-0.016 738 891 601 562 496 530 26 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 26(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
-0.016 738 891 601 562 496 530 26(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:

|-0.016 738 891 601 562 496 530 26| = 0.016 738 891 601 562 496 530 26


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: 0.016 738 891 601 562 496 530 26.

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) 0.016 738 891 601 562 496 530 26 × 2 = 0 + 0.033 477 783 203 124 993 060 52;
  • 2) 0.033 477 783 203 124 993 060 52 × 2 = 0 + 0.066 955 566 406 249 986 121 04;
  • 3) 0.066 955 566 406 249 986 121 04 × 2 = 0 + 0.133 911 132 812 499 972 242 08;
  • 4) 0.133 911 132 812 499 972 242 08 × 2 = 0 + 0.267 822 265 624 999 944 484 16;
  • 5) 0.267 822 265 624 999 944 484 16 × 2 = 0 + 0.535 644 531 249 999 888 968 32;
  • 6) 0.535 644 531 249 999 888 968 32 × 2 = 1 + 0.071 289 062 499 999 777 936 64;
  • 7) 0.071 289 062 499 999 777 936 64 × 2 = 0 + 0.142 578 124 999 999 555 873 28;
  • 8) 0.142 578 124 999 999 555 873 28 × 2 = 0 + 0.285 156 249 999 999 111 746 56;
  • 9) 0.285 156 249 999 999 111 746 56 × 2 = 0 + 0.570 312 499 999 998 223 493 12;
  • 10) 0.570 312 499 999 998 223 493 12 × 2 = 1 + 0.140 624 999 999 996 446 986 24;
  • 11) 0.140 624 999 999 996 446 986 24 × 2 = 0 + 0.281 249 999 999 992 893 972 48;
  • 12) 0.281 249 999 999 992 893 972 48 × 2 = 0 + 0.562 499 999 999 985 787 944 96;
  • 13) 0.562 499 999 999 985 787 944 96 × 2 = 1 + 0.124 999 999 999 971 575 889 92;
  • 14) 0.124 999 999 999 971 575 889 92 × 2 = 0 + 0.249 999 999 999 943 151 779 84;
  • 15) 0.249 999 999 999 943 151 779 84 × 2 = 0 + 0.499 999 999 999 886 303 559 68;
  • 16) 0.499 999 999 999 886 303 559 68 × 2 = 0 + 0.999 999 999 999 772 607 119 36;
  • 17) 0.999 999 999 999 772 607 119 36 × 2 = 1 + 0.999 999 999 999 545 214 238 72;
  • 18) 0.999 999 999 999 545 214 238 72 × 2 = 1 + 0.999 999 999 999 090 428 477 44;
  • 19) 0.999 999 999 999 090 428 477 44 × 2 = 1 + 0.999 999 999 998 180 856 954 88;
  • 20) 0.999 999 999 998 180 856 954 88 × 2 = 1 + 0.999 999 999 996 361 713 909 76;
  • 21) 0.999 999 999 996 361 713 909 76 × 2 = 1 + 0.999 999 999 992 723 427 819 52;
  • 22) 0.999 999 999 992 723 427 819 52 × 2 = 1 + 0.999 999 999 985 446 855 639 04;
  • 23) 0.999 999 999 985 446 855 639 04 × 2 = 1 + 0.999 999 999 970 893 711 278 08;
  • 24) 0.999 999 999 970 893 711 278 08 × 2 = 1 + 0.999 999 999 941 787 422 556 16;
  • 25) 0.999 999 999 941 787 422 556 16 × 2 = 1 + 0.999 999 999 883 574 845 112 32;
  • 26) 0.999 999 999 883 574 845 112 32 × 2 = 1 + 0.999 999 999 767 149 690 224 64;
  • 27) 0.999 999 999 767 149 690 224 64 × 2 = 1 + 0.999 999 999 534 299 380 449 28;
  • 28) 0.999 999 999 534 299 380 449 28 × 2 = 1 + 0.999 999 999 068 598 760 898 56;
  • 29) 0.999 999 999 068 598 760 898 56 × 2 = 1 + 0.999 999 998 137 197 521 797 12;
