-0.016 738 891 601 562 496 532 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 532 06(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
-0.016 738 891 601 562 496 532 06(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:

|-0.016 738 891 601 562 496 532 06| = 0.016 738 891 601 562 496 532 06


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: 0.016 738 891 601 562 496 532 06.

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) 0.016 738 891 601 562 496 532 06 × 2 = 0 + 0.033 477 783 203 124 993 064 12;
  • 2) 0.033 477 783 203 124 993 064 12 × 2 = 0 + 0.066 955 566 406 249 986 128 24;
  • 3) 0.066 955 566 406 249 986 128 24 × 2 = 0 + 0.133 911 132 812 499 972 256 48;
  • 4) 0.133 911 132 812 499 972 256 48 × 2 = 0 + 0.267 822 265 624 999 944 512 96;
  • 5) 0.267 822 265 624 999 944 512 96 × 2 = 0 + 0.535 644 531 249 999 889 025 92;
  • 6) 0.535 644 531 249 999 889 025 92 × 2 = 1 + 0.071 289 062 499 999 778 051 84;
  • 7) 0.071 289 062 499 999 778 051 84 × 2 = 0 + 0.142 578 124 999 999 556 103 68;
  • 8) 0.142 578 124 999 999 556 103 68 × 2 = 0 + 0.285 156 249 999 999 112 207 36;
  • 9) 0.285 156 249 999 999 112 207 36 × 2 = 0 + 0.570 312 499 999 998 224 414 72;
  • 10) 0.570 312 499 999 998 224 414 72 × 2 = 1 + 0.140 624 999 999 996 448 829 44;
  • 11) 0.140 624 999 999 996 448 829 44 × 2 = 0 + 0.281 249 999 999 992 897 658 88;
  • 12) 0.281 249 999 999 992 897 658 88 × 2 = 0 + 0.562 499 999 999 985 795 317 76;
  • 13) 0.562 499 999 999 985 795 317 76 × 2 = 1 + 0.124 999 999 999 971 590 635 52;
  • 14) 0.124 999 999 999 971 590 635 52 × 2 = 0 + 0.249 999 999 999 943 181 271 04;
  • 15) 0.249 999 999 999 943 181 271 04 × 2 = 0 + 0.499 999 999 999 886 362 542 08;
  • 16) 0.499 999 999 999 886 362 542 08 × 2 = 0 + 0.999 999 999 999 772 725 084 16;
  • 17) 0.999 999 999 999 772 725 084 16 × 2 = 1 + 0.999 999 999 999 545 450 168 32;
  • 18) 0.999 999 999 999 545 450 168 32 × 2 = 1 + 0.999 999 999 999 090 900 336 64;
  • 19) 0.999 999 999 999 090 900 336 64 × 2 = 1 + 0.999 999 999 998 181 800 673 28;
  • 20) 0.999 999 999 998 181 800 673 28 × 2 = 1 + 0.999 999 999 996 363 601 346 56;
  • 21) 0.999 999 999 996 363 601 346 56 × 2 = 1 + 0.999 999 999 992 727 202 693 12;
  • 22) 0.999 999 999 992 727 202 693 12 × 2 = 1 + 0.999 999 999 985 454 405 386 24;
  • 23) 0.999 999 999 985 454 405 386 24 × 2 = 1 + 0.999 999 999 970 908 810 772 48;
  • 24) 0.999 999 999 970 908 810 772 48 × 2 = 1 + 0.999 999 999 941 817 621 544 96;
  • 25) 0.999 999 999 941 817 621 544 96 × 2 = 1 + 0.999 999 999 883 635 243 089 92;
  • 26) 0.999 999 999 883 635 243 089 92 × 2 = 1 + 0.999 999 999 767 270 486 179 84;
  • 27) 0.999 999 999 767 270 486 179 84 × 2 = 1 + 0.999 999 999 534 540 972 359 68;
  • 28) 0.999 999 999 534 540 972 359 68 × 2 = 1 + 0.999 999 999 069 081 944 719 36;
  • 29) 0.999 999 999 069 081 944 719 36 × 2 = 1 + 0.999 999 998 138 163 889 438 72;
  • 30) 0.999 999 998 138 163 889 438 72 × 2 = 1 + 0.999 999 996 276 327 778 877 44;
