-0.000 282 612 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 612(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
-0.000 282 612(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:

|-0.000 282 612| = 0.000 282 612


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: 0.000 282 612.

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) 0.000 282 612 × 2 = 0 + 0.000 565 224;
  • 2) 0.000 565 224 × 2 = 0 + 0.001 130 448;
  • 3) 0.001 130 448 × 2 = 0 + 0.002 260 896;
  • 4) 0.002 260 896 × 2 = 0 + 0.004 521 792;
  • 5) 0.004 521 792 × 2 = 0 + 0.009 043 584;
  • 6) 0.009 043 584 × 2 = 0 + 0.018 087 168;
  • 7) 0.018 087 168 × 2 = 0 + 0.036 174 336;
  • 8) 0.036 174 336 × 2 = 0 + 0.072 348 672;
  • 9) 0.072 348 672 × 2 = 0 + 0.144 697 344;
  • 10) 0.144 697 344 × 2 = 0 + 0.289 394 688;
  • 11) 0.289 394 688 × 2 = 0 + 0.578 789 376;
  • 12) 0.578 789 376 × 2 = 1 + 0.157 578 752;
  • 13) 0.157 578 752 × 2 = 0 + 0.315 157 504;
  • 14) 0.315 157 504 × 2 = 0 + 0.630 315 008;
  • 15) 0.630 315 008 × 2 = 1 + 0.260 630 016;
  • 16) 0.260 630 016 × 2 = 0 + 0.521 260 032;
  • 17) 0.521 260 032 × 2 = 1 + 0.042 520 064;
  • 18) 0.042 520 064 × 2 = 0 + 0.085 040 128;
  • 19) 0.085 040 128 × 2 = 0 + 0.170 080 256;
  • 20) 0.170 080 256 × 2 = 0 + 0.340 160 512;
  • 21) 0.340 160 512 × 2 = 0 + 0.680 321 024;
  • 22) 0.680 321 024 × 2 = 1 + 0.360 642 048;
  • 23) 0.360 642 048 × 2 = 0 + 0.721 284 096;
  • 24) 0.721 284 096 × 2 = 1 + 0.442 568 192;
  • 25) 0.442 568 192 × 2 = 0 + 0.885 136 384;
  • 26) 0.885 136 384 × 2 = 1 + 0.770 272 768;
  • 27) 0.770 272 768 × 2 = 1 + 0.540 545 536;
  • 28) 0.540 545 536 × 2 = 1 + 0.081 091 072;
  • 29) 0.081 091 072 × 2 = 0 + 0.162 182 144;
  • 30) 0.162 182 144 × 2 = 0 + 0.324 364 288;
  • 31) 0.324 364 288 × 2 = 0 + 0.648 728 576;
  • 32) 0.648 728 576 × 2 = 1 + 0.297 457 152;
  • 33) 0.297 457 152 × 2 = 0 + 0.594 914 304;
  • 34) 0.594 914 304 × 2 = 1 + 0.189 828 608;
  • 35) 0.189 828 608 × 2 = 0 + 0.379 657 216;
  • 36) 0.379 657 216 × 2 = 0 + 0.759 314 432;
  • 37) 0.759 314 432 × 2 = 1 + 0.518 628 864;
  • 38) 0.518 628 864 × 2 = 1 + 0.037 257 728;
  • 39) 0.037 257 728 × 2 = 0 + 0.074 515 456;
  • 40) 0.074 515 456 × 2 = 0 + 0.149 030 912;
  • 41) 0.149 030 912 × 2 = 0 + 0.298 061 824;
  • 42) 0.298 061 824 × 2 = 0 + 0.596 123 648;
  • 43) 0.596 123 648 × 2 = 1 + 0.192 247 296;
  • 44) 0.192 247 296 × 2 = 0 + 0.384 494 592;
  • 45) 0.384 494 592 × 2 = 0 + 0.768 989 184;
  • 46) 0.768 989 184 × 2 = 1 + 0.537 978 368;
  • 47) 0.537 978 368 × 2 = 1 + 0.075 956 736;
  • 48) 0.075 956 736 × 2 = 0 + 0.151 913 472;
  • 49) 0.151 913 472 × 2 = 0 + 0.303 826 944;
  • 50) 0.303 826 944 × 2 = 0 + 0.607 653 888;
  • 51) 0.607 653 888 × 2 = 1 + 0.215 307 776;
  • 52) 0.215 307 776 × 2 = 0 + 0.430 615 552;
  • 53) 0.430 615 552 × 2 = 0 + 0.861 231 104;
  • 54) 0.861 231 104 × 2 = 1 + 0.722 462 208;
  • 55) 0.722 462 208 × 2 = 1 + 0.444 924 416;
  • 56) 0.444 924 416 × 2 = 0 + 0.889 848 832;
  • 57) 0.889 848 832 × 2 = 1 + 0.779 697 664;
  • 58) 0.779 697 664 × 2 = 1 + 0.559 395 328;
  • 59) 0.559 395 328 × 2 = 1 + 0.118 790 656;
  • 60) 0.118 790 656 × 2 = 0 + 0.237 581 312;
  • 61) 0.237 581 312 × 2 = 0 + 0.475 162 624;
  • 62) 0.475 162 624 × 2 = 0 + 0.950 325 248;
  • 63) 0.950 325 248 × 2 = 1 + 0.900 650 496;
  • 64) 0.900 650 496 × 2 = 1 + 0.801 300 992;

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:


0.000 282 612(10) =

0.0000 0000 0001 0010 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011(2)

6. Positive number before normalization:

0.000 282 612(10) =

0.0000 0000 0001 0010 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011(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:


0.000 282 612(10) =

0.0000 0000 0001 0010 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011(2) =

0.0000 0000 0001 0010 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011(2) × 20 =

1.0010 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011(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 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011


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 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011 =

0010 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011


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 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011


Decimal number -0.000 282 612 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0101 0111 0001 0100 1100 0010 0110 0010 0110 1110 0011

Share

» Convert decimal -0.000 282 608 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 minutes

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