-0.016 738 891 601 562 496 550 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 550 7(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
-0.016 738 891 601 562 496 550 7(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:

|-0.016 738 891 601 562 496 550 7| = 0.016 738 891 601 562 496 550 7


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: 0.016 738 891 601 562 496 550 7.

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) 0.016 738 891 601 562 496 550 7 × 2 = 0 + 0.033 477 783 203 124 993 101 4;
  • 2) 0.033 477 783 203 124 993 101 4 × 2 = 0 + 0.066 955 566 406 249 986 202 8;
  • 3) 0.066 955 566 406 249 986 202 8 × 2 = 0 + 0.133 911 132 812 499 972 405 6;
  • 4) 0.133 911 132 812 499 972 405 6 × 2 = 0 + 0.267 822 265 624 999 944 811 2;
  • 5) 0.267 822 265 624 999 944 811 2 × 2 = 0 + 0.535 644 531 249 999 889 622 4;
  • 6) 0.535 644 531 249 999 889 622 4 × 2 = 1 + 0.071 289 062 499 999 779 244 8;
  • 7) 0.071 289 062 499 999 779 244 8 × 2 = 0 + 0.142 578 124 999 999 558 489 6;
  • 8) 0.142 578 124 999 999 558 489 6 × 2 = 0 + 0.285 156 249 999 999 116 979 2;
  • 9) 0.285 156 249 999 999 116 979 2 × 2 = 0 + 0.570 312 499 999 998 233 958 4;
  • 10) 0.570 312 499 999 998 233 958 4 × 2 = 1 + 0.140 624 999 999 996 467 916 8;
  • 11) 0.140 624 999 999 996 467 916 8 × 2 = 0 + 0.281 249 999 999 992 935 833 6;
  • 12) 0.281 249 999 999 992 935 833 6 × 2 = 0 + 0.562 499 999 999 985 871 667 2;
  • 13) 0.562 499 999 999 985 871 667 2 × 2 = 1 + 0.124 999 999 999 971 743 334 4;
  • 14) 0.124 999 999 999 971 743 334 4 × 2 = 0 + 0.249 999 999 999 943 486 668 8;
  • 15) 0.249 999 999 999 943 486 668 8 × 2 = 0 + 0.499 999 999 999 886 973 337 6;
  • 16) 0.499 999 999 999 886 973 337 6 × 2 = 0 + 0.999 999 999 999 773 946 675 2;
  • 17) 0.999 999 999 999 773 946 675 2 × 2 = 1 + 0.999 999 999 999 547 893 350 4;
  • 18) 0.999 999 999 999 547 893 350 4 × 2 = 1 + 0.999 999 999 999 095 786 700 8;
  • 19) 0.999 999 999 999 095 786 700 8 × 2 = 1 + 0.999 999 999 998 191 573 401 6;
  • 20) 0.999 999 999 998 191 573 401 6 × 2 = 1 + 0.999 999 999 996 383 146 803 2;
  • 21) 0.999 999 999 996 383 146 803 2 × 2 = 1 + 0.999 999 999 992 766 293 606 4;
  • 22) 0.999 999 999 992 766 293 606 4 × 2 = 1 + 0.999 999 999 985 532 587 212 8;
  • 23) 0.999 999 999 985 532 587 212 8 × 2 = 1 + 0.999 999 999 971 065 174 425 6;
  • 24) 0.999 999 999 971 065 174 425 6 × 2 = 1 + 0.999 999 999 942 130 348 851 2;
  • 25) 0.999 999 999 942 130 348 851 2 × 2 = 1 + 0.999 999 999 884 260 697 702 4;
  • 26) 0.999 999 999 884 260 697 702 4 × 2 = 1 + 0.999 999 999 768 521 395 404 8;
  • 27) 0.999 999 999 768 521 395 404 8 × 2 = 1 + 0.999 999 999 537 042 790 809 6;
  • 28) 0.999 999 999 537 042 790 809 6 × 2 = 1 + 0.999 999 999 074 085 581 619 2;
  • 29) 0.999 999 999 074 085 581 619 2 × 2 = 1 + 0.999 999 998 148 171 163 238 4;
