-0.000 282 005 989 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 989(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
-0.000 282 005 989(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:

|-0.000 282 005 989| = 0.000 282 005 989


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: 0.000 282 005 989.

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) 0.000 282 005 989 × 2 = 0 + 0.000 564 011 978;
  • 2) 0.000 564 011 978 × 2 = 0 + 0.001 128 023 956;
  • 3) 0.001 128 023 956 × 2 = 0 + 0.002 256 047 912;
  • 4) 0.002 256 047 912 × 2 = 0 + 0.004 512 095 824;
  • 5) 0.004 512 095 824 × 2 = 0 + 0.009 024 191 648;
  • 6) 0.009 024 191 648 × 2 = 0 + 0.018 048 383 296;
  • 7) 0.018 048 383 296 × 2 = 0 + 0.036 096 766 592;
  • 8) 0.036 096 766 592 × 2 = 0 + 0.072 193 533 184;
  • 9) 0.072 193 533 184 × 2 = 0 + 0.144 387 066 368;
  • 10) 0.144 387 066 368 × 2 = 0 + 0.288 774 132 736;
  • 11) 0.288 774 132 736 × 2 = 0 + 0.577 548 265 472;
  • 12) 0.577 548 265 472 × 2 = 1 + 0.155 096 530 944;
  • 13) 0.155 096 530 944 × 2 = 0 + 0.310 193 061 888;
  • 14) 0.310 193 061 888 × 2 = 0 + 0.620 386 123 776;
  • 15) 0.620 386 123 776 × 2 = 1 + 0.240 772 247 552;
  • 16) 0.240 772 247 552 × 2 = 0 + 0.481 544 495 104;
  • 17) 0.481 544 495 104 × 2 = 0 + 0.963 088 990 208;
  • 18) 0.963 088 990 208 × 2 = 1 + 0.926 177 980 416;
  • 19) 0.926 177 980 416 × 2 = 1 + 0.852 355 960 832;
  • 20) 0.852 355 960 832 × 2 = 1 + 0.704 711 921 664;
  • 21) 0.704 711 921 664 × 2 = 1 + 0.409 423 843 328;
  • 22) 0.409 423 843 328 × 2 = 0 + 0.818 847 686 656;
  • 23) 0.818 847 686 656 × 2 = 1 + 0.637 695 373 312;
  • 24) 0.637 695 373 312 × 2 = 1 + 0.275 390 746 624;
  • 25) 0.275 390 746 624 × 2 = 0 + 0.550 781 493 248;
  • 26) 0.550 781 493 248 × 2 = 1 + 0.101 562 986 496;
  • 27) 0.101 562 986 496 × 2 = 0 + 0.203 125 972 992;
  • 28) 0.203 125 972 992 × 2 = 0 + 0.406 251 945 984;
  • 29) 0.406 251 945 984 × 2 = 0 + 0.812 503 891 968;
  • 30) 0.812 503 891 968 × 2 = 1 + 0.625 007 783 936;
  • 31) 0.625 007 783 936 × 2 = 1 + 0.250 015 567 872;
  • 32) 0.250 015 567 872 × 2 = 0 + 0.500 031 135 744;
  • 33) 0.500 031 135 744 × 2 = 1 + 0.000 062 271 488;
  • 34) 0.000 062 271 488 × 2 = 0 + 0.000 124 542 976;
  • 35) 0.000 124 542 976 × 2 = 0 + 0.000 249 085 952;
  • 36) 0.000 249 085 952 × 2 = 0 + 0.000 498 171 904;
  • 37) 0.000 498 171 904 × 2 = 0 + 0.000 996 343 808;
  • 38) 0.000 996 343 808 × 2 = 0 + 0.001 992 687 616;
  • 39) 0.001 992 687 616 × 2 = 0 + 0.003 985 375 232;
  • 40) 0.003 985 375 232 × 2 = 0 + 0.007 970 750 464;
  • 41) 0.007 970 750 464 × 2 = 0 + 0.015 941 500 928;
  • 42) 0.015 941 500 928 × 2 = 0 + 0.031 883 001 856;
  • 43) 0.031 883 001 856 × 2 = 0 + 0.063 766 003 712;
  • 44) 0.063 766 003 712 × 2 = 0 + 0.127 532 007 424;
  • 45) 0.127 532 007 424 × 2 = 0 + 0.255 064 014 848;
  • 46) 0.255 064 014 848 × 2 = 0 + 0.510 128 029 696;
  • 47) 0.510 128 029 696 × 2 = 1 + 0.020 256 059 392;
  • 48) 0.020 256 059 392 × 2 = 0 + 0.040 512 118 784;
  • 49) 0.040 512 118 784 × 2 = 0 + 0.081 024 237 568;
  • 50) 0.081 024 237 568 × 2 = 0 + 0.162 048 475 136;
  • 51) 0.162 048 475 136 × 2 = 0 + 0.324 096 950 272;
  • 52) 0.324 096 950 272 × 2 = 0 + 0.648 193 900 544;
  • 53) 0.648 193 900 544 × 2 = 1 + 0.296 387 801 088;
  • 54) 0.296 387 801 088 × 2 = 0 + 0.592 775 602 176;
  • 55) 0.592 775 602 176 × 2 = 1 + 0.185 551 204 352;
  • 56) 0.185 551 204 352 × 2 = 0 + 0.371 102 408 704;
  • 57) 0.371 102 408 704 × 2 = 0 + 0.742 204 817 408;
  • 58) 0.742 204 817 408 × 2 = 1 + 0.484 409 634 816;
  • 59) 0.484 409 634 816 × 2 = 0 + 0.968 819 269 632;
  • 60) 0.968 819 269 632 × 2 = 1 + 0.937 638 539 264;
  • 61) 0.937 638 539 264 × 2 = 1 + 0.875 277 078 528;
  • 62) 0.875 277 078 528 × 2 = 1 + 0.750 554 157 056;
  • 63) 0.750 554 157 056 × 2 = 1 + 0.501 108 314 112;
  • 64) 0.501 108 314 112 × 2 = 1 + 0.002 216 628 224;

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:


0.000 282 005 989(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111(2)

6. Positive number before normalization:

0.000 282 005 989(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 989(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111(2) × 20 =

1.0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111 =

0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111


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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111


Decimal number -0.000 282 005 989 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1000 0000 0000 0010 0000 1010 0101 1111

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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