-0.000 282 006 283 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 283(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
-0.000 282 006 283(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:

|-0.000 282 006 283| = 0.000 282 006 283


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: 0.000 282 006 283.

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) 0.000 282 006 283 × 2 = 0 + 0.000 564 012 566;
  • 2) 0.000 564 012 566 × 2 = 0 + 0.001 128 025 132;
  • 3) 0.001 128 025 132 × 2 = 0 + 0.002 256 050 264;
  • 4) 0.002 256 050 264 × 2 = 0 + 0.004 512 100 528;
  • 5) 0.004 512 100 528 × 2 = 0 + 0.009 024 201 056;
  • 6) 0.009 024 201 056 × 2 = 0 + 0.018 048 402 112;
  • 7) 0.018 048 402 112 × 2 = 0 + 0.036 096 804 224;
  • 8) 0.036 096 804 224 × 2 = 0 + 0.072 193 608 448;
  • 9) 0.072 193 608 448 × 2 = 0 + 0.144 387 216 896;
  • 10) 0.144 387 216 896 × 2 = 0 + 0.288 774 433 792;
  • 11) 0.288 774 433 792 × 2 = 0 + 0.577 548 867 584;
  • 12) 0.577 548 867 584 × 2 = 1 + 0.155 097 735 168;
  • 13) 0.155 097 735 168 × 2 = 0 + 0.310 195 470 336;
  • 14) 0.310 195 470 336 × 2 = 0 + 0.620 390 940 672;
  • 15) 0.620 390 940 672 × 2 = 1 + 0.240 781 881 344;
  • 16) 0.240 781 881 344 × 2 = 0 + 0.481 563 762 688;
  • 17) 0.481 563 762 688 × 2 = 0 + 0.963 127 525 376;
  • 18) 0.963 127 525 376 × 2 = 1 + 0.926 255 050 752;
  • 19) 0.926 255 050 752 × 2 = 1 + 0.852 510 101 504;
  • 20) 0.852 510 101 504 × 2 = 1 + 0.705 020 203 008;
  • 21) 0.705 020 203 008 × 2 = 1 + 0.410 040 406 016;
  • 22) 0.410 040 406 016 × 2 = 0 + 0.820 080 812 032;
  • 23) 0.820 080 812 032 × 2 = 1 + 0.640 161 624 064;
  • 24) 0.640 161 624 064 × 2 = 1 + 0.280 323 248 128;
  • 25) 0.280 323 248 128 × 2 = 0 + 0.560 646 496 256;
  • 26) 0.560 646 496 256 × 2 = 1 + 0.121 292 992 512;
  • 27) 0.121 292 992 512 × 2 = 0 + 0.242 585 985 024;
  • 28) 0.242 585 985 024 × 2 = 0 + 0.485 171 970 048;
  • 29) 0.485 171 970 048 × 2 = 0 + 0.970 343 940 096;
  • 30) 0.970 343 940 096 × 2 = 1 + 0.940 687 880 192;
  • 31) 0.940 687 880 192 × 2 = 1 + 0.881 375 760 384;
  • 32) 0.881 375 760 384 × 2 = 1 + 0.762 751 520 768;
  • 33) 0.762 751 520 768 × 2 = 1 + 0.525 503 041 536;
  • 34) 0.525 503 041 536 × 2 = 1 + 0.051 006 083 072;
  • 35) 0.051 006 083 072 × 2 = 0 + 0.102 012 166 144;
  • 36) 0.102 012 166 144 × 2 = 0 + 0.204 024 332 288;
  • 37) 0.204 024 332 288 × 2 = 0 + 0.408 048 664 576;
  • 38) 0.408 048 664 576 × 2 = 0 + 0.816 097 329 152;
  • 39) 0.816 097 329 152 × 2 = 1 + 0.632 194 658 304;
  • 40) 0.632 194 658 304 × 2 = 1 + 0.264 389 316 608;
  • 41) 0.264 389 316 608 × 2 = 0 + 0.528 778 633 216;
  • 42) 0.528 778 633 216 × 2 = 1 + 0.057 557 266 432;
  • 43) 0.057 557 266 432 × 2 = 0 + 0.115 114 532 864;
  • 44) 0.115 114 532 864 × 2 = 0 + 0.230 229 065 728;
  • 45) 0.230 229 065 728 × 2 = 0 + 0.460 458 131 456;
  • 46) 0.460 458 131 456 × 2 = 0 + 0.920 916 262 912;
  • 47) 0.920 916 262 912 × 2 = 1 + 0.841 832 525 824;
  • 48) 0.841 832 525 824 × 2 = 1 + 0.683 665 051 648;
  • 49) 0.683 665 051 648 × 2 = 1 + 0.367 330 103 296;
  • 50) 0.367 330 103 296 × 2 = 0 + 0.734 660 206 592;
  • 51) 0.734 660 206 592 × 2 = 1 + 0.469 320 413 184;
  • 52) 0.469 320 413 184 × 2 = 0 + 0.938 640 826 368;
  • 53) 0.938 640 826 368 × 2 = 1 + 0.877 281 652 736;
  • 54) 0.877 281 652 736 × 2 = 1 + 0.754 563 305 472;
  • 55) 0.754 563 305 472 × 2 = 1 + 0.509 126 610 944;
  • 56) 0.509 126 610 944 × 2 = 1 + 0.018 253 221 888;
  • 57) 0.018 253 221 888 × 2 = 0 + 0.036 506 443 776;
  • 58) 0.036 506 443 776 × 2 = 0 + 0.073 012 887 552;
  • 59) 0.073 012 887 552 × 2 = 0 + 0.146 025 775 104;
  • 60) 0.146 025 775 104 × 2 = 0 + 0.292 051 550 208;
  • 61) 0.292 051 550 208 × 2 = 0 + 0.584 103 100 416;
  • 62) 0.584 103 100 416 × 2 = 1 + 0.168 206 200 832;
  • 63) 0.168 206 200 832 × 2 = 0 + 0.336 412 401 664;
  • 64) 0.336 412 401 664 × 2 = 0 + 0.672 824 803 328;

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:


0.000 282 006 283(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 0011 0100 0011 1010 1111 0000 0100(2)

6. Positive number before normalization:

0.000 282 006 283(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 0011 0100 0011 1010 1111 0000 0100(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:


0.000 282 006 283(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 0011 0100 0011 1010 1111 0000 0100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 0011 0100 0011 1010 1111 0000 0100(2) × 20 =

1.0010 0111 1011 0100 0111 1100 0011 0100 0011 1010 1111 0000 0100(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 0111 1100 0011 0100 0011 1010 1111 0000 0100


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 0111 1100 0011 0100 0011 1010 1111 0000 0100 =

0010 0111 1011 0100 0111 1100 0011 0100 0011 1010 1111 0000 0100


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 0111 1100 0011 0100 0011 1010 1111 0000 0100


Decimal number -0.000 282 006 283 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1100 0011 0100 0011 1010 1111 0000 0100

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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