-0.016 738 891 601 562 496 529 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 529 17(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
-0.016 738 891 601 562 496 529 17(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:

|-0.016 738 891 601 562 496 529 17| = 0.016 738 891 601 562 496 529 17


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: 0.016 738 891 601 562 496 529 17.

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) 0.016 738 891 601 562 496 529 17 × 2 = 0 + 0.033 477 783 203 124 993 058 34;
  • 2) 0.033 477 783 203 124 993 058 34 × 2 = 0 + 0.066 955 566 406 249 986 116 68;
  • 3) 0.066 955 566 406 249 986 116 68 × 2 = 0 + 0.133 911 132 812 499 972 233 36;
  • 4) 0.133 911 132 812 499 972 233 36 × 2 = 0 + 0.267 822 265 624 999 944 466 72;
  • 5) 0.267 822 265 624 999 944 466 72 × 2 = 0 + 0.535 644 531 249 999 888 933 44;
  • 6) 0.535 644 531 249 999 888 933 44 × 2 = 1 + 0.071 289 062 499 999 777 866 88;
  • 7) 0.071 289 062 499 999 777 866 88 × 2 = 0 + 0.142 578 124 999 999 555 733 76;
  • 8) 0.142 578 124 999 999 555 733 76 × 2 = 0 + 0.285 156 249 999 999 111 467 52;
  • 9) 0.285 156 249 999 999 111 467 52 × 2 = 0 + 0.570 312 499 999 998 222 935 04;
  • 10) 0.570 312 499 999 998 222 935 04 × 2 = 1 + 0.140 624 999 999 996 445 870 08;
  • 11) 0.140 624 999 999 996 445 870 08 × 2 = 0 + 0.281 249 999 999 992 891 740 16;
  • 12) 0.281 249 999 999 992 891 740 16 × 2 = 0 + 0.562 499 999 999 985 783 480 32;
  • 13) 0.562 499 999 999 985 783 480 32 × 2 = 1 + 0.124 999 999 999 971 566 960 64;
  • 14) 0.124 999 999 999 971 566 960 64 × 2 = 0 + 0.249 999 999 999 943 133 921 28;
  • 15) 0.249 999 999 999 943 133 921 28 × 2 = 0 + 0.499 999 999 999 886 267 842 56;
  • 16) 0.499 999 999 999 886 267 842 56 × 2 = 0 + 0.999 999 999 999 772 535 685 12;
  • 17) 0.999 999 999 999 772 535 685 12 × 2 = 1 + 0.999 999 999 999 545 071 370 24;
  • 18) 0.999 999 999 999 545 071 370 24 × 2 = 1 + 0.999 999 999 999 090 142 740 48;
  • 19) 0.999 999 999 999 090 142 740 48 × 2 = 1 + 0.999 999 999 998 180 285 480 96;
  • 20) 0.999 999 999 998 180 285 480 96 × 2 = 1 + 0.999 999 999 996 360 570 961 92;
  • 21) 0.999 999 999 996 360 570 961 92 × 2 = 1 + 0.999 999 999 992 721 141 923 84;
  • 22) 0.999 999 999 992 721 141 923 84 × 2 = 1 + 0.999 999 999 985 442 283 847 68;
  • 23) 0.999 999 999 985 442 283 847 68 × 2 = 1 + 0.999 999 999 970 884 567 695 36;
  • 24) 0.999 999 999 970 884 567 695 36 × 2 = 1 + 0.999 999 999 941 769 135 390 72;
  • 25) 0.999 999 999 941 769 135 390 72 × 2 = 1 + 0.999 999 999 883 538 270 781 44;
  • 26) 0.999 999 999 883 538 270 781 44 × 2 = 1 + 0.999 999 999 767 076 541 562 88;
  • 27) 0.999 999 999 767 076 541 562 88 × 2 = 1 + 0.999 999 999 534 153 083 125 76;
  • 28) 0.999 999 999 534 153 083 125 76 × 2 = 1 + 0.999 999 999 068 306 166 251 52;
  • 29) 0.999 999 999 068 306 166 251 52 × 2 = 1 + 0.999 999 998 136 612 332 503 04;
