-0.016 738 891 601 562 495 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 495 17(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
-0.016 738 891 601 562 495 17(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:

|-0.016 738 891 601 562 495 17| = 0.016 738 891 601 562 495 17


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: 0.016 738 891 601 562 495 17.

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) 0.016 738 891 601 562 495 17 × 2 = 0 + 0.033 477 783 203 124 990 34;
  • 2) 0.033 477 783 203 124 990 34 × 2 = 0 + 0.066 955 566 406 249 980 68;
  • 3) 0.066 955 566 406 249 980 68 × 2 = 0 + 0.133 911 132 812 499 961 36;
  • 4) 0.133 911 132 812 499 961 36 × 2 = 0 + 0.267 822 265 624 999 922 72;
  • 5) 0.267 822 265 624 999 922 72 × 2 = 0 + 0.535 644 531 249 999 845 44;
  • 6) 0.535 644 531 249 999 845 44 × 2 = 1 + 0.071 289 062 499 999 690 88;
  • 7) 0.071 289 062 499 999 690 88 × 2 = 0 + 0.142 578 124 999 999 381 76;
  • 8) 0.142 578 124 999 999 381 76 × 2 = 0 + 0.285 156 249 999 998 763 52;
  • 9) 0.285 156 249 999 998 763 52 × 2 = 0 + 0.570 312 499 999 997 527 04;
  • 10) 0.570 312 499 999 997 527 04 × 2 = 1 + 0.140 624 999 999 995 054 08;
  • 11) 0.140 624 999 999 995 054 08 × 2 = 0 + 0.281 249 999 999 990 108 16;
  • 12) 0.281 249 999 999 990 108 16 × 2 = 0 + 0.562 499 999 999 980 216 32;
  • 13) 0.562 499 999 999 980 216 32 × 2 = 1 + 0.124 999 999 999 960 432 64;
  • 14) 0.124 999 999 999 960 432 64 × 2 = 0 + 0.249 999 999 999 920 865 28;
  • 15) 0.249 999 999 999 920 865 28 × 2 = 0 + 0.499 999 999 999 841 730 56;
  • 16) 0.499 999 999 999 841 730 56 × 2 = 0 + 0.999 999 999 999 683 461 12;
  • 17) 0.999 999 999 999 683 461 12 × 2 = 1 + 0.999 999 999 999 366 922 24;
  • 18) 0.999 999 999 999 366 922 24 × 2 = 1 + 0.999 999 999 998 733 844 48;
  • 19) 0.999 999 999 998 733 844 48 × 2 = 1 + 0.999 999 999 997 467 688 96;
  • 20) 0.999 999 999 997 467 688 96 × 2 = 1 + 0.999 999 999 994 935 377 92;
  • 21) 0.999 999 999 994 935 377 92 × 2 = 1 + 0.999 999 999 989 870 755 84;
  • 22) 0.999 999 999 989 870 755 84 × 2 = 1 + 0.999 999 999 979 741 511 68;
  • 23) 0.999 999 999 979 741 511 68 × 2 = 1 + 0.999 999 999 959 483 023 36;
  • 24) 0.999 999 999 959 483 023 36 × 2 = 1 + 0.999 999 999 918 966 046 72;
  • 25) 0.999 999 999 918 966 046 72 × 2 = 1 + 0.999 999 999 837 932 093 44;
  • 26) 0.999 999 999 837 932 093 44 × 2 = 1 + 0.999 999 999 675 864 186 88;
  • 27) 0.999 999 999 675 864 186 88 × 2 = 1 + 0.999 999 999 351 728 373 76;
  • 28) 0.999 999 999 351 728 373 76 × 2 = 1 + 0.999 999 998 703 456 747 52;
  • 29) 0.999 999 998 703 456 747 52 × 2 = 1 + 0.999 999 997 406 913 495 04;
  • 30) 0.999 999 997 406 913 495 04 × 2 = 1 + 0.999 999 994 813 826 990 08;
  • 31) 0.999 999 994 813 826 990 08 × 2 = 1 + 0.999 999 989 627 653 980 16;
  • 32) 0.999 999 989 627 653 980 16 × 2 = 1 + 0.999 999 979 255 307 960 32;
  • 33) 0.999 999 979 255 307 960 32 × 2 = 1 + 0.999 999 958 510 615 920 64;
  • 34) 0.999 999 958 510 615 920 64 × 2 = 1 + 0.999 999 917 021 231 841 28;
  • 35) 0.999 999 917 021 231 841 28 × 2 = 1 + 0.999 999 834 042 463 682 56;
  • 36) 0.999 999 834 042 463 682 56 × 2 = 1 + 0.999 999 668 084 927 365 12;
  • 37) 0.999 999 668 084 927 365 12 × 2 = 1 + 0.999 999 336 169 854 730 24;
  • 38) 0.999 999 336 169 854 730 24 × 2 = 1 + 0.999 998 672 339 709 460 48;
  • 39) 0.999 998 672 339 709 460 48 × 2 = 1 + 0.999 997 344 679 418 920 96;
  • 40) 0.999 997 344 679 418 920 96 × 2 = 1 + 0.999 994 689 358 837 841 92;
  • 41) 0.999 994 689 358 837 841 92 × 2 = 1 + 0.999 989 378 717 675 683 84;
  • 42) 0.999 989 378 717 675 683 84 × 2 = 1 + 0.999 978 757 435 351 367 68;
  • 43) 0.999 978 757 435 351 367 68 × 2 = 1 + 0.999 957 514 870 702 735 36;
  • 44) 0.999 957 514 870 702 735 36 × 2 = 1 + 0.999 915 029 741 405 470 72;
  • 45) 0.999 915 029 741 405 470 72 × 2 = 1 + 0.999 830 059 482 810 941 44;
  • 46) 0.999 830 059 482 810 941 44 × 2 = 1 + 0.999 660 118 965 621 882 88;
  • 47) 0.999 660 118 965 621 882 88 × 2 = 1 + 0.999 320 237 931 243 765 76;
  • 48) 0.999 320 237 931 243 765 76 × 2 = 1 + 0.998 640 475 862 487 531 52;
  • 49) 0.998 640 475 862 487 531 52 × 2 = 1 + 0.997 280 951 724 975 063 04;
  • 50) 0.997 280 951 724 975 063 04 × 2 = 1 + 0.994 561 903 449 950 126 08;
  • 51) 0.994 561 903 449 950 126 08 × 2 = 1 + 0.989 123 806 899 900 252 16;
  • 52) 0.989 123 806 899 900 252 16 × 2 = 1 + 0.978 247 613 799 800 504 32;
  • 53) 0.978 247 613 799 800 504 32 × 2 = 1 + 0.956 495 227 599 601 008 64;
  • 54) 0.956 495 227 599 601 008 64 × 2 = 1 + 0.912 990 455 199 202 017 28;
  • 55) 0.912 990 455 199 202 017 28 × 2 = 1 + 0.825 980 910 398 404 034 56;
  • 56) 0.825 980 910 398 404 034 56 × 2 = 1 + 0.651 961 820 796 808 069 12;
  • 57) 0.651 961 820 796 808 069 12 × 2 = 1 + 0.303 923 641 593 616 138 24;
  • 58) 0.303 923 641 593 616 138 24 × 2 = 0 + 0.607 847 283 187 232 276 48;

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:


0.016 738 891 601 562 495 17(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 495 17(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:


0.016 738 891 601 562 495 17(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 -0.016 738 891 601 562 495 17 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