-0.000 282 006 033 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 033(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
-0.000 282 006 033(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:

|-0.000 282 006 033| = 0.000 282 006 033


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: 0.000 282 006 033.

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) 0.000 282 006 033 × 2 = 0 + 0.000 564 012 066;
  • 2) 0.000 564 012 066 × 2 = 0 + 0.001 128 024 132;
  • 3) 0.001 128 024 132 × 2 = 0 + 0.002 256 048 264;
  • 4) 0.002 256 048 264 × 2 = 0 + 0.004 512 096 528;
  • 5) 0.004 512 096 528 × 2 = 0 + 0.009 024 193 056;
  • 6) 0.009 024 193 056 × 2 = 0 + 0.018 048 386 112;
  • 7) 0.018 048 386 112 × 2 = 0 + 0.036 096 772 224;
  • 8) 0.036 096 772 224 × 2 = 0 + 0.072 193 544 448;
  • 9) 0.072 193 544 448 × 2 = 0 + 0.144 387 088 896;
  • 10) 0.144 387 088 896 × 2 = 0 + 0.288 774 177 792;
  • 11) 0.288 774 177 792 × 2 = 0 + 0.577 548 355 584;
  • 12) 0.577 548 355 584 × 2 = 1 + 0.155 096 711 168;
  • 13) 0.155 096 711 168 × 2 = 0 + 0.310 193 422 336;
  • 14) 0.310 193 422 336 × 2 = 0 + 0.620 386 844 672;
  • 15) 0.620 386 844 672 × 2 = 1 + 0.240 773 689 344;
  • 16) 0.240 773 689 344 × 2 = 0 + 0.481 547 378 688;
  • 17) 0.481 547 378 688 × 2 = 0 + 0.963 094 757 376;
  • 18) 0.963 094 757 376 × 2 = 1 + 0.926 189 514 752;
  • 19) 0.926 189 514 752 × 2 = 1 + 0.852 379 029 504;
  • 20) 0.852 379 029 504 × 2 = 1 + 0.704 758 059 008;
  • 21) 0.704 758 059 008 × 2 = 1 + 0.409 516 118 016;
  • 22) 0.409 516 118 016 × 2 = 0 + 0.819 032 236 032;
  • 23) 0.819 032 236 032 × 2 = 1 + 0.638 064 472 064;
  • 24) 0.638 064 472 064 × 2 = 1 + 0.276 128 944 128;
  • 25) 0.276 128 944 128 × 2 = 0 + 0.552 257 888 256;
  • 26) 0.552 257 888 256 × 2 = 1 + 0.104 515 776 512;
  • 27) 0.104 515 776 512 × 2 = 0 + 0.209 031 553 024;
  • 28) 0.209 031 553 024 × 2 = 0 + 0.418 063 106 048;
  • 29) 0.418 063 106 048 × 2 = 0 + 0.836 126 212 096;
  • 30) 0.836 126 212 096 × 2 = 1 + 0.672 252 424 192;
  • 31) 0.672 252 424 192 × 2 = 1 + 0.344 504 848 384;
  • 32) 0.344 504 848 384 × 2 = 0 + 0.689 009 696 768;
  • 33) 0.689 009 696 768 × 2 = 1 + 0.378 019 393 536;
  • 34) 0.378 019 393 536 × 2 = 0 + 0.756 038 787 072;
  • 35) 0.756 038 787 072 × 2 = 1 + 0.512 077 574 144;
  • 36) 0.512 077 574 144 × 2 = 1 + 0.024 155 148 288;
  • 37) 0.024 155 148 288 × 2 = 0 + 0.048 310 296 576;
  • 38) 0.048 310 296 576 × 2 = 0 + 0.096 620 593 152;
  • 39) 0.096 620 593 152 × 2 = 0 + 0.193 241 186 304;
  • 40) 0.193 241 186 304 × 2 = 0 + 0.386 482 372 608;
  • 41) 0.386 482 372 608 × 2 = 0 + 0.772 964 745 216;
  • 42) 0.772 964 745 216 × 2 = 1 + 0.545 929 490 432;
  • 43) 0.545 929 490 432 × 2 = 1 + 0.091 858 980 864;
  • 44) 0.091 858 980 864 × 2 = 0 + 0.183 717 961 728;
  • 45) 0.183 717 961 728 × 2 = 0 + 0.367 435 923 456;
  • 46) 0.367 435 923 456 × 2 = 0 + 0.734 871 846 912;
  • 47) 0.734 871 846 912 × 2 = 1 + 0.469 743 693 824;
  • 48) 0.469 743 693 824 × 2 = 0 + 0.939 487 387 648;
  • 49) 0.939 487 387 648 × 2 = 1 + 0.878 974 775 296;
  • 50) 0.878 974 775 296 × 2 = 1 + 0.757 949 550 592;
  • 51) 0.757 949 550 592 × 2 = 1 + 0.515 899 101 184;
  • 52) 0.515 899 101 184 × 2 = 1 + 0.031 798 202 368;
  • 53) 0.031 798 202 368 × 2 = 0 + 0.063 596 404 736;
  • 54) 0.063 596 404 736 × 2 = 0 + 0.127 192 809 472;
  • 55) 0.127 192 809 472 × 2 = 0 + 0.254 385 618 944;
  • 56) 0.254 385 618 944 × 2 = 0 + 0.508 771 237 888;
  • 57) 0.508 771 237 888 × 2 = 1 + 0.017 542 475 776;
  • 58) 0.017 542 475 776 × 2 = 0 + 0.035 084 951 552;
  • 59) 0.035 084 951 552 × 2 = 0 + 0.070 169 903 104;
  • 60) 0.070 169 903 104 × 2 = 0 + 0.140 339 806 208;
  • 61) 0.140 339 806 208 × 2 = 0 + 0.280 679 612 416;
  • 62) 0.280 679 612 416 × 2 = 0 + 0.561 359 224 832;
  • 63) 0.561 359 224 832 × 2 = 1 + 0.122 718 449 664;
  • 64) 0.122 718 449 664 × 2 = 0 + 0.245 436 899 328;

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:


0.000 282 006 033(10) =

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

6. Positive number before normalization:

0.000 282 006 033(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 0000 0110 0010 1111 0000 1000 0010(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:


0.000 282 006 033(10) =

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

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

1.0010 0111 1011 0100 0110 1011 0000 0110 0010 1111 0000 1000 0010(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 1011 0000 0110 0010 1111 0000 1000 0010


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 1011 0000 0110 0010 1111 0000 1000 0010 =

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


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 1011 0000 0110 0010 1111 0000 1000 0010


Decimal number -0.000 282 006 033 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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