-0.000 281 967 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 967(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
-0.000 281 967(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:

|-0.000 281 967| = 0.000 281 967


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: 0.000 281 967.

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) 0.000 281 967 × 2 = 0 + 0.000 563 934;
  • 2) 0.000 563 934 × 2 = 0 + 0.001 127 868;
  • 3) 0.001 127 868 × 2 = 0 + 0.002 255 736;
  • 4) 0.002 255 736 × 2 = 0 + 0.004 511 472;
  • 5) 0.004 511 472 × 2 = 0 + 0.009 022 944;
  • 6) 0.009 022 944 × 2 = 0 + 0.018 045 888;
  • 7) 0.018 045 888 × 2 = 0 + 0.036 091 776;
  • 8) 0.036 091 776 × 2 = 0 + 0.072 183 552;
  • 9) 0.072 183 552 × 2 = 0 + 0.144 367 104;
  • 10) 0.144 367 104 × 2 = 0 + 0.288 734 208;
  • 11) 0.288 734 208 × 2 = 0 + 0.577 468 416;
  • 12) 0.577 468 416 × 2 = 1 + 0.154 936 832;
  • 13) 0.154 936 832 × 2 = 0 + 0.309 873 664;
  • 14) 0.309 873 664 × 2 = 0 + 0.619 747 328;
  • 15) 0.619 747 328 × 2 = 1 + 0.239 494 656;
  • 16) 0.239 494 656 × 2 = 0 + 0.478 989 312;
  • 17) 0.478 989 312 × 2 = 0 + 0.957 978 624;
  • 18) 0.957 978 624 × 2 = 1 + 0.915 957 248;
  • 19) 0.915 957 248 × 2 = 1 + 0.831 914 496;
  • 20) 0.831 914 496 × 2 = 1 + 0.663 828 992;
  • 21) 0.663 828 992 × 2 = 1 + 0.327 657 984;
  • 22) 0.327 657 984 × 2 = 0 + 0.655 315 968;
  • 23) 0.655 315 968 × 2 = 1 + 0.310 631 936;
  • 24) 0.310 631 936 × 2 = 0 + 0.621 263 872;
  • 25) 0.621 263 872 × 2 = 1 + 0.242 527 744;
  • 26) 0.242 527 744 × 2 = 0 + 0.485 055 488;
  • 27) 0.485 055 488 × 2 = 0 + 0.970 110 976;
  • 28) 0.970 110 976 × 2 = 1 + 0.940 221 952;
  • 29) 0.940 221 952 × 2 = 1 + 0.880 443 904;
  • 30) 0.880 443 904 × 2 = 1 + 0.760 887 808;
  • 31) 0.760 887 808 × 2 = 1 + 0.521 775 616;
  • 32) 0.521 775 616 × 2 = 1 + 0.043 551 232;
  • 33) 0.043 551 232 × 2 = 0 + 0.087 102 464;
  • 34) 0.087 102 464 × 2 = 0 + 0.174 204 928;
  • 35) 0.174 204 928 × 2 = 0 + 0.348 409 856;
  • 36) 0.348 409 856 × 2 = 0 + 0.696 819 712;
  • 37) 0.696 819 712 × 2 = 1 + 0.393 639 424;
  • 38) 0.393 639 424 × 2 = 0 + 0.787 278 848;
  • 39) 0.787 278 848 × 2 = 1 + 0.574 557 696;
  • 40) 0.574 557 696 × 2 = 1 + 0.149 115 392;
  • 41) 0.149 115 392 × 2 = 0 + 0.298 230 784;
  • 42) 0.298 230 784 × 2 = 0 + 0.596 461 568;
  • 43) 0.596 461 568 × 2 = 1 + 0.192 923 136;
  • 44) 0.192 923 136 × 2 = 0 + 0.385 846 272;
  • 45) 0.385 846 272 × 2 = 0 + 0.771 692 544;
  • 46) 0.771 692 544 × 2 = 1 + 0.543 385 088;
  • 47) 0.543 385 088 × 2 = 1 + 0.086 770 176;
  • 48) 0.086 770 176 × 2 = 0 + 0.173 540 352;
  • 49) 0.173 540 352 × 2 = 0 + 0.347 080 704;
  • 50) 0.347 080 704 × 2 = 0 + 0.694 161 408;
  • 51) 0.694 161 408 × 2 = 1 + 0.388 322 816;
  • 52) 0.388 322 816 × 2 = 0 + 0.776 645 632;
  • 53) 0.776 645 632 × 2 = 1 + 0.553 291 264;
  • 54) 0.553 291 264 × 2 = 1 + 0.106 582 528;
  • 55) 0.106 582 528 × 2 = 0 + 0.213 165 056;
  • 56) 0.213 165 056 × 2 = 0 + 0.426 330 112;
  • 57) 0.426 330 112 × 2 = 0 + 0.852 660 224;
  • 58) 0.852 660 224 × 2 = 1 + 0.705 320 448;
  • 59) 0.705 320 448 × 2 = 1 + 0.410 640 896;
  • 60) 0.410 640 896 × 2 = 0 + 0.821 281 792;
  • 61) 0.821 281 792 × 2 = 1 + 0.642 563 584;
  • 62) 0.642 563 584 × 2 = 1 + 0.285 127 168;
  • 63) 0.285 127 168 × 2 = 0 + 0.570 254 336;
  • 64) 0.570 254 336 × 2 = 1 + 0.140 508 672;

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:


0.000 281 967(10) =

0.0000 0000 0001 0010 0111 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101(2)

6. Positive number before normalization:

0.000 281 967(10) =

0.0000 0000 0001 0010 0111 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101(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:


0.000 281 967(10) =

0.0000 0000 0001 0010 0111 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101(2) =

0.0000 0000 0001 0010 0111 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101(2) × 20 =

1.0010 0111 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101(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 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101


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 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101 =

0010 0111 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101


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 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101


Decimal number -0.000 281 967 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1010 1001 1111 0000 1011 0010 0110 0010 1100 0110 1101

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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