-0.016 738 891 601 562 490 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 490 9(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
-0.016 738 891 601 562 490 9(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:

|-0.016 738 891 601 562 490 9| = 0.016 738 891 601 562 490 9


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: 0.016 738 891 601 562 490 9.

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) 0.016 738 891 601 562 490 9 × 2 = 0 + 0.033 477 783 203 124 981 8;
  • 2) 0.033 477 783 203 124 981 8 × 2 = 0 + 0.066 955 566 406 249 963 6;
  • 3) 0.066 955 566 406 249 963 6 × 2 = 0 + 0.133 911 132 812 499 927 2;
  • 4) 0.133 911 132 812 499 927 2 × 2 = 0 + 0.267 822 265 624 999 854 4;
  • 5) 0.267 822 265 624 999 854 4 × 2 = 0 + 0.535 644 531 249 999 708 8;
  • 6) 0.535 644 531 249 999 708 8 × 2 = 1 + 0.071 289 062 499 999 417 6;
  • 7) 0.071 289 062 499 999 417 6 × 2 = 0 + 0.142 578 124 999 998 835 2;
  • 8) 0.142 578 124 999 998 835 2 × 2 = 0 + 0.285 156 249 999 997 670 4;
  • 9) 0.285 156 249 999 997 670 4 × 2 = 0 + 0.570 312 499 999 995 340 8;
  • 10) 0.570 312 499 999 995 340 8 × 2 = 1 + 0.140 624 999 999 990 681 6;
  • 11) 0.140 624 999 999 990 681 6 × 2 = 0 + 0.281 249 999 999 981 363 2;
  • 12) 0.281 249 999 999 981 363 2 × 2 = 0 + 0.562 499 999 999 962 726 4;
  • 13) 0.562 499 999 999 962 726 4 × 2 = 1 + 0.124 999 999 999 925 452 8;
  • 14) 0.124 999 999 999 925 452 8 × 2 = 0 + 0.249 999 999 999 850 905 6;
  • 15) 0.249 999 999 999 850 905 6 × 2 = 0 + 0.499 999 999 999 701 811 2;
  • 16) 0.499 999 999 999 701 811 2 × 2 = 0 + 0.999 999 999 999 403 622 4;
  • 17) 0.999 999 999 999 403 622 4 × 2 = 1 + 0.999 999 999 998 807 244 8;
  • 18) 0.999 999 999 998 807 244 8 × 2 = 1 + 0.999 999 999 997 614 489 6;
  • 19) 0.999 999 999 997 614 489 6 × 2 = 1 + 0.999 999 999 995 228 979 2;
  • 20) 0.999 999 999 995 228 979 2 × 2 = 1 + 0.999 999 999 990 457 958 4;
  • 21) 0.999 999 999 990 457 958 4 × 2 = 1 + 0.999 999 999 980 915 916 8;
  • 22) 0.999 999 999 980 915 916 8 × 2 = 1 + 0.999 999 999 961 831 833 6;
  • 23) 0.999 999 999 961 831 833 6 × 2 = 1 + 0.999 999 999 923 663 667 2;
  • 24) 0.999 999 999 923 663 667 2 × 2 = 1 + 0.999 999 999 847 327 334 4;
  • 25) 0.999 999 999 847 327 334 4 × 2 = 1 + 0.999 999 999 694 654 668 8;
  • 26) 0.999 999 999 694 654 668 8 × 2 = 1 + 0.999 999 999 389 309 337 6;
  • 27) 0.999 999 999 389 309 337 6 × 2 = 1 + 0.999 999 998 778 618 675 2;
  • 28) 0.999 999 998 778 618 675 2 × 2 = 1 + 0.999 999 997 557 237 350 4;
  • 29) 0.999 999 997 557 237 350 4 × 2 = 1 + 0.999 999 995 114 474 700 8;
  • 30) 0.999 999 995 114 474 700 8 × 2 = 1 + 0.999 999 990 228 949 401 6;
  • 31) 0.999 999 990 228 949 401 6 × 2 = 1 + 0.999 999 980 457 898 803 2;
  • 32) 0.999 999 980 457 898 803 2 × 2 = 1 + 0.999 999 960 915 797 606 4;
  • 33) 0.999 999 960 915 797 606 4 × 2 = 1 + 0.999 999 921 831 595 212 8;
  • 34) 0.999 999 921 831 595 212 8 × 2 = 1 + 0.999 999 843 663 190 425 6;
  • 35) 0.999 999 843 663 190 425 6 × 2 = 1 + 0.999 999 687 326 380 851 2;
  • 36) 0.999 999 687 326 380 851 2 × 2 = 1 + 0.999 999 374 652 761 702 4;
  • 37) 0.999 999 374 652 761 702 4 × 2 = 1 + 0.999 998 749 305 523 404 8;
  • 38) 0.999 998 749 305 523 404 8 × 2 = 1 + 0.999 997 498 611 046 809 6;
  • 39) 0.999 997 498 611 046 809 6 × 2 = 1 + 0.999 994 997 222 093 619 2;
  • 40) 0.999 994 997 222 093 619 2 × 2 = 1 + 0.999 989 994 444 187 238 4;
  • 41) 0.999 989 994 444 187 238 4 × 2 = 1 + 0.999 979 988 888 374 476 8;
  • 42) 0.999 979 988 888 374 476 8 × 2 = 1 + 0.999 959 977 776 748 953 6;
  • 43) 0.999 959 977 776 748 953 6 × 2 = 1 + 0.999 919 955 553 497 907 2;
  • 44) 0.999 919 955 553 497 907 2 × 2 = 1 + 0.999 839 911 106 995 814 4;
  • 45) 0.999 839 911 106 995 814 4 × 2 = 1 + 0.999 679 822 213 991 628 8;
  • 46) 0.999 679 822 213 991 628 8 × 2 = 1 + 0.999 359 644 427 983 257 6;
  • 47) 0.999 359 644 427 983 257 6 × 2 = 1 + 0.998 719 288 855 966 515 2;
  • 48) 0.998 719 288 855 966 515 2 × 2 = 1 + 0.997 438 577 711 933 030 4;
  • 49) 0.997 438 577 711 933 030 4 × 2 = 1 + 0.994 877 155 423 866 060 8;
  • 50) 0.994 877 155 423 866 060 8 × 2 = 1 + 0.989 754 310 847 732 121 6;
  • 51) 0.989 754 310 847 732 121 6 × 2 = 1 + 0.979 508 621 695 464 243 2;
  • 52) 0.979 508 621 695 464 243 2 × 2 = 1 + 0.959 017 243 390 928 486 4;
  • 53) 0.959 017 243 390 928 486 4 × 2 = 1 + 0.918 034 486 781 856 972 8;
  • 54) 0.918 034 486 781 856 972 8 × 2 = 1 + 0.836 068 973 563 713 945 6;
  • 55) 0.836 068 973 563 713 945 6 × 2 = 1 + 0.672 137 947 127 427 891 2;
  • 56) 0.672 137 947 127 427 891 2 × 2 = 1 + 0.344 275 894 254 855 782 4;
  • 57) 0.344 275 894 254 855 782 4 × 2 = 0 + 0.688 551 788 509 711 564 8;
  • 58) 0.688 551 788 509 711 564 8 × 2 = 1 + 0.377 103 577 019 423 129 6;

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:


0.016 738 891 601 562 490 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.016 738 891 601 562 490 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(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:


0.016 738 891 601 562 490 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101(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 1101


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 1101 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101


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 1101


Decimal number -0.016 738 891 601 562 490 9 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 1101

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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