-0.000 282 005 901 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 901(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
-0.000 282 005 901(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:

|-0.000 282 005 901| = 0.000 282 005 901


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: 0.000 282 005 901.

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) 0.000 282 005 901 × 2 = 0 + 0.000 564 011 802;
  • 2) 0.000 564 011 802 × 2 = 0 + 0.001 128 023 604;
  • 3) 0.001 128 023 604 × 2 = 0 + 0.002 256 047 208;
  • 4) 0.002 256 047 208 × 2 = 0 + 0.004 512 094 416;
  • 5) 0.004 512 094 416 × 2 = 0 + 0.009 024 188 832;
  • 6) 0.009 024 188 832 × 2 = 0 + 0.018 048 377 664;
  • 7) 0.018 048 377 664 × 2 = 0 + 0.036 096 755 328;
  • 8) 0.036 096 755 328 × 2 = 0 + 0.072 193 510 656;
  • 9) 0.072 193 510 656 × 2 = 0 + 0.144 387 021 312;
  • 10) 0.144 387 021 312 × 2 = 0 + 0.288 774 042 624;
  • 11) 0.288 774 042 624 × 2 = 0 + 0.577 548 085 248;
  • 12) 0.577 548 085 248 × 2 = 1 + 0.155 096 170 496;
  • 13) 0.155 096 170 496 × 2 = 0 + 0.310 192 340 992;
  • 14) 0.310 192 340 992 × 2 = 0 + 0.620 384 681 984;
  • 15) 0.620 384 681 984 × 2 = 1 + 0.240 769 363 968;
  • 16) 0.240 769 363 968 × 2 = 0 + 0.481 538 727 936;
  • 17) 0.481 538 727 936 × 2 = 0 + 0.963 077 455 872;
  • 18) 0.963 077 455 872 × 2 = 1 + 0.926 154 911 744;
  • 19) 0.926 154 911 744 × 2 = 1 + 0.852 309 823 488;
  • 20) 0.852 309 823 488 × 2 = 1 + 0.704 619 646 976;
  • 21) 0.704 619 646 976 × 2 = 1 + 0.409 239 293 952;
  • 22) 0.409 239 293 952 × 2 = 0 + 0.818 478 587 904;
  • 23) 0.818 478 587 904 × 2 = 1 + 0.636 957 175 808;
  • 24) 0.636 957 175 808 × 2 = 1 + 0.273 914 351 616;
  • 25) 0.273 914 351 616 × 2 = 0 + 0.547 828 703 232;
  • 26) 0.547 828 703 232 × 2 = 1 + 0.095 657 406 464;
  • 27) 0.095 657 406 464 × 2 = 0 + 0.191 314 812 928;
  • 28) 0.191 314 812 928 × 2 = 0 + 0.382 629 625 856;
  • 29) 0.382 629 625 856 × 2 = 0 + 0.765 259 251 712;
  • 30) 0.765 259 251 712 × 2 = 1 + 0.530 518 503 424;
  • 31) 0.530 518 503 424 × 2 = 1 + 0.061 037 006 848;
  • 32) 0.061 037 006 848 × 2 = 0 + 0.122 074 013 696;
  • 33) 0.122 074 013 696 × 2 = 0 + 0.244 148 027 392;
  • 34) 0.244 148 027 392 × 2 = 0 + 0.488 296 054 784;
  • 35) 0.488 296 054 784 × 2 = 0 + 0.976 592 109 568;
  • 36) 0.976 592 109 568 × 2 = 1 + 0.953 184 219 136;
  • 37) 0.953 184 219 136 × 2 = 1 + 0.906 368 438 272;
  • 38) 0.906 368 438 272 × 2 = 1 + 0.812 736 876 544;
  • 39) 0.812 736 876 544 × 2 = 1 + 0.625 473 753 088;
  • 40) 0.625 473 753 088 × 2 = 1 + 0.250 947 506 176;
  • 41) 0.250 947 506 176 × 2 = 0 + 0.501 895 012 352;
  • 42) 0.501 895 012 352 × 2 = 1 + 0.003 790 024 704;
  • 43) 0.003 790 024 704 × 2 = 0 + 0.007 580 049 408;
  • 44) 0.007 580 049 408 × 2 = 0 + 0.015 160 098 816;
  • 45) 0.015 160 098 816 × 2 = 0 + 0.030 320 197 632;
  • 46) 0.030 320 197 632 × 2 = 0 + 0.060 640 395 264;
  • 47) 0.060 640 395 264 × 2 = 0 + 0.121 280 790 528;
  • 48) 0.121 280 790 528 × 2 = 0 + 0.242 561 581 056;
  • 49) 0.242 561 581 056 × 2 = 0 + 0.485 123 162 112;
  • 50) 0.485 123 162 112 × 2 = 0 + 0.970 246 324 224;
  • 51) 0.970 246 324 224 × 2 = 1 + 0.940 492 648 448;
  • 52) 0.940 492 648 448 × 2 = 1 + 0.880 985 296 896;
  • 53) 0.880 985 296 896 × 2 = 1 + 0.761 970 593 792;
  • 54) 0.761 970 593 792 × 2 = 1 + 0.523 941 187 584;
  • 55) 0.523 941 187 584 × 2 = 1 + 0.047 882 375 168;
  • 56) 0.047 882 375 168 × 2 = 0 + 0.095 764 750 336;
  • 57) 0.095 764 750 336 × 2 = 0 + 0.191 529 500 672;
  • 58) 0.191 529 500 672 × 2 = 0 + 0.383 059 001 344;
  • 59) 0.383 059 001 344 × 2 = 0 + 0.766 118 002 688;
  • 60) 0.766 118 002 688 × 2 = 1 + 0.532 236 005 376;
  • 61) 0.532 236 005 376 × 2 = 1 + 0.064 472 010 752;
  • 62) 0.064 472 010 752 × 2 = 0 + 0.128 944 021 504;
  • 63) 0.128 944 021 504 × 2 = 0 + 0.257 888 043 008;
  • 64) 0.257 888 043 008 × 2 = 0 + 0.515 776 086 016;

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:


0.000 282 005 901(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1111 0100 0000 0011 1110 0001 1000(2)

6. Positive number before normalization:

0.000 282 005 901(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1111 0100 0000 0011 1110 0001 1000(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:


0.000 282 005 901(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1111 0100 0000 0011 1110 0001 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1111 0100 0000 0011 1110 0001 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1111 0100 0000 0011 1110 0001 1000(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 0001 1111 0100 0000 0011 1110 0001 1000


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 0001 1111 0100 0000 0011 1110 0001 1000 =

0010 0111 1011 0100 0110 0001 1111 0100 0000 0011 1110 0001 1000


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 0001 1111 0100 0000 0011 1110 0001 1000


Decimal number -0.000 282 005 901 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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