-0.000 282 006 252 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 252(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
-0.000 282 006 252(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:

|-0.000 282 006 252| = 0.000 282 006 252


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: 0.000 282 006 252.

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) 0.000 282 006 252 × 2 = 0 + 0.000 564 012 504;
  • 2) 0.000 564 012 504 × 2 = 0 + 0.001 128 025 008;
  • 3) 0.001 128 025 008 × 2 = 0 + 0.002 256 050 016;
  • 4) 0.002 256 050 016 × 2 = 0 + 0.004 512 100 032;
  • 5) 0.004 512 100 032 × 2 = 0 + 0.009 024 200 064;
  • 6) 0.009 024 200 064 × 2 = 0 + 0.018 048 400 128;
  • 7) 0.018 048 400 128 × 2 = 0 + 0.036 096 800 256;
  • 8) 0.036 096 800 256 × 2 = 0 + 0.072 193 600 512;
  • 9) 0.072 193 600 512 × 2 = 0 + 0.144 387 201 024;
  • 10) 0.144 387 201 024 × 2 = 0 + 0.288 774 402 048;
  • 11) 0.288 774 402 048 × 2 = 0 + 0.577 548 804 096;
  • 12) 0.577 548 804 096 × 2 = 1 + 0.155 097 608 192;
  • 13) 0.155 097 608 192 × 2 = 0 + 0.310 195 216 384;
  • 14) 0.310 195 216 384 × 2 = 0 + 0.620 390 432 768;
  • 15) 0.620 390 432 768 × 2 = 1 + 0.240 780 865 536;
  • 16) 0.240 780 865 536 × 2 = 0 + 0.481 561 731 072;
  • 17) 0.481 561 731 072 × 2 = 0 + 0.963 123 462 144;
  • 18) 0.963 123 462 144 × 2 = 1 + 0.926 246 924 288;
  • 19) 0.926 246 924 288 × 2 = 1 + 0.852 493 848 576;
  • 20) 0.852 493 848 576 × 2 = 1 + 0.704 987 697 152;
  • 21) 0.704 987 697 152 × 2 = 1 + 0.409 975 394 304;
  • 22) 0.409 975 394 304 × 2 = 0 + 0.819 950 788 608;
  • 23) 0.819 950 788 608 × 2 = 1 + 0.639 901 577 216;
  • 24) 0.639 901 577 216 × 2 = 1 + 0.279 803 154 432;
  • 25) 0.279 803 154 432 × 2 = 0 + 0.559 606 308 864;
  • 26) 0.559 606 308 864 × 2 = 1 + 0.119 212 617 728;
  • 27) 0.119 212 617 728 × 2 = 0 + 0.238 425 235 456;
  • 28) 0.238 425 235 456 × 2 = 0 + 0.476 850 470 912;
  • 29) 0.476 850 470 912 × 2 = 0 + 0.953 700 941 824;
  • 30) 0.953 700 941 824 × 2 = 1 + 0.907 401 883 648;
  • 31) 0.907 401 883 648 × 2 = 1 + 0.814 803 767 296;
  • 32) 0.814 803 767 296 × 2 = 1 + 0.629 607 534 592;
  • 33) 0.629 607 534 592 × 2 = 1 + 0.259 215 069 184;
  • 34) 0.259 215 069 184 × 2 = 0 + 0.518 430 138 368;
  • 35) 0.518 430 138 368 × 2 = 1 + 0.036 860 276 736;
  • 36) 0.036 860 276 736 × 2 = 0 + 0.073 720 553 472;
  • 37) 0.073 720 553 472 × 2 = 0 + 0.147 441 106 944;
  • 38) 0.147 441 106 944 × 2 = 0 + 0.294 882 213 888;
  • 39) 0.294 882 213 888 × 2 = 0 + 0.589 764 427 776;
  • 40) 0.589 764 427 776 × 2 = 1 + 0.179 528 855 552;
  • 41) 0.179 528 855 552 × 2 = 0 + 0.359 057 711 104;
  • 42) 0.359 057 711 104 × 2 = 0 + 0.718 115 422 208;
  • 43) 0.718 115 422 208 × 2 = 1 + 0.436 230 844 416;
  • 44) 0.436 230 844 416 × 2 = 0 + 0.872 461 688 832;
  • 45) 0.872 461 688 832 × 2 = 1 + 0.744 923 377 664;
  • 46) 0.744 923 377 664 × 2 = 1 + 0.489 846 755 328;
  • 47) 0.489 846 755 328 × 2 = 0 + 0.979 693 510 656;
  • 48) 0.979 693 510 656 × 2 = 1 + 0.959 387 021 312;
  • 49) 0.959 387 021 312 × 2 = 1 + 0.918 774 042 624;
  • 50) 0.918 774 042 624 × 2 = 1 + 0.837 548 085 248;
  • 51) 0.837 548 085 248 × 2 = 1 + 0.675 096 170 496;
  • 52) 0.675 096 170 496 × 2 = 1 + 0.350 192 340 992;
  • 53) 0.350 192 340 992 × 2 = 0 + 0.700 384 681 984;
  • 54) 0.700 384 681 984 × 2 = 1 + 0.400 769 363 968;
  • 55) 0.400 769 363 968 × 2 = 0 + 0.801 538 727 936;
  • 56) 0.801 538 727 936 × 2 = 1 + 0.603 077 455 872;
  • 57) 0.603 077 455 872 × 2 = 1 + 0.206 154 911 744;
  • 58) 0.206 154 911 744 × 2 = 0 + 0.412 309 823 488;
  • 59) 0.412 309 823 488 × 2 = 0 + 0.824 619 646 976;
  • 60) 0.824 619 646 976 × 2 = 1 + 0.649 239 293 952;
  • 61) 0.649 239 293 952 × 2 = 1 + 0.298 478 587 904;
  • 62) 0.298 478 587 904 × 2 = 0 + 0.596 957 175 808;
  • 63) 0.596 957 175 808 × 2 = 1 + 0.193 914 351 616;
  • 64) 0.193 914 351 616 × 2 = 0 + 0.387 828 703 232;

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:


0.000 282 006 252(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 0001 0010 1101 1111 0101 1001 1010(2)

6. Positive number before normalization:

0.000 282 006 252(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 0001 0010 1101 1111 0101 1001 1010(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:


0.000 282 006 252(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 0001 0010 1101 1111 0101 1001 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 0001 0010 1101 1111 0101 1001 1010(2) × 20 =

1.0010 0111 1011 0100 0111 1010 0001 0010 1101 1111 0101 1001 1010(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 0111 1010 0001 0010 1101 1111 0101 1001 1010


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 0111 1010 0001 0010 1101 1111 0101 1001 1010 =

0010 0111 1011 0100 0111 1010 0001 0010 1101 1111 0101 1001 1010


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 0111 1010 0001 0010 1101 1111 0101 1001 1010


Decimal number -0.000 282 006 252 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1010 0001 0010 1101 1111 0101 1001 1010

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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