509 44 × 2 = 1 + 0.884 136 222 302 913 665 562 449 018 88;
  • 26) 0.884 136 222 302 913 665 562 449 018 88 × 2 = 1 + 0.768 272 444 605 827 331 124 898 037 76;
  • 27) 0.768 272 444 605 827 331 124 898 037 76 × 2 = 1 + 0.536 544 889 211 654 662 249 796 075 52;
  • 28) 0.536 544 889 211 654 662 249 796 075 52 × 2 = 1 + 0.073 089 778 423 309 324 499 592 151 04;
  • 29) 0.073 089 778 423 309 324 499 592 151 04 × 2 = 0 + 0.146 179 556 846 618 648 999 184 302 08;
  • 30) 0.146 179 556 846 618 648 999 184 302 08 × 2 = 0 + 0.292 359 113 693 237 297 998 368 604 16;
  • 31) 0.292 359 113 693 237 297 998 368 604 16 × 2 = 0 + 0.584 718 227 386 474 595 996 737 208 32;
  • 32) 0.584 718 227 386 474 595 996 737 208 32 × 2 = 1 + 0.169 436 454 772 949 191 993 474 416 64;
  • 33) 0.169 436 454 772 949 191 993 474 416 64 × 2 = 0 + 0.338 872 909 545 898 383 986 948 833 28;
  • 34) 0.338 872 909 545 898 383 986 948 833 28 × 2 = 0 + 0.677 745 819 091 796 767 973 897 666 56;
  • 35) 0.677 745 819 091 796 767 973 897 666 56 × 2 = 1 + 0.355 491 638 183 593 535 947 795 333 12;
  • 36) 0.355 491 638 183 593 535 947 795 333 12 × 2 = 0 + 0.710 983 276 367 187 071 895 590 666 24;
  • 37) 0.710 983 276 367 187 071 895 590 666 24 × 2 = 1 + 0.421 966 552 734 374 143 791 181 332 48;
  • 38) 0.421 966 552 734 374 143 791 181 332 48 × 2 = 0 + 0.843 933 105 468 748 287 582 362 664 96;
  • 39) 0.843 933 105 468 748 287 582 362 664 96 × 2 = 1 + 0.687 866 210 937 496 575 164 725 329 92;
  • 40) 0.687 866 210 937 496 575 164 725 329 92 × 2 = 1 + 0.375 732 421 874 993 150 329 450 659 84;
  • 41) 0.375 732 421 874 993 150 329 450 659 84 × 2 = 0 + 0.751 464 843 749 986 300 658 901 319 68;
  • 42) 0.751 464 843 749 986 300 658 901 319 68 × 2 = 1 + 0.502 929 687 499 972 601 317 802 639 36;
  • 43) 0.502 929 687 499 972 601 317 802 639 36 × 2 = 1 + 0.005 859 374 999 945 202 635 605 278 72;
  • 44) 0.005 859 374 999 945 202 635 605 278 72 × 2 = 0 + 0.011 718 749 999 890 405 271 210 557 44;
  • 45) 0.011 718 749 999 890 405 271 210 557 44 × 2 = 0 + 0.023 437 499 999 780 810 542 421 114 88;
  • 46) 0.023 437 499 999 780 810 542 421 114 88 × 2 = 0 + 0.046 874 999 999 561 621 084 842 229 76;
  • 47) 0.046 874 999 999 561 621 084 842 229 76 × 2 = 0 + 0.093 749 999 999 123 242 169 684 459 52;
  • 48) 0.093 749 999 999 123 242 169 684 459 52 × 2 = 0 + 0.187 499 999 998 246 484 339 368 919 04;
  • 49) 0.187 499 999 998 246 484 339 368 919 04 × 2 = 0 + 0.374 999 999 996 492 968 678 737 838 08;
  • 50) 0.374 999 999 996 492 968 678 737 838 08 × 2 = 0 + 0.749 999 999 992 985 937 357 475 676 16;
  • 51) 0.749 999 999 992 985 937 357 475 676 16 × 2 = 1 + 0.499 999 999 985 971 874 714 951 352 32;
  • 52) 0.499 999 999 985 971 874 714 951 352 32 × 2 = 0 + 0.999 999 999 971 943 749 429 902 704 64;
  • 53) 0.999 999 999 971 943 749 429 902 704 64 × 2 = 1 + 0.999 999 999 943 887 498 859 805 409 28;
  • 54) 0.999 999 999 943 887 498 859 805 409 28 × 2 = 1 + 0.999 999 999 887 774 997 719 610 818 56;
  • 55) 0.999 999 999 887 774 997 719 610 818 56 × 2 = 1 + 0.999 999 999 775 549 995 439 221 637 12;
  • 56) 0.999 999 999 775 549 995 439 221 637 12 × 2 = 1 + 0.999 999 999 551 099 990 878 443 274 24;
  • 57) 0.999 999 999 551 099 990 878 443 274 24 × 2 = 1 + 0.999 999 999 102 199 981 756 886 548 48;

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:


0.048 632 677 916 771 838 155 796 097 59(10) =

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

6. Positive number before normalization:

0.048 632 677 916 771 838 155 796 097 59(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:


0.048 632 677 916 771 838 155 796 097 59(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 -0.048 632 677 916 771 838 155 796 097 59 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