-0.016 738 891 601 562 494 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 494 91(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
-0.016 738 891 601 562 494 91(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:

|-0.016 738 891 601 562 494 91| = 0.016 738 891 601 562 494 91


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: 0.016 738 891 601 562 494 91.

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) 0.016 738 891 601 562 494 91 × 2 = 0 + 0.033 477 783 203 124 989 82;
  • 2) 0.033 477 783 203 124 989 82 × 2 = 0 + 0.066 955 566 406 249 979 64;
  • 3) 0.066 955 566 406 249 979 64 × 2 = 0 + 0.133 911 132 812 499 959 28;
  • 4) 0.133 911 132 812 499 959 28 × 2 = 0 + 0.267 822 265 624 999 918 56;
  • 5) 0.267 822 265 624 999 918 56 × 2 = 0 + 0.535 644 531 249 999 837 12;
  • 6) 0.535 644 531 249 999 837 12 × 2 = 1 + 0.071 289 062 499 999 674 24;
  • 7) 0.071 289 062 499 999 674 24 × 2 = 0 + 0.142 578 124 999 999 348 48;
  • 8) 0.142 578 124 999 999 348 48 × 2 = 0 + 0.285 156 249 999 998 696 96;
  • 9) 0.285 156 249 999 998 696 96 × 2 = 0 + 0.570 312 499 999 997 393 92;
  • 10) 0.570 312 499 999 997 393 92 × 2 = 1 + 0.140 624 999 999 994 787 84;
  • 11) 0.140 624 999 999 994 787 84 × 2 = 0 + 0.281 249 999 999 989 575 68;
  • 12) 0.281 249 999 999 989 575 68 × 2 = 0 + 0.562 499 999 999 979 151 36;
  • 13) 0.562 499 999 999 979 151 36 × 2 = 1 + 0.124 999 999 999 958 302 72;
  • 14) 0.124 999 999 999 958 302 72 × 2 = 0 + 0.249 999 999 999 916 605 44;
  • 15) 0.249 999 999 999 916 605 44 × 2 = 0 + 0.499 999 999 999 833 210 88;
  • 16) 0.499 999 999 999 833 210 88 × 2 = 0 + 0.999 999 999 999 666 421 76;
  • 17) 0.999 999 999 999 666 421 76 × 2 = 1 + 0.999 999 999 999 332 843 52;
  • 18) 0.999 999 999 999 332 843 52 × 2 = 1 + 0.999 999 999 998 665 687 04;
  • 19) 0.999 999 999 998 665 687 04 × 2 = 1 + 0.999 999 999 997 331 374 08;
  • 20) 0.999 999 999 997 331 374 08 × 2 = 1 + 0.999 999 999 994 662 748 16;
  • 21) 0.999 999 999 994 662 748 16 × 2 = 1 + 0.999 999 999 989 325 496 32;
  • 22) 0.999 999 999 989 325 496 32 × 2 = 1 + 0.999 999 999 978 650 992 64;
  • 23) 0.999 999 999 978 650 992 64 × 2 = 1 + 0.999 999 999 957 301 985 28;
  • 24) 0.999 999 999 957 301 985 28 × 2 = 1 + 0.999 999 999 914 603 970 56;
  • 25) 0.999 999 999 914 603 970 56 × 2 = 1 + 0.999 999 999 829 207 941 12;
  • 26) 0.999 999 999 829 207 941 12 × 2 = 1 + 0.999 999 999 658 415 882 24;
  • 27) 0.999 999 999 658 415 882 24 × 2 = 1 + 0.999 999 999 316 831 764 48;
  • 28) 0.999 999 999 316 831 764 48 × 2 = 1 + 0.999 999 998 633 663 528 96;
  • 29) 0.999 999 998 633 663 528 96 × 2 = 1 + 0.999 999 997 267 327 057 92;
  • 30) 0.999 999 997 267 327 057 92 × 2 = 1 + 0.999 999 994 534 654 115 84;
  • 31) 0.999 999 994 534 654 115 84 × 2 = 1 + 0.999 999 989 069 308 231 68;
  • 32) 0.999 999 989 069 308 231 68 × 2 = 1 + 0.999 999 978 138 616 463 36;
  • 33) 0.999 999 978 138 616 463 36 × 2 = 1 + 0.999 999 956 277 232 926 72;
  • 34) 0.999 999 956 277 232 926 72 × 2 = 1 + 0.999 999 912 554 465 853 44;
  • 35) 0.999 999 912 554 465 853 44 × 2 = 1 + 0.999 999 825 108 931 706 88;
  • 36) 0.999 999 825 108 931 706 88 × 2 = 1 + 0.999 999 650 217 863 413 76;
  • 37) 0.999 999 650 217 863 413 76 × 2 = 1 + 0.999 999 300 435 726 827 52;
  • 38) 0.999 999 300 435 726 827 52 × 2 = 1 + 0.999 998 600 871 453 655 04;
  • 39) 0.999 998 600 871 453 655 04 × 2 = 1 + 0.999 997 201 742 907 310 08;
  • 40) 0.999 997 201 742 907 310 08 × 2 = 1 + 0.999 994 403 485 814 620 16;
  • 41) 0.999 994 403 485 814 620 16 × 2 = 1 + 0.999 988 806 971 629 240 32;
  • 42) 0.999 988 806 971 629 240 32 × 2 = 1 + 0.999 977 613 943 258 480 64;
  • 43) 0.999 977 613 943 258 480 64 × 2 = 1 + 0.999 955 227 886 516 961 28;
  • 44) 0.999 955 227 886 516 961 28 × 2 = 1 + 0.999 910 455 773 033 922 56;
  • 45) 0.999 910 455 773 033 922 56 × 2 = 1 + 0.999 820 911 546 067 845 12;
  • 46) 0.999 820 911 546 067 845 12 × 2 = 1 + 0.999 641 823 092 135 690 24;
  • 47) 0.999 641 823 092 135 690 24 × 2 = 1 + 0.999 283 646 184 271 380 48;
  • 48) 0.999 283 646 184 271 380 48 × 2 = 1 + 0.998 567 292 368 542 760 96;
  • 49) 0.998 567 292 368 542 760 96 × 2 = 1 + 0.997 134 584 737 085 521 92;
  • 50) 0.997 134 584 737 085 521 92 × 2 = 1 + 0.994 269 169 474 171 043 84;
  • 51) 0.994 269 169 474 171 043 84 × 2 = 1 + 0.988 538 338 948 342 087 68;
  • 52) 0.988 538 338 948 342 087 68 × 2 = 1 + 0.977 076 677 896 684 175 36;
  • 53) 0.977 076 677 896 684 175 36 × 2 = 1 + 0.954 153 355 793 368 350 72;
  • 54) 0.954 153 355 793 368 350 72 × 2 = 1 + 0.908 306 711 586 736 701 44;
  • 55) 0.908 306 711 586 736 701 44 × 2 = 1 + 0.816 613 423 173 473 402 88;
  • 56) 0.816 613 423 173 473 402 88 × 2 = 1 + 0.633 226 846 346 946 805 76;
  • 57) 0.633 226 846 346 946 805 76 × 2 = 1 + 0.266 453 692 693 893 611 52;
  • 58) 0.266 453 692 693 893 611 52 × 2 = 0 + 0.532 907 385 387 787 223 04;

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:


0.016 738 891 601 562 494 91(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 494 91(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:


0.016 738 891 601 562 494 91(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 -0.016 738 891 601 562 494 91 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