-0.016 738 891 601 562 496 531 98 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 531 98(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
-0.016 738 891 601 562 496 531 98(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:

|-0.016 738 891 601 562 496 531 98| = 0.016 738 891 601 562 496 531 98


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: 0.016 738 891 601 562 496 531 98.

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) 0.016 738 891 601 562 496 531 98 × 2 = 0 + 0.033 477 783 203 124 993 063 96;
  • 2) 0.033 477 783 203 124 993 063 96 × 2 = 0 + 0.066 955 566 406 249 986 127 92;
  • 3) 0.066 955 566 406 249 986 127 92 × 2 = 0 + 0.133 911 132 812 499 972 255 84;
  • 4) 0.133 911 132 812 499 972 255 84 × 2 = 0 + 0.267 822 265 624 999 944 511 68;
  • 5) 0.267 822 265 624 999 944 511 68 × 2 = 0 + 0.535 644 531 249 999 889 023 36;
  • 6) 0.535 644 531 249 999 889 023 36 × 2 = 1 + 0.071 289 062 499 999 778 046 72;
  • 7) 0.071 289 062 499 999 778 046 72 × 2 = 0 + 0.142 578 124 999 999 556 093 44;
  • 8) 0.142 578 124 999 999 556 093 44 × 2 = 0 + 0.285 156 249 999 999 112 186 88;
  • 9) 0.285 156 249 999 999 112 186 88 × 2 = 0 + 0.570 312 499 999 998 224 373 76;
  • 10) 0.570 312 499 999 998 224 373 76 × 2 = 1 + 0.140 624 999 999 996 448 747 52;
  • 11) 0.140 624 999 999 996 448 747 52 × 2 = 0 + 0.281 249 999 999 992 897 495 04;
  • 12) 0.281 249 999 999 992 897 495 04 × 2 = 0 + 0.562 499 999 999 985 794 990 08;
  • 13) 0.562 499 999 999 985 794 990 08 × 2 = 1 + 0.124 999 999 999 971 589 980 16;
  • 14) 0.124 999 999 999 971 589 980 16 × 2 = 0 + 0.249 999 999 999 943 179 960 32;
  • 15) 0.249 999 999 999 943 179 960 32 × 2 = 0 + 0.499 999 999 999 886 359 920 64;
  • 16) 0.499 999 999 999 886 359 920 64 × 2 = 0 + 0.999 999 999 999 772 719 841 28;
  • 17) 0.999 999 999 999 772 719 841 28 × 2 = 1 + 0.999 999 999 999 545 439 682 56;
  • 18) 0.999 999 999 999 545 439 682 56 × 2 = 1 + 0.999 999 999 999 090 879 365 12;
  • 19) 0.999 999 999 999 090 879 365 12 × 2 = 1 + 0.999 999 999 998 181 758 730 24;
  • 20) 0.999 999 999 998 181 758 730 24 × 2 = 1 + 0.999 999 999 996 363 517 460 48;
  • 21) 0.999 999 999 996 363 517 460 48 × 2 = 1 + 0.999 999 999 992 727 034 920 96;
  • 22) 0.999 999 999 992 727 034 920 96 × 2 = 1 + 0.999 999 999 985 454 069 841 92;
  • 23) 0.999 999 999 985 454 069 841 92 × 2 = 1 + 0.999 999 999 970 908 139 683 84;
  • 24) 0.999 999 999 970 908 139 683 84 × 2 = 1 + 0.999 999 999 941 816 279 367 68;
  • 25) 0.999 999 999 941 816 279 367 68 × 2 = 1 + 0.999 999 999 883 632 558 735 36;
  • 26) 0.999 999 999 883 632 558 735 36 × 2 = 1 + 0.999 999 999 767 265 117 470 72;
  • 27) 0.999 999 999 767 265 117 470 72 × 2 = 1 + 0.999 999 999 534 530 234 941 44;
  • 28) 0.999 999 999 534 530 234 941 44 × 2 = 1 + 0.999 999 999 069 060 469 882 88;
  • 29) 0.999 999 999 069 060 469 882 88 × 2 = 1 + 0.999 999 998 138 120 939 765 76;
