-0.016 738 891 601 562 496 549 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 549 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
-0.016 738 891 601 562 496 549 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:

|-0.016 738 891 601 562 496 549 7| = 0.016 738 891 601 562 496 549 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: 0.016 738 891 601 562 496 549 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) 0.016 738 891 601 562 496 549 7 × 2 = 0 + 0.033 477 783 203 124 993 099 4;
  • 2) 0.033 477 783 203 124 993 099 4 × 2 = 0 + 0.066 955 566 406 249 986 198 8;
  • 3) 0.066 955 566 406 249 986 198 8 × 2 = 0 + 0.133 911 132 812 499 972 397 6;
  • 4) 0.133 911 132 812 499 972 397 6 × 2 = 0 + 0.267 822 265 624 999 944 795 2;
  • 5) 0.267 822 265 624 999 944 795 2 × 2 = 0 + 0.535 644 531 249 999 889 590 4;
  • 6) 0.535 644 531 249 999 889 590 4 × 2 = 1 + 0.071 289 062 499 999 779 180 8;
  • 7) 0.071 289 062 499 999 779 180 8 × 2 = 0 + 0.142 578 124 999 999 558 361 6;
  • 8) 0.142 578 124 999 999 558 361 6 × 2 = 0 + 0.285 156 249 999 999 116 723 2;
  • 9) 0.285 156 249 999 999 116 723 2 × 2 = 0 + 0.570 312 499 999 998 233 446 4;
  • 10) 0.570 312 499 999 998 233 446 4 × 2 = 1 + 0.140 624 999 999 996 466 892 8;
  • 11) 0.140 624 999 999 996 466 892 8 × 2 = 0 + 0.281 249 999 999 992 933 785 6;
  • 12) 0.281 249 999 999 992 933 785 6 × 2 = 0 + 0.562 499 999 999 985 867 571 2;
  • 13) 0.562 499 999 999 985 867 571 2 × 2 = 1 + 0.124 999 999 999 971 735 142 4;
  • 14) 0.124 999 999 999 971 735 142 4 × 2 = 0 + 0.249 999 999 999 943 470 284 8;
  • 15) 0.249 999 999 999 943 470 284 8 × 2 = 0 + 0.499 999 999 999 886 940 569 6;
  • 16) 0.499 999 999 999 886 940 569 6 × 2 = 0 + 0.999 999 999 999 773 881 139 2;
  • 17) 0.999 999 999 999 773 881 139 2 × 2 = 1 + 0.999 999 999 999 547 762 278 4;
  • 18) 0.999 999 999 999 547 762 278 4 × 2 = 1 + 0.999 999 999 999 095 524 556 8;
  • 19) 0.999 999 999 999 095 524 556 8 × 2 = 1 + 0.999 999 999 998 191 049 113 6;
  • 20) 0.999 999 999 998 191 049 113 6 × 2 = 1 + 0.999 999 999 996 382 098 227 2;
  • 21) 0.999 999 999 996 382 098 227 2 × 2 = 1 + 0.999 999 999 992 764 196 454 4;
  • 22) 0.999 999 999 992 764 196 454 4 × 2 = 1 + 0.999 999 999 985 528 392 908 8;
  • 23) 0.999 999 999 985 528 392 908 8 × 2 = 1 + 0.999 999 999 971 056 785 817 6;
  • 24) 0.999 999 999 971 056 785 817 6 × 2 = 1 + 0.999 999 999 942 113 571 635 2;
  • 25) 0.999 999 999 942 113 571 635 2 × 2 = 1 + 0.999 999 999 884 227 143 270 4;
  • 26) 0.999 999 999 884 227 143 270 4 × 2 = 1 + 0.999 999 999 768 454 286 540 8;
  • 27) 0.999 999 999 768 454 286 540 8 × 2 = 1 + 0.999 999 999 536 908 573 081 6;
  • 28) 0.999 999 999 536 908 573 081 6 × 2 = 1 + 0.999 999 999 073 817 146 163 2;
  • 29) 0.999 999 999 073 817 146 163 2 × 2 = 1 + 0.999 999 998 147 634 292 326 4;