  • 30) 0.999 999 998 136 612 332 503 04 × 2 = 1 + 0.999 999 996 273 224 665 006 08;
  • 31) 0.999 999 996 273 224 665 006 08 × 2 = 1 + 0.999 999 992 546 449 330 012 16;
  • 32) 0.999 999 992 546 449 330 012 16 × 2 = 1 + 0.999 999 985 092 898 660 024 32;
  • 33) 0.999 999 985 092 898 660 024 32 × 2 = 1 + 0.999 999 970 185 797 320 048 64;
  • 34) 0.999 999 970 185 797 320 048 64 × 2 = 1 + 0.999 999 940 371 594 640 097 28;
  • 35) 0.999 999 940 371 594 640 097 28 × 2 = 1 + 0.999 999 880 743 189 280 194 56;
  • 36) 0.999 999 880 743 189 280 194 56 × 2 = 1 + 0.999 999 761 486 378 560 389 12;
  • 37) 0.999 999 761 486 378 560 389 12 × 2 = 1 + 0.999 999 522 972 757 120 778 24;
  • 38) 0.999 999 522 972 757 120 778 24 × 2 = 1 + 0.999 999 045 945 514 241 556 48;
  • 39) 0.999 999 045 945 514 241 556 48 × 2 = 1 + 0.999 998 091 891 028 483 112 96;
  • 40) 0.999 998 091 891 028 483 112 96 × 2 = 1 + 0.999 996 183 782 056 966 225 92;
  • 41) 0.999 996 183 782 056 966 225 92 × 2 = 1 + 0.999 992 367 564 113 932 451 84;
  • 42) 0.999 992 367 564 113 932 451 84 × 2 = 1 + 0.999 984 735 128 227 864 903 68;
  • 43) 0.999 984 735 128 227 864 903 68 × 2 = 1 + 0.999 969 470 256 455 729 807 36;
  • 44) 0.999 969 470 256 455 729 807 36 × 2 = 1 + 0.999 938 940 512 911 459 614 72;
  • 45) 0.999 938 940 512 911 459 614 72 × 2 = 1 + 0.999 877 881 025 822 919 229 44;
  • 46) 0.999 877 881 025 822 919 229 44 × 2 = 1 + 0.999 755 762 051 645 838 458 88;
  • 47) 0.999 755 762 051 645 838 458 88 × 2 = 1 + 0.999 511 524 103 291 676 917 76;
  • 48) 0.999 511 524 103 291 676 917 76 × 2 = 1 + 0.999 023 048 206 583 353 835 52;
  • 49) 0.999 023 048 206 583 353 835 52 × 2 = 1 + 0.998 046 096 413 166 707 671 04;
  • 50) 0.998 046 096 413 166 707 671 04 × 2 = 1 + 0.996 092 192 826 333 415 342 08;
  • 51) 0.996 092 192 826 333 415 342 08 × 2 = 1 + 0.992 184 385 652 666 830 684 16;
  • 52) 0.992 184 385 652 666 830 684 16 × 2 = 1 + 0.984 368 771 305 333 661 368 32;
  • 53) 0.984 368 771 305 333 661 368 32 × 2 = 1 + 0.968 737 542 610 667 322 736 64;
  • 54) 0.968 737 542 610 667 322 736 64 × 2 = 1 + 0.937 475 085 221 334 645 473 28;
  • 55) 0.937 475 085 221 334 645 473 28 × 2 = 1 + 0.874 950 170 442 669 290 946 56;
  • 56) 0.874 950 170 442 669 290 946 56 × 2 = 1 + 0.749 900 340 885 338 581 893 12;
  • 57) 0.749 900 340 885 338 581 893 12 × 2 = 1 + 0.499 800 681 770 677 163 786 24;
  • 58) 0.499 800 681 770 677 163 786 24 × 2 = 0 + 0.999 601 363 541 354 327 572 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:


0.016 738 891 601 562 496 529 17(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 529 17(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 529 17(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 529 17 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 1110

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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