-0.016 738 891 601 562 494 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 494 72(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 72(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 72| = 0.016 738 891 601 562 494 72


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 72.

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 72 × 2 = 0 + 0.033 477 783 203 124 989 44;
  • 2) 0.033 477 783 203 124 989 44 × 2 = 0 + 0.066 955 566 406 249 978 88;
  • 3) 0.066 955 566 406 249 978 88 × 2 = 0 + 0.133 911 132 812 499 957 76;
  • 4) 0.133 911 132 812 499 957 76 × 2 = 0 + 0.267 822 265 624 999 915 52;
  • 5) 0.267 822 265 624 999 915 52 × 2 = 0 + 0.535 644 531 249 999 831 04;
  • 6) 0.535 644 531 249 999 831 04 × 2 = 1 + 0.071 289 062 499 999 662 08;
  • 7) 0.071 289 062 499 999 662 08 × 2 = 0 + 0.142 578 124 999 999 324 16;
  • 8) 0.142 578 124 999 999 324 16 × 2 = 0 + 0.285 156 249 999 998 648 32;
  • 9) 0.285 156 249 999 998 648 32 × 2 = 0 + 0.570 312 499 999 997 296 64;
  • 10) 0.570 312 499 999 997 296 64 × 2 = 1 + 0.140 624 999 999 994 593 28;
  • 11) 0.140 624 999 999 994 593 28 × 2 = 0 + 0.281 249 999 999 989 186 56;
  • 12) 0.281 249 999 999 989 186 56 × 2 = 0 + 0.562 499 999 999 978 373 12;
  • 13) 0.562 499 999 999 978 373 12 × 2 = 1 + 0.124 999 999 999 956 746 24;
  • 14) 0.124 999 999 999 956 746 24 × 2 = 0 + 0.249 999 999 999 913 492 48;
  • 15) 0.249 999 999 999 913 492 48 × 2 = 0 + 0.499 999 999 999 826 984 96;
  • 16) 0.499 999 999 999 826 984 96 × 2 = 0 + 0.999 999 999 999 653 969 92;
  • 17) 0.999 999 999 999 653 969 92 × 2 = 1 + 0.999 999 999 999 307 939 84;
  • 18) 0.999 999 999 999 307 939 84 × 2 = 1 + 0.999 999 999 998 615 879 68;
  • 19) 0.999 999 999 998 615 879 68 × 2 = 1 + 0.999 999 999 997 231 759 36;
  • 20) 0.999 999 999 997 231 759 36 × 2 = 1 + 0.999 999 999 994 463 518 72;
  • 21) 0.999 999 999 994 463 518 72 × 2 = 1 + 0.999 999 999 988 927 037 44;
  • 22) 0.999 999 999 988 927 037 44 × 2 = 1 + 0.999 999 999 977 854 074 88;
  • 23) 0.999 999 999 977 854 074 88 × 2 = 1 + 0.999 999 999 955 708 149 76;
  • 24) 0.999 999 999 955 708 149 76 × 2 = 1 + 0.999 999 999 911 416 299 52;
  • 25) 0.999 999 999 911 416 299 52 × 2 = 1 + 0.999 999 999 822 832 599 04;
  • 26) 0.999 999 999 822 832 599 04 × 2 = 1 + 0.999 999 999 645 665 198 08;
  • 27) 0.999 999 999 645 665 198 08 × 2 = 1 + 0.999 999 999 291 330 396 16;
  • 28) 0.999 999 999 291 330 396 16 × 2 = 1 + 0.999 999 998 582 660 792 32;
  • 29) 0.999 999 998 582 660 792 32 × 2 = 1 + 0.999 999 997 165 321 584 64;
  • 30) 0.999 999 997 165 321 584 64 × 2 = 1 + 0.999 999 994 330 643 169 28;
  • 31) 0.999 999 994 330 643 169 28 × 2 = 1 + 0.999 999 988 661 286 338 56;
  • 32) 0.999 999 988 661 286 338 56 × 2 = 1 + 0.999 999 977 322 572 677 12;
  • 33) 0.999 999 977 322 572 677 12 × 2 = 1 + 0.999 999 954 645 145 354 24;
  • 34) 0.999 999 954 645 145 354 24 × 2 = 1 + 0.999 999 909 290 290 708 48;
  • 35) 0.999 999 909 290 290 708 48 × 2 = 1 + 0.999 999 818 580 581 416 96;
  • 36) 0.999 999 818 580 581 416 96 × 2 = 1 + 0.999 999 637 161 162 833 92;
  • 37) 0.999 999 637 161 162 833 92 × 2 = 1 + 0.999 999 274 322 325 667 84;
  • 38) 0.999 999 274 322 325 667 84 × 2 = 1 + 0.999 998 548 644 651 335 68;
  • 39) 0.999 998 548 644 651 335 68 × 2 = 1 + 0.999 997 097 289 302 671 36;
  • 40) 0.999 997 097 289 302 671 36 × 2 = 1 + 0.999 994 194 578 605 342 72;
  • 41) 0.999 994 194 578 605 342 72 × 2 = 1 + 0.999 988 389 157 210 685 44;
  • 42) 0.999 988 389 157 210 685 44 × 2 = 1 + 0.999 976 778 314 421 370 88;
  • 43) 0.999 976 778 314 421 370 88 × 2 = 1 + 0.999 953 556 628 842 741 76;
  • 44) 0.999 953 556 628 842 741 76 × 2 = 1 + 0.999 907 113 257 685 483 52;
  • 45) 0.999 907 113 257 685 483 52 × 2 = 1 + 0.999 814 226 515 370 967 04;
  • 46) 0.999 814 226 515 370 967 04 × 2 = 1 + 0.999 628 453 030 741 934 08;
  • 47) 0.999 628 453 030 741 934 08 × 2 = 1 + 0.999 256 906 061 483 868 16;
  • 48) 0.999 256 906 061 483 868 16 × 2 = 1 + 0.998 513 812 122 967 736 32;
  • 49) 0.998 513 812 122 967 736 32 × 2 = 1 + 0.997 027 624 245 935 472 64;
  • 50) 0.997 027 624 245 935 472 64 × 2 = 1 + 0.994 055 248 491 870 945 28;
  • 51) 0.994 055 248 491 870 945 28 × 2 = 1 + 0.988 110 496 983 741 890 56;
  • 52) 0.988 110 496 983 741 890 56 × 2 = 1 + 0.976 220 993 967 483 781 12;
  • 53) 0.976 220 993 967 483 781 12 × 2 = 1 + 0.952 441 987 934 967 562 24;
  • 54) 0.952 441 987 934 967 562 24 × 2 = 1 + 0.904 883 975 869 935 124 48;
  • 55) 0.904 883 975 869 935 124 48 × 2 = 1 + 0.809 767 951 739 870 248 96;
  • 56) 0.809 767 951 739 870 248 96 × 2 = 1 + 0.619 535 903 479 740 497 92;
  • 57) 0.619 535 903 479 740 497 92 × 2 = 1 + 0.239 071 806 959 480 995 84;
  • 58) 0.239 071 806 959 480 995 84 × 2 = 0 + 0.478 143 613 918 961 991 68;

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 72(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 72(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 72(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 72 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