  • 30) 0.999 999 998 147 634 292 326 4 × 2 = 1 + 0.999 999 996 295 268 584 652 8;
  • 31) 0.999 999 996 295 268 584 652 8 × 2 = 1 + 0.999 999 992 590 537 169 305 6;
  • 32) 0.999 999 992 590 537 169 305 6 × 2 = 1 + 0.999 999 985 181 074 338 611 2;
  • 33) 0.999 999 985 181 074 338 611 2 × 2 = 1 + 0.999 999 970 362 148 677 222 4;
  • 34) 0.999 999 970 362 148 677 222 4 × 2 = 1 + 0.999 999 940 724 297 354 444 8;
  • 35) 0.999 999 940 724 297 354 444 8 × 2 = 1 + 0.999 999 881 448 594 708 889 6;
  • 36) 0.999 999 881 448 594 708 889 6 × 2 = 1 + 0.999 999 762 897 189 417 779 2;
  • 37) 0.999 999 762 897 189 417 779 2 × 2 = 1 + 0.999 999 525 794 378 835 558 4;
  • 38) 0.999 999 525 794 378 835 558 4 × 2 = 1 + 0.999 999 051 588 757 671 116 8;
  • 39) 0.999 999 051 588 757 671 116 8 × 2 = 1 + 0.999 998 103 177 515 342 233 6;
  • 40) 0.999 998 103 177 515 342 233 6 × 2 = 1 + 0.999 996 206 355 030 684 467 2;
  • 41) 0.999 996 206 355 030 684 467 2 × 2 = 1 + 0.999 992 412 710 061 368 934 4;
  • 42) 0.999 992 412 710 061 368 934 4 × 2 = 1 + 0.999 984 825 420 122 737 868 8;
  • 43) 0.999 984 825 420 122 737 868 8 × 2 = 1 + 0.999 969 650 840 245 475 737 6;
  • 44) 0.999 969 650 840 245 475 737 6 × 2 = 1 + 0.999 939 301 680 490 951 475 2;
  • 45) 0.999 939 301 680 490 951 475 2 × 2 = 1 + 0.999 878 603 360 981 902 950 4;
  • 46) 0.999 878 603 360 981 902 950 4 × 2 = 1 + 0.999 757 206 721 963 805 900 8;
  • 47) 0.999 757 206 721 963 805 900 8 × 2 = 1 + 0.999 514 413 443 927 611 801 6;
  • 48) 0.999 514 413 443 927 611 801 6 × 2 = 1 + 0.999 028 826 887 855 223 603 2;
  • 49) 0.999 028 826 887 855 223 603 2 × 2 = 1 + 0.998 057 653 775 710 447 206 4;
  • 50) 0.998 057 653 775 710 447 206 4 × 2 = 1 + 0.996 115 307 551 420 894 412 8;
  • 51) 0.996 115 307 551 420 894 412 8 × 2 = 1 + 0.992 230 615 102 841 788 825 6;
  • 52) 0.992 230 615 102 841 788 825 6 × 2 = 1 + 0.984 461 230 205 683 577 651 2;
  • 53) 0.984 461 230 205 683 577 651 2 × 2 = 1 + 0.968 922 460 411 367 155 302 4;
  • 54) 0.968 922 460 411 367 155 302 4 × 2 = 1 + 0.937 844 920 822 734 310 604 8;
  • 55) 0.937 844 920 822 734 310 604 8 × 2 = 1 + 0.875 689 841 645 468 621 209 6;
  • 56) 0.875 689 841 645 468 621 209 6 × 2 = 1 + 0.751 379 683 290 937 242 419 2;
  • 57) 0.751 379 683 290 937 242 419 2 × 2 = 1 + 0.502 759 366 581 874 484 838 4;
  • 58) 0.502 759 366 581 874 484 838 4 × 2 = 1 + 0.005 518 733 163 748 969 676 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:


0.016 738 891 601 562 496 549 7(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 549 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:


0.016 738 891 601 562 496 549 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 -0.016 738 891 601 562 496 549 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