-0.000 007 769 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 769(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
-0.000 007 769(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:

|-0.000 007 769| = 0.000 007 769


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: 0.000 007 769.

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) 0.000 007 769 × 2 = 0 + 0.000 015 538;
  • 2) 0.000 015 538 × 2 = 0 + 0.000 031 076;
  • 3) 0.000 031 076 × 2 = 0 + 0.000 062 152;
  • 4) 0.000 062 152 × 2 = 0 + 0.000 124 304;
  • 5) 0.000 124 304 × 2 = 0 + 0.000 248 608;
  • 6) 0.000 248 608 × 2 = 0 + 0.000 497 216;
  • 7) 0.000 497 216 × 2 = 0 + 0.000 994 432;
  • 8) 0.000 994 432 × 2 = 0 + 0.001 988 864;
  • 9) 0.001 988 864 × 2 = 0 + 0.003 977 728;
  • 10) 0.003 977 728 × 2 = 0 + 0.007 955 456;
  • 11) 0.007 955 456 × 2 = 0 + 0.015 910 912;
  • 12) 0.015 910 912 × 2 = 0 + 0.031 821 824;
  • 13) 0.031 821 824 × 2 = 0 + 0.063 643 648;
  • 14) 0.063 643 648 × 2 = 0 + 0.127 287 296;
  • 15) 0.127 287 296 × 2 = 0 + 0.254 574 592;
  • 16) 0.254 574 592 × 2 = 0 + 0.509 149 184;
  • 17) 0.509 149 184 × 2 = 1 + 0.018 298 368;
  • 18) 0.018 298 368 × 2 = 0 + 0.036 596 736;
  • 19) 0.036 596 736 × 2 = 0 + 0.073 193 472;
  • 20) 0.073 193 472 × 2 = 0 + 0.146 386 944;
  • 21) 0.146 386 944 × 2 = 0 + 0.292 773 888;
  • 22) 0.292 773 888 × 2 = 0 + 0.585 547 776;
  • 23) 0.585 547 776 × 2 = 1 + 0.171 095 552;
  • 24) 0.171 095 552 × 2 = 0 + 0.342 191 104;
  • 25) 0.342 191 104 × 2 = 0 + 0.684 382 208;
  • 26) 0.684 382 208 × 2 = 1 + 0.368 764 416;
  • 27) 0.368 764 416 × 2 = 0 + 0.737 528 832;
  • 28) 0.737 528 832 × 2 = 1 + 0.475 057 664;
  • 29) 0.475 057 664 × 2 = 0 + 0.950 115 328;
  • 30) 0.950 115 328 × 2 = 1 + 0.900 230 656;
  • 31) 0.900 230 656 × 2 = 1 + 0.800 461 312;
  • 32) 0.800 461 312 × 2 = 1 + 0.600 922 624;
  • 33) 0.600 922 624 × 2 = 1 + 0.201 845 248;
  • 34) 0.201 845 248 × 2 = 0 + 0.403 690 496;
  • 35) 0.403 690 496 × 2 = 0 + 0.807 380 992;
  • 36) 0.807 380 992 × 2 = 1 + 0.614 761 984;
  • 37) 0.614 761 984 × 2 = 1 + 0.229 523 968;
  • 38) 0.229 523 968 × 2 = 0 + 0.459 047 936;
  • 39) 0.459 047 936 × 2 = 0 + 0.918 095 872;
  • 40) 0.918 095 872 × 2 = 1 + 0.836 191 744;
  • 41) 0.836 191 744 × 2 = 1 + 0.672 383 488;
  • 42) 0.672 383 488 × 2 = 1 + 0.344 766 976;
  • 43) 0.344 766 976 × 2 = 0 + 0.689 533 952;
  • 44) 0.689 533 952 × 2 = 1 + 0.379 067 904;
  • 45) 0.379 067 904 × 2 = 0 + 0.758 135 808;
  • 46) 0.758 135 808 × 2 = 1 + 0.516 271 616;
  • 47) 0.516 271 616 × 2 = 1 + 0.032 543 232;
  • 48) 0.032 543 232 × 2 = 0 + 0.065 086 464;
  • 49) 0.065 086 464 × 2 = 0 + 0.130 172 928;
  • 50) 0.130 172 928 × 2 = 0 + 0.260 345 856;
  • 51) 0.260 345 856 × 2 = 0 + 0.520 691 712;
  • 52) 0.520 691 712 × 2 = 1 + 0.041 383 424;
  • 53) 0.041 383 424 × 2 = 0 + 0.082 766 848;
  • 54) 0.082 766 848 × 2 = 0 + 0.165 533 696;
  • 55) 0.165 533 696 × 2 = 0 + 0.331 067 392;
  • 56) 0.331 067 392 × 2 = 0 + 0.662 134 784;
  • 57) 0.662 134 784 × 2 = 1 + 0.324 269 568;
  • 58) 0.324 269 568 × 2 = 0 + 0.648 539 136;
  • 59) 0.648 539 136 × 2 = 1 + 0.297 078 272;
  • 60) 0.297 078 272 × 2 = 0 + 0.594 156 544;
  • 61) 0.594 156 544 × 2 = 1 + 0.188 313 088;
  • 62) 0.188 313 088 × 2 = 0 + 0.376 626 176;
  • 63) 0.376 626 176 × 2 = 0 + 0.753 252 352;
  • 64) 0.753 252 352 × 2 = 1 + 0.506 504 704;
  • 65) 0.506 504 704 × 2 = 1 + 0.013 009 408;
  • 66) 0.013 009 408 × 2 = 0 + 0.026 018 816;
  • 67) 0.026 018 816 × 2 = 0 + 0.052 037 632;
  • 68) 0.052 037 632 × 2 = 0 + 0.104 075 264;
  • 69) 0.104 075 264 × 2 = 0 + 0.208 150 528;

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:


0.000 007 769(10) =

0.0000 0000 0000 0000 1000 0010 0101 0111 1001 1001 1101 0110 0001 0000 1010 1001 1000 0(2)

6. Positive number before normalization:

0.000 007 769(10) =

0.0000 0000 0000 0000 1000 0010 0101 0111 1001 1001 1101 0110 0001 0000 1010 1001 1000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 007 769(10) =

0.0000 0000 0000 0000 1000 0010 0101 0111 1001 1001 1101 0110 0001 0000 1010 1001 1000 0(2) =

0.0000 0000 0000 0000 1000 0010 0101 0111 1001 1001 1101 0110 0001 0000 1010 1001 1000 0(2) × 20 =

1.0000 0100 1010 1111 0011 0011 1010 1100 0010 0001 0101 0011 0000(2) × 2-17


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): -17

Mantissa (not normalized):
1.0000 0100 1010 1111 0011 0011 1010 1100 0010 0001 0101 0011 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 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) =

1006(10) =

011 1110 1110(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. 0000 0100 1010 1111 0011 0011 1010 1100 0010 0001 0101 0011 0000 =

0000 0100 1010 1111 0011 0011 1010 1100 0010 0001 0101 0011 0000


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 1110

Mantissa (52 bits) =
0000 0100 1010 1111 0011 0011 1010 1100 0010 0001 0101 0011 0000


Decimal number -0.000 007 769 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0100 1010 1111 0011 0011 1010 1100 0010 0001 0101 0011 0000

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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