-0.000 282 006 353 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 353(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
-0.000 282 006 353(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:

|-0.000 282 006 353| = 0.000 282 006 353


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: 0.000 282 006 353.

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) 0.000 282 006 353 × 2 = 0 + 0.000 564 012 706;
  • 2) 0.000 564 012 706 × 2 = 0 + 0.001 128 025 412;
  • 3) 0.001 128 025 412 × 2 = 0 + 0.002 256 050 824;
  • 4) 0.002 256 050 824 × 2 = 0 + 0.004 512 101 648;
  • 5) 0.004 512 101 648 × 2 = 0 + 0.009 024 203 296;
  • 6) 0.009 024 203 296 × 2 = 0 + 0.018 048 406 592;
  • 7) 0.018 048 406 592 × 2 = 0 + 0.036 096 813 184;
  • 8) 0.036 096 813 184 × 2 = 0 + 0.072 193 626 368;
  • 9) 0.072 193 626 368 × 2 = 0 + 0.144 387 252 736;
  • 10) 0.144 387 252 736 × 2 = 0 + 0.288 774 505 472;
  • 11) 0.288 774 505 472 × 2 = 0 + 0.577 549 010 944;
  • 12) 0.577 549 010 944 × 2 = 1 + 0.155 098 021 888;
  • 13) 0.155 098 021 888 × 2 = 0 + 0.310 196 043 776;
  • 14) 0.310 196 043 776 × 2 = 0 + 0.620 392 087 552;
  • 15) 0.620 392 087 552 × 2 = 1 + 0.240 784 175 104;
  • 16) 0.240 784 175 104 × 2 = 0 + 0.481 568 350 208;
  • 17) 0.481 568 350 208 × 2 = 0 + 0.963 136 700 416;
  • 18) 0.963 136 700 416 × 2 = 1 + 0.926 273 400 832;
  • 19) 0.926 273 400 832 × 2 = 1 + 0.852 546 801 664;
  • 20) 0.852 546 801 664 × 2 = 1 + 0.705 093 603 328;
  • 21) 0.705 093 603 328 × 2 = 1 + 0.410 187 206 656;
  • 22) 0.410 187 206 656 × 2 = 0 + 0.820 374 413 312;
  • 23) 0.820 374 413 312 × 2 = 1 + 0.640 748 826 624;
  • 24) 0.640 748 826 624 × 2 = 1 + 0.281 497 653 248;
  • 25) 0.281 497 653 248 × 2 = 0 + 0.562 995 306 496;
  • 26) 0.562 995 306 496 × 2 = 1 + 0.125 990 612 992;
  • 27) 0.125 990 612 992 × 2 = 0 + 0.251 981 225 984;
  • 28) 0.251 981 225 984 × 2 = 0 + 0.503 962 451 968;
  • 29) 0.503 962 451 968 × 2 = 1 + 0.007 924 903 936;
  • 30) 0.007 924 903 936 × 2 = 0 + 0.015 849 807 872;
  • 31) 0.015 849 807 872 × 2 = 0 + 0.031 699 615 744;
  • 32) 0.031 699 615 744 × 2 = 0 + 0.063 399 231 488;
  • 33) 0.063 399 231 488 × 2 = 0 + 0.126 798 462 976;
  • 34) 0.126 798 462 976 × 2 = 0 + 0.253 596 925 952;
  • 35) 0.253 596 925 952 × 2 = 0 + 0.507 193 851 904;
  • 36) 0.507 193 851 904 × 2 = 1 + 0.014 387 703 808;
  • 37) 0.014 387 703 808 × 2 = 0 + 0.028 775 407 616;
  • 38) 0.028 775 407 616 × 2 = 0 + 0.057 550 815 232;
  • 39) 0.057 550 815 232 × 2 = 0 + 0.115 101 630 464;
  • 40) 0.115 101 630 464 × 2 = 0 + 0.230 203 260 928;
  • 41) 0.230 203 260 928 × 2 = 0 + 0.460 406 521 856;
  • 42) 0.460 406 521 856 × 2 = 0 + 0.920 813 043 712;
  • 43) 0.920 813 043 712 × 2 = 1 + 0.841 626 087 424;
  • 44) 0.841 626 087 424 × 2 = 1 + 0.683 252 174 848;
  • 45) 0.683 252 174 848 × 2 = 1 + 0.366 504 349 696;
  • 46) 0.366 504 349 696 × 2 = 0 + 0.733 008 699 392;
  • 47) 0.733 008 699 392 × 2 = 1 + 0.466 017 398 784;
  • 48) 0.466 017 398 784 × 2 = 0 + 0.932 034 797 568;
  • 49) 0.932 034 797 568 × 2 = 1 + 0.864 069 595 136;
  • 50) 0.864 069 595 136 × 2 = 1 + 0.728 139 190 272;
  • 51) 0.728 139 190 272 × 2 = 1 + 0.456 278 380 544;
  • 52) 0.456 278 380 544 × 2 = 0 + 0.912 556 761 088;
  • 53) 0.912 556 761 088 × 2 = 1 + 0.825 113 522 176;
  • 54) 0.825 113 522 176 × 2 = 1 + 0.650 227 044 352;
  • 55) 0.650 227 044 352 × 2 = 1 + 0.300 454 088 704;
  • 56) 0.300 454 088 704 × 2 = 0 + 0.600 908 177 408;
  • 57) 0.600 908 177 408 × 2 = 1 + 0.201 816 354 816;
  • 58) 0.201 816 354 816 × 2 = 0 + 0.403 632 709 632;
  • 59) 0.403 632 709 632 × 2 = 0 + 0.807 265 419 264;
  • 60) 0.807 265 419 264 × 2 = 1 + 0.614 530 838 528;
  • 61) 0.614 530 838 528 × 2 = 1 + 0.229 061 677 056;
  • 62) 0.229 061 677 056 × 2 = 0 + 0.458 123 354 112;
  • 63) 0.458 123 354 112 × 2 = 0 + 0.916 246 708 224;
  • 64) 0.916 246 708 224 × 2 = 1 + 0.832 493 416 448;

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:


0.000 282 006 353(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 0000 0011 1010 1110 1110 1001 1001(2)

6. Positive number before normalization:

0.000 282 006 353(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 0000 0011 1010 1110 1110 1001 1001(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:


0.000 282 006 353(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 0000 0011 1010 1110 1110 1001 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 0000 0011 1010 1110 1110 1001 1001(2) × 20 =

1.0010 0111 1011 0100 1000 0001 0000 0011 1010 1110 1110 1001 1001(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 1000 0001 0000 0011 1010 1110 1110 1001 1001


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 1000 0001 0000 0011 1010 1110 1110 1001 1001 =

0010 0111 1011 0100 1000 0001 0000 0011 1010 1110 1110 1001 1001


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 1000 0001 0000 0011 1010 1110 1110 1001 1001


Decimal number -0.000 282 006 353 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0001 0000 0011 1010 1110 1110 1001 1001

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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