-0.000 282 006 464 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 464(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
-0.000 282 006 464(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:

|-0.000 282 006 464| = 0.000 282 006 464


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: 0.000 282 006 464.

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) 0.000 282 006 464 × 2 = 0 + 0.000 564 012 928;
  • 2) 0.000 564 012 928 × 2 = 0 + 0.001 128 025 856;
  • 3) 0.001 128 025 856 × 2 = 0 + 0.002 256 051 712;
  • 4) 0.002 256 051 712 × 2 = 0 + 0.004 512 103 424;
  • 5) 0.004 512 103 424 × 2 = 0 + 0.009 024 206 848;
  • 6) 0.009 024 206 848 × 2 = 0 + 0.018 048 413 696;
  • 7) 0.018 048 413 696 × 2 = 0 + 0.036 096 827 392;
  • 8) 0.036 096 827 392 × 2 = 0 + 0.072 193 654 784;
  • 9) 0.072 193 654 784 × 2 = 0 + 0.144 387 309 568;
  • 10) 0.144 387 309 568 × 2 = 0 + 0.288 774 619 136;
  • 11) 0.288 774 619 136 × 2 = 0 + 0.577 549 238 272;
  • 12) 0.577 549 238 272 × 2 = 1 + 0.155 098 476 544;
  • 13) 0.155 098 476 544 × 2 = 0 + 0.310 196 953 088;
  • 14) 0.310 196 953 088 × 2 = 0 + 0.620 393 906 176;
  • 15) 0.620 393 906 176 × 2 = 1 + 0.240 787 812 352;
  • 16) 0.240 787 812 352 × 2 = 0 + 0.481 575 624 704;
  • 17) 0.481 575 624 704 × 2 = 0 + 0.963 151 249 408;
  • 18) 0.963 151 249 408 × 2 = 1 + 0.926 302 498 816;
  • 19) 0.926 302 498 816 × 2 = 1 + 0.852 604 997 632;
  • 20) 0.852 604 997 632 × 2 = 1 + 0.705 209 995 264;
  • 21) 0.705 209 995 264 × 2 = 1 + 0.410 419 990 528;
  • 22) 0.410 419 990 528 × 2 = 0 + 0.820 839 981 056;
  • 23) 0.820 839 981 056 × 2 = 1 + 0.641 679 962 112;
  • 24) 0.641 679 962 112 × 2 = 1 + 0.283 359 924 224;
  • 25) 0.283 359 924 224 × 2 = 0 + 0.566 719 848 448;
  • 26) 0.566 719 848 448 × 2 = 1 + 0.133 439 696 896;
  • 27) 0.133 439 696 896 × 2 = 0 + 0.266 879 393 792;
  • 28) 0.266 879 393 792 × 2 = 0 + 0.533 758 787 584;
  • 29) 0.533 758 787 584 × 2 = 1 + 0.067 517 575 168;
  • 30) 0.067 517 575 168 × 2 = 0 + 0.135 035 150 336;
  • 31) 0.135 035 150 336 × 2 = 0 + 0.270 070 300 672;
  • 32) 0.270 070 300 672 × 2 = 0 + 0.540 140 601 344;
  • 33) 0.540 140 601 344 × 2 = 1 + 0.080 281 202 688;
  • 34) 0.080 281 202 688 × 2 = 0 + 0.160 562 405 376;
  • 35) 0.160 562 405 376 × 2 = 0 + 0.321 124 810 752;
  • 36) 0.321 124 810 752 × 2 = 0 + 0.642 249 621 504;
  • 37) 0.642 249 621 504 × 2 = 1 + 0.284 499 243 008;
  • 38) 0.284 499 243 008 × 2 = 0 + 0.568 998 486 016;
  • 39) 0.568 998 486 016 × 2 = 1 + 0.137 996 972 032;
  • 40) 0.137 996 972 032 × 2 = 0 + 0.275 993 944 064;
  • 41) 0.275 993 944 064 × 2 = 0 + 0.551 987 888 128;
  • 42) 0.551 987 888 128 × 2 = 1 + 0.103 975 776 256;
  • 43) 0.103 975 776 256 × 2 = 0 + 0.207 951 552 512;
  • 44) 0.207 951 552 512 × 2 = 0 + 0.415 903 105 024;
  • 45) 0.415 903 105 024 × 2 = 0 + 0.831 806 210 048;
  • 46) 0.831 806 210 048 × 2 = 1 + 0.663 612 420 096;
  • 47) 0.663 612 420 096 × 2 = 1 + 0.327 224 840 192;
  • 48) 0.327 224 840 192 × 2 = 0 + 0.654 449 680 384;
  • 49) 0.654 449 680 384 × 2 = 1 + 0.308 899 360 768;
  • 50) 0.308 899 360 768 × 2 = 0 + 0.617 798 721 536;
  • 51) 0.617 798 721 536 × 2 = 1 + 0.235 597 443 072;
  • 52) 0.235 597 443 072 × 2 = 0 + 0.471 194 886 144;
  • 53) 0.471 194 886 144 × 2 = 0 + 0.942 389 772 288;
  • 54) 0.942 389 772 288 × 2 = 1 + 0.884 779 544 576;
  • 55) 0.884 779 544 576 × 2 = 1 + 0.769 559 089 152;
  • 56) 0.769 559 089 152 × 2 = 1 + 0.539 118 178 304;
  • 57) 0.539 118 178 304 × 2 = 1 + 0.078 236 356 608;
  • 58) 0.078 236 356 608 × 2 = 0 + 0.156 472 713 216;
  • 59) 0.156 472 713 216 × 2 = 0 + 0.312 945 426 432;
  • 60) 0.312 945 426 432 × 2 = 0 + 0.625 890 852 864;
  • 61) 0.625 890 852 864 × 2 = 1 + 0.251 781 705 728;
  • 62) 0.251 781 705 728 × 2 = 0 + 0.503 563 411 456;
  • 63) 0.503 563 411 456 × 2 = 1 + 0.007 126 822 912;
  • 64) 0.007 126 822 912 × 2 = 0 + 0.014 253 645 824;

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:


0.000 282 006 464(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 1010 0100 0110 1010 0111 1000 1010(2)

6. Positive number before normalization:

0.000 282 006 464(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 1010 0100 0110 1010 0111 1000 1010(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:


0.000 282 006 464(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 1010 0100 0110 1010 0111 1000 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 1010 0100 0110 1010 0111 1000 1010(2) × 20 =

1.0010 0111 1011 0100 1000 1000 1010 0100 0110 1010 0111 1000 1010(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 1010 0100 0110 1010 0111 1000 1010


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 1010 0100 0110 1010 0111 1000 1010 =

0010 0111 1011 0100 1000 1000 1010 0100 0110 1010 0111 1000 1010


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 1010 0100 0110 1010 0111 1000 1010


Decimal number -0.000 282 006 464 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 1000 1010 0100 0110 1010 0111 1000 1010

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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