-0.000 282 006 456 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 456(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
-0.000 282 006 456(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:

|-0.000 282 006 456| = 0.000 282 006 456


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: 0.000 282 006 456.

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) 0.000 282 006 456 × 2 = 0 + 0.000 564 012 912;
  • 2) 0.000 564 012 912 × 2 = 0 + 0.001 128 025 824;
  • 3) 0.001 128 025 824 × 2 = 0 + 0.002 256 051 648;
  • 4) 0.002 256 051 648 × 2 = 0 + 0.004 512 103 296;
  • 5) 0.004 512 103 296 × 2 = 0 + 0.009 024 206 592;
  • 6) 0.009 024 206 592 × 2 = 0 + 0.018 048 413 184;
  • 7) 0.018 048 413 184 × 2 = 0 + 0.036 096 826 368;
  • 8) 0.036 096 826 368 × 2 = 0 + 0.072 193 652 736;
  • 9) 0.072 193 652 736 × 2 = 0 + 0.144 387 305 472;
  • 10) 0.144 387 305 472 × 2 = 0 + 0.288 774 610 944;
  • 11) 0.288 774 610 944 × 2 = 0 + 0.577 549 221 888;
  • 12) 0.577 549 221 888 × 2 = 1 + 0.155 098 443 776;
  • 13) 0.155 098 443 776 × 2 = 0 + 0.310 196 887 552;
  • 14) 0.310 196 887 552 × 2 = 0 + 0.620 393 775 104;
  • 15) 0.620 393 775 104 × 2 = 1 + 0.240 787 550 208;
  • 16) 0.240 787 550 208 × 2 = 0 + 0.481 575 100 416;
  • 17) 0.481 575 100 416 × 2 = 0 + 0.963 150 200 832;
  • 18) 0.963 150 200 832 × 2 = 1 + 0.926 300 401 664;
  • 19) 0.926 300 401 664 × 2 = 1 + 0.852 600 803 328;
  • 20) 0.852 600 803 328 × 2 = 1 + 0.705 201 606 656;
  • 21) 0.705 201 606 656 × 2 = 1 + 0.410 403 213 312;
  • 22) 0.410 403 213 312 × 2 = 0 + 0.820 806 426 624;
  • 23) 0.820 806 426 624 × 2 = 1 + 0.641 612 853 248;
  • 24) 0.641 612 853 248 × 2 = 1 + 0.283 225 706 496;
  • 25) 0.283 225 706 496 × 2 = 0 + 0.566 451 412 992;
  • 26) 0.566 451 412 992 × 2 = 1 + 0.132 902 825 984;
  • 27) 0.132 902 825 984 × 2 = 0 + 0.265 805 651 968;
  • 28) 0.265 805 651 968 × 2 = 0 + 0.531 611 303 936;
  • 29) 0.531 611 303 936 × 2 = 1 + 0.063 222 607 872;
  • 30) 0.063 222 607 872 × 2 = 0 + 0.126 445 215 744;
  • 31) 0.126 445 215 744 × 2 = 0 + 0.252 890 431 488;
  • 32) 0.252 890 431 488 × 2 = 0 + 0.505 780 862 976;
  • 33) 0.505 780 862 976 × 2 = 1 + 0.011 561 725 952;
  • 34) 0.011 561 725 952 × 2 = 0 + 0.023 123 451 904;
  • 35) 0.023 123 451 904 × 2 = 0 + 0.046 246 903 808;
  • 36) 0.046 246 903 808 × 2 = 0 + 0.092 493 807 616;
  • 37) 0.092 493 807 616 × 2 = 0 + 0.184 987 615 232;
  • 38) 0.184 987 615 232 × 2 = 0 + 0.369 975 230 464;
  • 39) 0.369 975 230 464 × 2 = 0 + 0.739 950 460 928;
  • 40) 0.739 950 460 928 × 2 = 1 + 0.479 900 921 856;
  • 41) 0.479 900 921 856 × 2 = 0 + 0.959 801 843 712;
  • 42) 0.959 801 843 712 × 2 = 1 + 0.919 603 687 424;
  • 43) 0.919 603 687 424 × 2 = 1 + 0.839 207 374 848;
  • 44) 0.839 207 374 848 × 2 = 1 + 0.678 414 749 696;
  • 45) 0.678 414 749 696 × 2 = 1 + 0.356 829 499 392;
  • 46) 0.356 829 499 392 × 2 = 0 + 0.713 658 998 784;
  • 47) 0.713 658 998 784 × 2 = 1 + 0.427 317 997 568;
  • 48) 0.427 317 997 568 × 2 = 0 + 0.854 635 995 136;
  • 49) 0.854 635 995 136 × 2 = 1 + 0.709 271 990 272;
  • 50) 0.709 271 990 272 × 2 = 1 + 0.418 543 980 544;
  • 51) 0.418 543 980 544 × 2 = 0 + 0.837 087 961 088;
  • 52) 0.837 087 961 088 × 2 = 1 + 0.674 175 922 176;
  • 53) 0.674 175 922 176 × 2 = 1 + 0.348 351 844 352;
  • 54) 0.348 351 844 352 × 2 = 0 + 0.696 703 688 704;
  • 55) 0.696 703 688 704 × 2 = 1 + 0.393 407 377 408;
  • 56) 0.393 407 377 408 × 2 = 0 + 0.786 814 754 816;
  • 57) 0.786 814 754 816 × 2 = 1 + 0.573 629 509 632;
  • 58) 0.573 629 509 632 × 2 = 1 + 0.147 259 019 264;
  • 59) 0.147 259 019 264 × 2 = 0 + 0.294 518 038 528;
  • 60) 0.294 518 038 528 × 2 = 0 + 0.589 036 077 056;
  • 61) 0.589 036 077 056 × 2 = 1 + 0.178 072 154 112;
  • 62) 0.178 072 154 112 × 2 = 0 + 0.356 144 308 224;
  • 63) 0.356 144 308 224 × 2 = 0 + 0.712 288 616 448;
  • 64) 0.712 288 616 448 × 2 = 1 + 0.424 577 232 896;

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:


0.000 282 006 456(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0001 0111 1010 1101 1010 1100 1001(2)

6. Positive number before normalization:

0.000 282 006 456(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0001 0111 1010 1101 1010 1100 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:


0.000 282 006 456(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0001 0111 1010 1101 1010 1100 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0001 0111 1010 1101 1010 1100 1001(2) × 20 =

1.0010 0111 1011 0100 1000 1000 0001 0111 1010 1101 1010 1100 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 1000 0001 0111 1010 1101 1010 1100 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 1000 0001 0111 1010 1101 1010 1100 1001 =

0010 0111 1011 0100 1000 1000 0001 0111 1010 1101 1010 1100 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 1000 0001 0111 1010 1101 1010 1100 1001


Decimal number -0.000 282 006 456 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 1000 0001 0111 1010 1101 1010 1100 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