  • 30) 0.999 999 998 138 120 939 765 76 × 2 = 1 + 0.999 999 996 276 241 879 531 52;
  • 31) 0.999 999 996 276 241 879 531 52 × 2 = 1 + 0.999 999 992 552 483 759 063 04;
  • 32) 0.999 999 992 552 483 759 063 04 × 2 = 1 + 0.999 999 985 104 967 518 126 08;
  • 33) 0.999 999 985 104 967 518 126 08 × 2 = 1 + 0.999 999 970 209 935 036 252 16;
  • 34) 0.999 999 970 209 935 036 252 16 × 2 = 1 + 0.999 999 940 419 870 072 504 32;
  • 35) 0.999 999 940 419 870 072 504 32 × 2 = 1 + 0.999 999 880 839 740 145 008 64;
  • 36) 0.999 999 880 839 740 145 008 64 × 2 = 1 + 0.999 999 761 679 480 290 017 28;
  • 37) 0.999 999 761 679 480 290 017 28 × 2 = 1 + 0.999 999 523 358 960 580 034 56;
  • 38) 0.999 999 523 358 960 580 034 56 × 2 = 1 + 0.999 999 046 717 921 160 069 12;
  • 39) 0.999 999 046 717 921 160 069 12 × 2 = 1 + 0.999 998 093 435 842 320 138 24;
  • 40) 0.999 998 093 435 842 320 138 24 × 2 = 1 + 0.999 996 186 871 684 640 276 48;
  • 41) 0.999 996 186 871 684 640 276 48 × 2 = 1 + 0.999 992 373 743 369 280 552 96;
  • 42) 0.999 992 373 743 369 280 552 96 × 2 = 1 + 0.999 984 747 486 738 561 105 92;
  • 43) 0.999 984 747 486 738 561 105 92 × 2 = 1 + 0.999 969 494 973 477 122 211 84;
  • 44) 0.999 969 494 973 477 122 211 84 × 2 = 1 + 0.999 938 989 946 954 244 423 68;
  • 45) 0.999 938 989 946 954 244 423 68 × 2 = 1 + 0.999 877 979 893 908 488 847 36;
  • 46) 0.999 877 979 893 908 488 847 36 × 2 = 1 + 0.999 755 959 787 816 977 694 72;
  • 47) 0.999 755 959 787 816 977 694 72 × 2 = 1 + 0.999 511 919 575 633 955 389 44;
  • 48) 0.999 511 919 575 633 955 389 44 × 2 = 1 + 0.999 023 839 151 267 910 778 88;
  • 49) 0.999 023 839 151 267 910 778 88 × 2 = 1 + 0.998 047 678 302 535 821 557 76;
  • 50) 0.998 047 678 302 535 821 557 76 × 2 = 1 + 0.996 095 356 605 071 643 115 52;
  • 51) 0.996 095 356 605 071 643 115 52 × 2 = 1 + 0.992 190 713 210 143 286 231 04;
  • 52) 0.992 190 713 210 143 286 231 04 × 2 = 1 + 0.984 381 426 420 286 572 462 08;
  • 53) 0.984 381 426 420 286 572 462 08 × 2 = 1 + 0.968 762 852 840 573 144 924 16;
  • 54) 0.968 762 852 840 573 144 924 16 × 2 = 1 + 0.937 525 705 681 146 289 848 32;
  • 55) 0.937 525 705 681 146 289 848 32 × 2 = 1 + 0.875 051 411 362 292 579 696 64;
  • 56) 0.875 051 411 362 292 579 696 64 × 2 = 1 + 0.750 102 822 724 585 159 393 28;
  • 57) 0.750 102 822 724 585 159 393 28 × 2 = 1 + 0.500 205 645 449 170 318 786 56;
  • 58) 0.500 205 645 449 170 318 786 56 × 2 = 1 + 0.000 411 290 898 340 637 573 12;

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:


0.016 738 891 601 562 496 531 98(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 531 98(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:


0.016 738 891 601 562 496 531 98(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 -0.016 738 891 601 562 496 531 98 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