  • 30) 0.999 999 998 137 197 521 797 12 × 2 = 1 + 0.999 999 996 274 395 043 594 24;
  • 31) 0.999 999 996 274 395 043 594 24 × 2 = 1 + 0.999 999 992 548 790 087 188 48;
  • 32) 0.999 999 992 548 790 087 188 48 × 2 = 1 + 0.999 999 985 097 580 174 376 96;
  • 33) 0.999 999 985 097 580 174 376 96 × 2 = 1 + 0.999 999 970 195 160 348 753 92;
  • 34) 0.999 999 970 195 160 348 753 92 × 2 = 1 + 0.999 999 940 390 320 697 507 84;
  • 35) 0.999 999 940 390 320 697 507 84 × 2 = 1 + 0.999 999 880 780 641 395 015 68;
  • 36) 0.999 999 880 780 641 395 015 68 × 2 = 1 + 0.999 999 761 561 282 790 031 36;
  • 37) 0.999 999 761 561 282 790 031 36 × 2 = 1 + 0.999 999 523 122 565 580 062 72;
  • 38) 0.999 999 523 122 565 580 062 72 × 2 = 1 + 0.999 999 046 245 131 160 125 44;
  • 39) 0.999 999 046 245 131 160 125 44 × 2 = 1 + 0.999 998 092 490 262 320 250 88;
  • 40) 0.999 998 092 490 262 320 250 88 × 2 = 1 + 0.999 996 184 980 524 640 501 76;
  • 41) 0.999 996 184 980 524 640 501 76 × 2 = 1 + 0.999 992 369 961 049 281 003 52;
  • 42) 0.999 992 369 961 049 281 003 52 × 2 = 1 + 0.999 984 739 922 098 562 007 04;
  • 43) 0.999 984 739 922 098 562 007 04 × 2 = 1 + 0.999 969 479 844 197 124 014 08;
  • 44) 0.999 969 479 844 197 124 014 08 × 2 = 1 + 0.999 938 959 688 394 248 028 16;
  • 45) 0.999 938 959 688 394 248 028 16 × 2 = 1 + 0.999 877 919 376 788 496 056 32;
  • 46) 0.999 877 919 376 788 496 056 32 × 2 = 1 + 0.999 755 838 753 576 992 112 64;
  • 47) 0.999 755 838 753 576 992 112 64 × 2 = 1 + 0.999 511 677 507 153 984 225 28;
  • 48) 0.999 511 677 507 153 984 225 28 × 2 = 1 + 0.999 023 355 014 307 968 450 56;
  • 49) 0.999 023 355 014 307 968 450 56 × 2 = 1 + 0.998 046 710 028 615 936 901 12;
  • 50) 0.998 046 710 028 615 936 901 12 × 2 = 1 + 0.996 093 420 057 231 873 802 24;
  • 51) 0.996 093 420 057 231 873 802 24 × 2 = 1 + 0.992 186 840 114 463 747 604 48;
  • 52) 0.992 186 840 114 463 747 604 48 × 2 = 1 + 0.984 373 680 228 927 495 208 96;
  • 53) 0.984 373 680 228 927 495 208 96 × 2 = 1 + 0.968 747 360 457 854 990 417 92;
  • 54) 0.968 747 360 457 854 990 417 92 × 2 = 1 + 0.937 494 720 915 709 980 835 84;
  • 55) 0.937 494 720 915 709 980 835 84 × 2 = 1 + 0.874 989 441 831 419 961 671 68;
  • 56) 0.874 989 441 831 419 961 671 68 × 2 = 1 + 0.749 978 883 662 839 923 343 36;
  • 57) 0.749 978 883 662 839 923 343 36 × 2 = 1 + 0.499 957 767 325 679 846 686 72;
  • 58) 0.499 957 767 325 679 846 686 72 × 2 = 0 + 0.999 915 534 651 359 693 373 44;

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:


0.016 738 891 601 562 496 530 26(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 26(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:


0.016 738 891 601 562 496 530 26(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 -0.016 738 891 601 562 496 530 26 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