-0.000 282 079 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 079(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
-0.000 282 079(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:

|-0.000 282 079| = 0.000 282 079


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: 0.000 282 079.

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) 0.000 282 079 × 2 = 0 + 0.000 564 158;
  • 2) 0.000 564 158 × 2 = 0 + 0.001 128 316;
  • 3) 0.001 128 316 × 2 = 0 + 0.002 256 632;
  • 4) 0.002 256 632 × 2 = 0 + 0.004 513 264;
  • 5) 0.004 513 264 × 2 = 0 + 0.009 026 528;
  • 6) 0.009 026 528 × 2 = 0 + 0.018 053 056;
  • 7) 0.018 053 056 × 2 = 0 + 0.036 106 112;
  • 8) 0.036 106 112 × 2 = 0 + 0.072 212 224;
  • 9) 0.072 212 224 × 2 = 0 + 0.144 424 448;
  • 10) 0.144 424 448 × 2 = 0 + 0.288 848 896;
  • 11) 0.288 848 896 × 2 = 0 + 0.577 697 792;
  • 12) 0.577 697 792 × 2 = 1 + 0.155 395 584;
  • 13) 0.155 395 584 × 2 = 0 + 0.310 791 168;
  • 14) 0.310 791 168 × 2 = 0 + 0.621 582 336;
  • 15) 0.621 582 336 × 2 = 1 + 0.243 164 672;
  • 16) 0.243 164 672 × 2 = 0 + 0.486 329 344;
  • 17) 0.486 329 344 × 2 = 0 + 0.972 658 688;
  • 18) 0.972 658 688 × 2 = 1 + 0.945 317 376;
  • 19) 0.945 317 376 × 2 = 1 + 0.890 634 752;
  • 20) 0.890 634 752 × 2 = 1 + 0.781 269 504;
  • 21) 0.781 269 504 × 2 = 1 + 0.562 539 008;
  • 22) 0.562 539 008 × 2 = 1 + 0.125 078 016;
  • 23) 0.125 078 016 × 2 = 0 + 0.250 156 032;
  • 24) 0.250 156 032 × 2 = 0 + 0.500 312 064;
  • 25) 0.500 312 064 × 2 = 1 + 0.000 624 128;
  • 26) 0.000 624 128 × 2 = 0 + 0.001 248 256;
  • 27) 0.001 248 256 × 2 = 0 + 0.002 496 512;
  • 28) 0.002 496 512 × 2 = 0 + 0.004 993 024;
  • 29) 0.004 993 024 × 2 = 0 + 0.009 986 048;
  • 30) 0.009 986 048 × 2 = 0 + 0.019 972 096;
  • 31) 0.019 972 096 × 2 = 0 + 0.039 944 192;
  • 32) 0.039 944 192 × 2 = 0 + 0.079 888 384;
  • 33) 0.079 888 384 × 2 = 0 + 0.159 776 768;
  • 34) 0.159 776 768 × 2 = 0 + 0.319 553 536;
  • 35) 0.319 553 536 × 2 = 0 + 0.639 107 072;
  • 36) 0.639 107 072 × 2 = 1 + 0.278 214 144;
  • 37) 0.278 214 144 × 2 = 0 + 0.556 428 288;
  • 38) 0.556 428 288 × 2 = 1 + 0.112 856 576;
  • 39) 0.112 856 576 × 2 = 0 + 0.225 713 152;
  • 40) 0.225 713 152 × 2 = 0 + 0.451 426 304;
  • 41) 0.451 426 304 × 2 = 0 + 0.902 852 608;
  • 42) 0.902 852 608 × 2 = 1 + 0.805 705 216;
  • 43) 0.805 705 216 × 2 = 1 + 0.611 410 432;
  • 44) 0.611 410 432 × 2 = 1 + 0.222 820 864;
  • 45) 0.222 820 864 × 2 = 0 + 0.445 641 728;
  • 46) 0.445 641 728 × 2 = 0 + 0.891 283 456;
  • 47) 0.891 283 456 × 2 = 1 + 0.782 566 912;
  • 48) 0.782 566 912 × 2 = 1 + 0.565 133 824;
  • 49) 0.565 133 824 × 2 = 1 + 0.130 267 648;
  • 50) 0.130 267 648 × 2 = 0 + 0.260 535 296;
  • 51) 0.260 535 296 × 2 = 0 + 0.521 070 592;
  • 52) 0.521 070 592 × 2 = 1 + 0.042 141 184;
  • 53) 0.042 141 184 × 2 = 0 + 0.084 282 368;
  • 54) 0.084 282 368 × 2 = 0 + 0.168 564 736;
  • 55) 0.168 564 736 × 2 = 0 + 0.337 129 472;
  • 56) 0.337 129 472 × 2 = 0 + 0.674 258 944;
  • 57) 0.674 258 944 × 2 = 1 + 0.348 517 888;
  • 58) 0.348 517 888 × 2 = 0 + 0.697 035 776;
  • 59) 0.697 035 776 × 2 = 1 + 0.394 071 552;
  • 60) 0.394 071 552 × 2 = 0 + 0.788 143 104;
  • 61) 0.788 143 104 × 2 = 1 + 0.576 286 208;
  • 62) 0.576 286 208 × 2 = 1 + 0.152 572 416;
  • 63) 0.152 572 416 × 2 = 0 + 0.305 144 832;
  • 64) 0.305 144 832 × 2 = 0 + 0.610 289 664;

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:


0.000 282 079(10) =

0.0000 0000 0001 0010 0111 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100(2)

6. Positive number before normalization:

0.000 282 079(10) =

0.0000 0000 0001 0010 0111 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100(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:


0.000 282 079(10) =

0.0000 0000 0001 0010 0111 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100(2) =

0.0000 0000 0001 0010 0111 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100(2) × 20 =

1.0010 0111 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100(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 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100


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 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100 =

0010 0111 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100


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 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100


Decimal number -0.000 282 079 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 1000 0000 0001 0100 0111 0011 1001 0000 1010 1100

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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