-0.000 000 111 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 111 7(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
-0.000 000 111 7(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:

|-0.000 000 111 7| = 0.000 000 111 7


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: 0.000 000 111 7.

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) 0.000 000 111 7 × 2 = 0 + 0.000 000 223 4;
  • 2) 0.000 000 223 4 × 2 = 0 + 0.000 000 446 8;
  • 3) 0.000 000 446 8 × 2 = 0 + 0.000 000 893 6;
  • 4) 0.000 000 893 6 × 2 = 0 + 0.000 001 787 2;
  • 5) 0.000 001 787 2 × 2 = 0 + 0.000 003 574 4;
  • 6) 0.000 003 574 4 × 2 = 0 + 0.000 007 148 8;
  • 7) 0.000 007 148 8 × 2 = 0 + 0.000 014 297 6;
  • 8) 0.000 014 297 6 × 2 = 0 + 0.000 028 595 2;
  • 9) 0.000 028 595 2 × 2 = 0 + 0.000 057 190 4;
  • 10) 0.000 057 190 4 × 2 = 0 + 0.000 114 380 8;
  • 11) 0.000 114 380 8 × 2 = 0 + 0.000 228 761 6;
  • 12) 0.000 228 761 6 × 2 = 0 + 0.000 457 523 2;
  • 13) 0.000 457 523 2 × 2 = 0 + 0.000 915 046 4;
  • 14) 0.000 915 046 4 × 2 = 0 + 0.001 830 092 8;
  • 15) 0.001 830 092 8 × 2 = 0 + 0.003 660 185 6;
  • 16) 0.003 660 185 6 × 2 = 0 + 0.007 320 371 2;
  • 17) 0.007 320 371 2 × 2 = 0 + 0.014 640 742 4;
  • 18) 0.014 640 742 4 × 2 = 0 + 0.029 281 484 8;
  • 19) 0.029 281 484 8 × 2 = 0 + 0.058 562 969 6;
  • 20) 0.058 562 969 6 × 2 = 0 + 0.117 125 939 2;
  • 21) 0.117 125 939 2 × 2 = 0 + 0.234 251 878 4;
  • 22) 0.234 251 878 4 × 2 = 0 + 0.468 503 756 8;
  • 23) 0.468 503 756 8 × 2 = 0 + 0.937 007 513 6;
  • 24) 0.937 007 513 6 × 2 = 1 + 0.874 015 027 2;
  • 25) 0.874 015 027 2 × 2 = 1 + 0.748 030 054 4;
  • 26) 0.748 030 054 4 × 2 = 1 + 0.496 060 108 8;
  • 27) 0.496 060 108 8 × 2 = 0 + 0.992 120 217 6;
  • 28) 0.992 120 217 6 × 2 = 1 + 0.984 240 435 2;
  • 29) 0.984 240 435 2 × 2 = 1 + 0.968 480 870 4;
  • 30) 0.968 480 870 4 × 2 = 1 + 0.936 961 740 8;
  • 31) 0.936 961 740 8 × 2 = 1 + 0.873 923 481 6;
  • 32) 0.873 923 481 6 × 2 = 1 + 0.747 846 963 2;
  • 33) 0.747 846 963 2 × 2 = 1 + 0.495 693 926 4;
  • 34) 0.495 693 926 4 × 2 = 0 + 0.991 387 852 8;
  • 35) 0.991 387 852 8 × 2 = 1 + 0.982 775 705 6;
  • 36) 0.982 775 705 6 × 2 = 1 + 0.965 551 411 2;
  • 37) 0.965 551 411 2 × 2 = 1 + 0.931 102 822 4;
  • 38) 0.931 102 822 4 × 2 = 1 + 0.862 205 644 8;
  • 39) 0.862 205 644 8 × 2 = 1 + 0.724 411 289 6;
  • 40) 0.724 411 289 6 × 2 = 1 + 0.448 822 579 2;
  • 41) 0.448 822 579 2 × 2 = 0 + 0.897 645 158 4;
  • 42) 0.897 645 158 4 × 2 = 1 + 0.795 290 316 8;
  • 43) 0.795 290 316 8 × 2 = 1 + 0.590 580 633 6;
  • 44) 0.590 580 633 6 × 2 = 1 + 0.181 161 267 2;
  • 45) 0.181 161 267 2 × 2 = 0 + 0.362 322 534 4;
  • 46) 0.362 322 534 4 × 2 = 0 + 0.724 645 068 8;
  • 47) 0.724 645 068 8 × 2 = 1 + 0.449 290 137 6;
  • 48) 0.449 290 137 6 × 2 = 0 + 0.898 580 275 2;
  • 49) 0.898 580 275 2 × 2 = 1 + 0.797 160 550 4;
