-0.000 010 459 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 010 459(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
-0.000 010 459(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:

|-0.000 010 459| = 0.000 010 459


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: 0.000 010 459.

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) 0.000 010 459 × 2 = 0 + 0.000 020 918;
  • 2) 0.000 020 918 × 2 = 0 + 0.000 041 836;
  • 3) 0.000 041 836 × 2 = 0 + 0.000 083 672;
  • 4) 0.000 083 672 × 2 = 0 + 0.000 167 344;
  • 5) 0.000 167 344 × 2 = 0 + 0.000 334 688;
  • 6) 0.000 334 688 × 2 = 0 + 0.000 669 376;
  • 7) 0.000 669 376 × 2 = 0 + 0.001 338 752;
  • 8) 0.001 338 752 × 2 = 0 + 0.002 677 504;
  • 9) 0.002 677 504 × 2 = 0 + 0.005 355 008;
  • 10) 0.005 355 008 × 2 = 0 + 0.010 710 016;
  • 11) 0.010 710 016 × 2 = 0 + 0.021 420 032;
  • 12) 0.021 420 032 × 2 = 0 + 0.042 840 064;
  • 13) 0.042 840 064 × 2 = 0 + 0.085 680 128;
  • 14) 0.085 680 128 × 2 = 0 + 0.171 360 256;
  • 15) 0.171 360 256 × 2 = 0 + 0.342 720 512;
  • 16) 0.342 720 512 × 2 = 0 + 0.685 441 024;
  • 17) 0.685 441 024 × 2 = 1 + 0.370 882 048;
  • 18) 0.370 882 048 × 2 = 0 + 0.741 764 096;
  • 19) 0.741 764 096 × 2 = 1 + 0.483 528 192;
  • 20) 0.483 528 192 × 2 = 0 + 0.967 056 384;
  • 21) 0.967 056 384 × 2 = 1 + 0.934 112 768;
  • 22) 0.934 112 768 × 2 = 1 + 0.868 225 536;
  • 23) 0.868 225 536 × 2 = 1 + 0.736 451 072;
  • 24) 0.736 451 072 × 2 = 1 + 0.472 902 144;
  • 25) 0.472 902 144 × 2 = 0 + 0.945 804 288;
  • 26) 0.945 804 288 × 2 = 1 + 0.891 608 576;
  • 27) 0.891 608 576 × 2 = 1 + 0.783 217 152;
  • 28) 0.783 217 152 × 2 = 1 + 0.566 434 304;
  • 29) 0.566 434 304 × 2 = 1 + 0.132 868 608;
  • 30) 0.132 868 608 × 2 = 0 + 0.265 737 216;
  • 31) 0.265 737 216 × 2 = 0 + 0.531 474 432;
  • 32) 0.531 474 432 × 2 = 1 + 0.062 948 864;
  • 33) 0.062 948 864 × 2 = 0 + 0.125 897 728;
  • 34) 0.125 897 728 × 2 = 0 + 0.251 795 456;
  • 35) 0.251 795 456 × 2 = 0 + 0.503 590 912;
  • 36) 0.503 590 912 × 2 = 1 + 0.007 181 824;
  • 37) 0.007 181 824 × 2 = 0 + 0.014 363 648;
  • 38) 0.014 363 648 × 2 = 0 + 0.028 727 296;
  • 39) 0.028 727 296 × 2 = 0 + 0.057 454 592;
  • 40) 0.057 454 592 × 2 = 0 + 0.114 909 184;
  • 41) 0.114 909 184 × 2 = 0 + 0.229 818 368;
  • 42) 0.229 818 368 × 2 = 0 + 0.459 636 736;
  • 43) 0.459 636 736 × 2 = 0 + 0.919 273 472;
  • 44) 0.919 273 472 × 2 = 1 + 0.838 546 944;
  • 45) 0.838 546 944 × 2 = 1 + 0.677 093 888;
  • 46) 0.677 093 888 × 2 = 1 + 0.354 187 776;
  • 47) 0.354 187 776 × 2 = 0 + 0.708 375 552;
  • 48) 0.708 375 552 × 2 = 1 + 0.416 751 104;
  • 49) 0.416 751 104 × 2 = 0 + 0.833 502 208;
  • 50) 0.833 502 208 × 2 = 1 + 0.667 004 416;
  • 51) 0.667 004 416 × 2 = 1 + 0.334 008 832;
  • 52) 0.334 008 832 × 2 = 0 + 0.668 017 664;
  • 53) 0.668 017 664 × 2 = 1 + 0.336 035 328;
  • 54) 0.336 035 328 × 2 = 0 + 0.672 070 656;
  • 55) 0.672 070 656 × 2 = 1 + 0.344 141 312;
  • 56) 0.344 141 312 × 2 = 0 + 0.688 282 624;
  • 57) 0.688 282 624 × 2 = 1 + 0.376 565 248;
  • 58) 0.376 565 248 × 2 = 0 + 0.753 130 496;
  • 59) 0.753 130 496 × 2 = 1 + 0.506 260 992;
  • 60) 0.506 260 992 × 2 = 1 + 0.012 521 984;
  • 61) 0.012 521 984 × 2 = 0 + 0.025 043 968;
  • 62) 0.025 043 968 × 2 = 0 + 0.050 087 936;
  • 63) 0.050 087 936 × 2 = 0 + 0.100 175 872;
  • 64) 0.100 175 872 × 2 = 0 + 0.200 351 744;
  • 65) 0.200 351 744 × 2 = 0 + 0.400 703 488;
  • 66) 0.400 703 488 × 2 = 0 + 0.801 406 976;
  • 67) 0.801 406 976 × 2 = 1 + 0.602 813 952;
  • 68) 0.602 813 952 × 2 = 1 + 0.205 627 904;
  • 69) 0.205 627 904 × 2 = 0 + 0.411 255 808;

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:


0.000 010 459(10) =

0.0000 0000 0000 0000 1010 1111 0111 1001 0001 0000 0001 1101 0110 1010 1011 0000 0011 0(2)

6. Positive number before normalization:

0.000 010 459(10) =

0.0000 0000 0000 0000 1010 1111 0111 1001 0001 0000 0001 1101 0110 1010 1011 0000 0011 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:


0.000 010 459(10) =

0.0000 0000 0000 0000 1010 1111 0111 1001 0001 0000 0001 1101 0110 1010 1011 0000 0011 0(2) =

0.0000 0000 0000 0000 1010 1111 0111 1001 0001 0000 0001 1101 0110 1010 1011 0000 0011 0(2) × 20 =

1.0101 1110 1111 0010 0010 0000 0011 1010 1101 0101 0110 0000 0110(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):
1.0101 1110 1111 0010 0010 0000 0011 1010 1101 0101 0110 0000 0110


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. 0101 1110 1111 0010 0010 0000 0011 1010 1101 0101 0110 0000 0110 =

0101 1110 1111 0010 0010 0000 0011 1010 1101 0101 0110 0000 0110


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) =
0101 1110 1111 0010 0010 0000 0011 1010 1101 0101 0110 0000 0110


Decimal number -0.000 010 459 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0101 1110 1111 0010 0010 0000 0011 1010 1101 0101 0110 0000 0110

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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