  • 30) 0.999 999 998 148 171 163 238 4 × 2 = 1 + 0.999 999 996 296 342 326 476 8;
  • 31) 0.999 999 996 296 342 326 476 8 × 2 = 1 + 0.999 999 992 592 684 652 953 6;
  • 32) 0.999 999 992 592 684 652 953 6 × 2 = 1 + 0.999 999 985 185 369 305 907 2;
  • 33) 0.999 999 985 185 369 305 907 2 × 2 = 1 + 0.999 999 970 370 738 611 814 4;
  • 34) 0.999 999 970 370 738 611 814 4 × 2 = 1 + 0.999 999 940 741 477 223 628 8;
  • 35) 0.999 999 940 741 477 223 628 8 × 2 = 1 + 0.999 999 881 482 954 447 257 6;
  • 36) 0.999 999 881 482 954 447 257 6 × 2 = 1 + 0.999 999 762 965 908 894 515 2;
  • 37) 0.999 999 762 965 908 894 515 2 × 2 = 1 + 0.999 999 525 931 817 789 030 4;
  • 38) 0.999 999 525 931 817 789 030 4 × 2 = 1 + 0.999 999 051 863 635 578 060 8;
  • 39) 0.999 999 051 863 635 578 060 8 × 2 = 1 + 0.999 998 103 727 271 156 121 6;
  • 40) 0.999 998 103 727 271 156 121 6 × 2 = 1 + 0.999 996 207 454 542 312 243 2;
  • 41) 0.999 996 207 454 542 312 243 2 × 2 = 1 + 0.999 992 414 909 084 624 486 4;
  • 42) 0.999 992 414 909 084 624 486 4 × 2 = 1 + 0.999 984 829 818 169 248 972 8;
  • 43) 0.999 984 829 818 169 248 972 8 × 2 = 1 + 0.999 969 659 636 338 497 945 6;
  • 44) 0.999 969 659 636 338 497 945 6 × 2 = 1 + 0.999 939 319 272 676 995 891 2;
  • 45) 0.999 939 319 272 676 995 891 2 × 2 = 1 + 0.999 878 638 545 353 991 782 4;
  • 46) 0.999 878 638 545 353 991 782 4 × 2 = 1 + 0.999 757 277 090 707 983 564 8;
  • 47) 0.999 757 277 090 707 983 564 8 × 2 = 1 + 0.999 514 554 181 415 967 129 6;
  • 48) 0.999 514 554 181 415 967 129 6 × 2 = 1 + 0.999 029 108 362 831 934 259 2;
  • 49) 0.999 029 108 362 831 934 259 2 × 2 = 1 + 0.998 058 216 725 663 868 518 4;
  • 50) 0.998 058 216 725 663 868 518 4 × 2 = 1 + 0.996 116 433 451 327 737 036 8;
  • 51) 0.996 116 433 451 327 737 036 8 × 2 = 1 + 0.992 232 866 902 655 474 073 6;
  • 52) 0.992 232 866 902 655 474 073 6 × 2 = 1 + 0.984 465 733 805 310 948 147 2;
  • 53) 0.984 465 733 805 310 948 147 2 × 2 = 1 + 0.968 931 467 610 621 896 294 4;
  • 54) 0.968 931 467 610 621 896 294 4 × 2 = 1 + 0.937 862 935 221 243 792 588 8;
  • 55) 0.937 862 935 221 243 792 588 8 × 2 = 1 + 0.875 725 870 442 487 585 177 6;
  • 56) 0.875 725 870 442 487 585 177 6 × 2 = 1 + 0.751 451 740 884 975 170 355 2;
  • 57) 0.751 451 740 884 975 170 355 2 × 2 = 1 + 0.502 903 481 769 950 340 710 4;
  • 58) 0.502 903 481 769 950 340 710 4 × 2 = 1 + 0.005 806 963 539 900 681 420 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:


0.016 738 891 601 562 496 550 7(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 550 7(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:


0.016 738 891 601 562 496 550 7(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 -0.016 738 891 601 562 496 550 7 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