  • 50) 0.797 160 550 4 × 2 = 1 + 0.594 321 100 8;
  • 51) 0.594 321 100 8 × 2 = 1 + 0.188 642 201 6;
  • 52) 0.188 642 201 6 × 2 = 0 + 0.377 284 403 2;
  • 53) 0.377 284 403 2 × 2 = 0 + 0.754 568 806 4;
  • 54) 0.754 568 806 4 × 2 = 1 + 0.509 137 612 8;
  • 55) 0.509 137 612 8 × 2 = 1 + 0.018 275 225 6;
  • 56) 0.018 275 225 6 × 2 = 0 + 0.036 550 451 2;
  • 57) 0.036 550 451 2 × 2 = 0 + 0.073 100 902 4;
  • 58) 0.073 100 902 4 × 2 = 0 + 0.146 201 804 8;
  • 59) 0.146 201 804 8 × 2 = 0 + 0.292 403 609 6;
  • 60) 0.292 403 609 6 × 2 = 0 + 0.584 807 219 2;
  • 61) 0.584 807 219 2 × 2 = 1 + 0.169 614 438 4;
  • 62) 0.169 614 438 4 × 2 = 0 + 0.339 228 876 8;
  • 63) 0.339 228 876 8 × 2 = 0 + 0.678 457 753 6;
  • 64) 0.678 457 753 6 × 2 = 1 + 0.356 915 507 2;
  • 65) 0.356 915 507 2 × 2 = 0 + 0.713 831 014 4;
  • 66) 0.713 831 014 4 × 2 = 1 + 0.427 662 028 8;
  • 67) 0.427 662 028 8 × 2 = 0 + 0.855 324 057 6;
  • 68) 0.855 324 057 6 × 2 = 1 + 0.710 648 115 2;
  • 69) 0.710 648 115 2 × 2 = 1 + 0.421 296 230 4;
  • 70) 0.421 296 230 4 × 2 = 0 + 0.842 592 460 8;
  • 71) 0.842 592 460 8 × 2 = 1 + 0.685 184 921 6;
  • 72) 0.685 184 921 6 × 2 = 1 + 0.370 369 843 2;
  • 73) 0.370 369 843 2 × 2 = 0 + 0.740 739 686 4;
  • 74) 0.740 739 686 4 × 2 = 1 + 0.481 479 372 8;
  • 75) 0.481 479 372 8 × 2 = 0 + 0.962 958 745 6;
  • 76) 0.962 958 745 6 × 2 = 1 + 0.925 917 491 2;

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:


0.000 000 111 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101(2)

6. Positive number before normalization:

0.000 000 111 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 111 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101(2) × 20 =

1.1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101(2) × 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): -24

Mantissa (not normalized):
1.1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-24 + 2(11-1) - 1 =

(-24 + 1 023)(10) =

999(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;
  • 999 ÷ 2 = 499 + 1;
  • 499 ÷ 2 = 249 + 1;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

999(10) =

011 1110 0111(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. 1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101 =

1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101


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 1110 0111

Mantissa (52 bits) =
1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101


Decimal number -0.000 000 111 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0111 - 1101 1111 1011 1111 0111 0010 1110 0110 0000 1001 0101 1011 0101

» Convert decimal -0.000 000 101 9 to 64 bit double precision IEEE 754 binary floating point representation standard

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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