-0.000 282 005 968 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 968(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
-0.000 282 005 968(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:

|-0.000 282 005 968| = 0.000 282 005 968


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: 0.000 282 005 968.

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) 0.000 282 005 968 × 2 = 0 + 0.000 564 011 936;
  • 2) 0.000 564 011 936 × 2 = 0 + 0.001 128 023 872;
  • 3) 0.001 128 023 872 × 2 = 0 + 0.002 256 047 744;
  • 4) 0.002 256 047 744 × 2 = 0 + 0.004 512 095 488;
  • 5) 0.004 512 095 488 × 2 = 0 + 0.009 024 190 976;
  • 6) 0.009 024 190 976 × 2 = 0 + 0.018 048 381 952;
  • 7) 0.018 048 381 952 × 2 = 0 + 0.036 096 763 904;
  • 8) 0.036 096 763 904 × 2 = 0 + 0.072 193 527 808;
  • 9) 0.072 193 527 808 × 2 = 0 + 0.144 387 055 616;
  • 10) 0.144 387 055 616 × 2 = 0 + 0.288 774 111 232;
  • 11) 0.288 774 111 232 × 2 = 0 + 0.577 548 222 464;
  • 12) 0.577 548 222 464 × 2 = 1 + 0.155 096 444 928;
  • 13) 0.155 096 444 928 × 2 = 0 + 0.310 192 889 856;
  • 14) 0.310 192 889 856 × 2 = 0 + 0.620 385 779 712;
  • 15) 0.620 385 779 712 × 2 = 1 + 0.240 771 559 424;
  • 16) 0.240 771 559 424 × 2 = 0 + 0.481 543 118 848;
  • 17) 0.481 543 118 848 × 2 = 0 + 0.963 086 237 696;
  • 18) 0.963 086 237 696 × 2 = 1 + 0.926 172 475 392;
  • 19) 0.926 172 475 392 × 2 = 1 + 0.852 344 950 784;
  • 20) 0.852 344 950 784 × 2 = 1 + 0.704 689 901 568;
  • 21) 0.704 689 901 568 × 2 = 1 + 0.409 379 803 136;
  • 22) 0.409 379 803 136 × 2 = 0 + 0.818 759 606 272;
  • 23) 0.818 759 606 272 × 2 = 1 + 0.637 519 212 544;
  • 24) 0.637 519 212 544 × 2 = 1 + 0.275 038 425 088;
  • 25) 0.275 038 425 088 × 2 = 0 + 0.550 076 850 176;
  • 26) 0.550 076 850 176 × 2 = 1 + 0.100 153 700 352;
  • 27) 0.100 153 700 352 × 2 = 0 + 0.200 307 400 704;
  • 28) 0.200 307 400 704 × 2 = 0 + 0.400 614 801 408;
  • 29) 0.400 614 801 408 × 2 = 0 + 0.801 229 602 816;
  • 30) 0.801 229 602 816 × 2 = 1 + 0.602 459 205 632;
  • 31) 0.602 459 205 632 × 2 = 1 + 0.204 918 411 264;
  • 32) 0.204 918 411 264 × 2 = 0 + 0.409 836 822 528;
  • 33) 0.409 836 822 528 × 2 = 0 + 0.819 673 645 056;
  • 34) 0.819 673 645 056 × 2 = 1 + 0.639 347 290 112;
  • 35) 0.639 347 290 112 × 2 = 1 + 0.278 694 580 224;
  • 36) 0.278 694 580 224 × 2 = 0 + 0.557 389 160 448;
  • 37) 0.557 389 160 448 × 2 = 1 + 0.114 778 320 896;
  • 38) 0.114 778 320 896 × 2 = 0 + 0.229 556 641 792;
  • 39) 0.229 556 641 792 × 2 = 0 + 0.459 113 283 584;
  • 40) 0.459 113 283 584 × 2 = 0 + 0.918 226 567 168;
  • 41) 0.918 226 567 168 × 2 = 1 + 0.836 453 134 336;
  • 42) 0.836 453 134 336 × 2 = 1 + 0.672 906 268 672;
  • 43) 0.672 906 268 672 × 2 = 1 + 0.345 812 537 344;
  • 44) 0.345 812 537 344 × 2 = 0 + 0.691 625 074 688;
  • 45) 0.691 625 074 688 × 2 = 1 + 0.383 250 149 376;
  • 46) 0.383 250 149 376 × 2 = 0 + 0.766 500 298 752;
  • 47) 0.766 500 298 752 × 2 = 1 + 0.533 000 597 504;
  • 48) 0.533 000 597 504 × 2 = 1 + 0.066 001 195 008;
  • 49) 0.066 001 195 008 × 2 = 0 + 0.132 002 390 016;
  • 50) 0.132 002 390 016 × 2 = 0 + 0.264 004 780 032;
  • 51) 0.264 004 780 032 × 2 = 0 + 0.528 009 560 064;
  • 52) 0.528 009 560 064 × 2 = 1 + 0.056 019 120 128;
  • 53) 0.056 019 120 128 × 2 = 0 + 0.112 038 240 256;
  • 54) 0.112 038 240 256 × 2 = 0 + 0.224 076 480 512;
  • 55) 0.224 076 480 512 × 2 = 0 + 0.448 152 961 024;
  • 56) 0.448 152 961 024 × 2 = 0 + 0.896 305 922 048;
  • 57) 0.896 305 922 048 × 2 = 1 + 0.792 611 844 096;
  • 58) 0.792 611 844 096 × 2 = 1 + 0.585 223 688 192;
  • 59) 0.585 223 688 192 × 2 = 1 + 0.170 447 376 384;
  • 60) 0.170 447 376 384 × 2 = 0 + 0.340 894 752 768;
  • 61) 0.340 894 752 768 × 2 = 0 + 0.681 789 505 536;
  • 62) 0.681 789 505 536 × 2 = 1 + 0.363 579 011 072;
  • 63) 0.363 579 011 072 × 2 = 0 + 0.727 158 022 144;
  • 64) 0.727 158 022 144 × 2 = 1 + 0.454 316 044 288;

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:


0.000 282 005 968(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 1000 1110 1011 0001 0000 1110 0101(2)

6. Positive number before normalization:

0.000 282 005 968(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 1000 1110 1011 0001 0000 1110 0101(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:


0.000 282 005 968(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 1000 1110 1011 0001 0000 1110 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 1000 1110 1011 0001 0000 1110 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0110 1000 1110 1011 0001 0000 1110 0101(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 0110 1000 1110 1011 0001 0000 1110 0101


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 0110 1000 1110 1011 0001 0000 1110 0101 =

0010 0111 1011 0100 0110 0110 1000 1110 1011 0001 0000 1110 0101


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 0110 1000 1110 1011 0001 0000 1110 0101


Decimal number -0.000 282 005 968 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0110 1000 1110 1011 0001 0000 1110 0101

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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