  • 31) 0.999 999 996 276 327 778 877 44 × 2 = 1 + 0.999 999 992 552 655 557 754 88;
  • 32) 0.999 999 992 552 655 557 754 88 × 2 = 1 + 0.999 999 985 105 311 115 509 76;
  • 33) 0.999 999 985 105 311 115 509 76 × 2 = 1 + 0.999 999 970 210 622 231 019 52;
  • 34) 0.999 999 970 210 622 231 019 52 × 2 = 1 + 0.999 999 940 421 244 462 039 04;
  • 35) 0.999 999 940 421 244 462 039 04 × 2 = 1 + 0.999 999 880 842 488 924 078 08;
  • 36) 0.999 999 880 842 488 924 078 08 × 2 = 1 + 0.999 999 761 684 977 848 156 16;
  • 37) 0.999 999 761 684 977 848 156 16 × 2 = 1 + 0.999 999 523 369 955 696 312 32;
  • 38) 0.999 999 523 369 955 696 312 32 × 2 = 1 + 0.999 999 046 739 911 392 624 64;
  • 39) 0.999 999 046 739 911 392 624 64 × 2 = 1 + 0.999 998 093 479 822 785 249 28;
  • 40) 0.999 998 093 479 822 785 249 28 × 2 = 1 + 0.999 996 186 959 645 570 498 56;
  • 41) 0.999 996 186 959 645 570 498 56 × 2 = 1 + 0.999 992 373 919 291 140 997 12;
  • 42) 0.999 992 373 919 291 140 997 12 × 2 = 1 + 0.999 984 747 838 582 281 994 24;
  • 43) 0.999 984 747 838 582 281 994 24 × 2 = 1 + 0.999 969 495 677 164 563 988 48;
  • 44) 0.999 969 495 677 164 563 988 48 × 2 = 1 + 0.999 938 991 354 329 127 976 96;
  • 45) 0.999 938 991 354 329 127 976 96 × 2 = 1 + 0.999 877 982 708 658 255 953 92;
  • 46) 0.999 877 982 708 658 255 953 92 × 2 = 1 + 0.999 755 965 417 316 511 907 84;
  • 47) 0.999 755 965 417 316 511 907 84 × 2 = 1 + 0.999 511 930 834 633 023 815 68;
  • 48) 0.999 511 930 834 633 023 815 68 × 2 = 1 + 0.999 023 861 669 266 047 631 36;
  • 49) 0.999 023 861 669 266 047 631 36 × 2 = 1 + 0.998 047 723 338 532 095 262 72;
  • 50) 0.998 047 723 338 532 095 262 72 × 2 = 1 + 0.996 095 446 677 064 190 525 44;
  • 51) 0.996 095 446 677 064 190 525 44 × 2 = 1 + 0.992 190 893 354 128 381 050 88;
  • 52) 0.992 190 893 354 128 381 050 88 × 2 = 1 + 0.984 381 786 708 256 762 101 76;
  • 53) 0.984 381 786 708 256 762 101 76 × 2 = 1 + 0.968 763 573 416 513 524 203 52;
  • 54) 0.968 763 573 416 513 524 203 52 × 2 = 1 + 0.937 527 146 833 027 048 407 04;
  • 55) 0.937 527 146 833 027 048 407 04 × 2 = 1 + 0.875 054 293 666 054 096 814 08;
  • 56) 0.875 054 293 666 054 096 814 08 × 2 = 1 + 0.750 108 587 332 108 193 628 16;
  • 57) 0.750 108 587 332 108 193 628 16 × 2 = 1 + 0.500 217 174 664 216 387 256 32;
  • 58) 0.500 217 174 664 216 387 256 32 × 2 = 1 + 0.000 434 349 328 432 774 512 64;

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:


0.016 738 891 601 562 496 532 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 532 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 532 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 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) =

1017(10) =

011 1111 1001(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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number -0.016 738 891 601 562 496 532 06 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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