-0.000 282 006 079 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 079(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
-0.000 282 006 079(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:

|-0.000 282 006 079| = 0.000 282 006 079


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: 0.000 282 006 079.

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) 0.000 282 006 079 × 2 = 0 + 0.000 564 012 158;
  • 2) 0.000 564 012 158 × 2 = 0 + 0.001 128 024 316;
  • 3) 0.001 128 024 316 × 2 = 0 + 0.002 256 048 632;
  • 4) 0.002 256 048 632 × 2 = 0 + 0.004 512 097 264;
  • 5) 0.004 512 097 264 × 2 = 0 + 0.009 024 194 528;
  • 6) 0.009 024 194 528 × 2 = 0 + 0.018 048 389 056;
  • 7) 0.018 048 389 056 × 2 = 0 + 0.036 096 778 112;
  • 8) 0.036 096 778 112 × 2 = 0 + 0.072 193 556 224;
  • 9) 0.072 193 556 224 × 2 = 0 + 0.144 387 112 448;
  • 10) 0.144 387 112 448 × 2 = 0 + 0.288 774 224 896;
  • 11) 0.288 774 224 896 × 2 = 0 + 0.577 548 449 792;
  • 12) 0.577 548 449 792 × 2 = 1 + 0.155 096 899 584;
  • 13) 0.155 096 899 584 × 2 = 0 + 0.310 193 799 168;
  • 14) 0.310 193 799 168 × 2 = 0 + 0.620 387 598 336;
  • 15) 0.620 387 598 336 × 2 = 1 + 0.240 775 196 672;
  • 16) 0.240 775 196 672 × 2 = 0 + 0.481 550 393 344;
  • 17) 0.481 550 393 344 × 2 = 0 + 0.963 100 786 688;
  • 18) 0.963 100 786 688 × 2 = 1 + 0.926 201 573 376;
  • 19) 0.926 201 573 376 × 2 = 1 + 0.852 403 146 752;
  • 20) 0.852 403 146 752 × 2 = 1 + 0.704 806 293 504;
  • 21) 0.704 806 293 504 × 2 = 1 + 0.409 612 587 008;
  • 22) 0.409 612 587 008 × 2 = 0 + 0.819 225 174 016;
  • 23) 0.819 225 174 016 × 2 = 1 + 0.638 450 348 032;
  • 24) 0.638 450 348 032 × 2 = 1 + 0.276 900 696 064;
  • 25) 0.276 900 696 064 × 2 = 0 + 0.553 801 392 128;
  • 26) 0.553 801 392 128 × 2 = 1 + 0.107 602 784 256;
  • 27) 0.107 602 784 256 × 2 = 0 + 0.215 205 568 512;
  • 28) 0.215 205 568 512 × 2 = 0 + 0.430 411 137 024;
  • 29) 0.430 411 137 024 × 2 = 0 + 0.860 822 274 048;
  • 30) 0.860 822 274 048 × 2 = 1 + 0.721 644 548 096;
  • 31) 0.721 644 548 096 × 2 = 1 + 0.443 289 096 192;
  • 32) 0.443 289 096 192 × 2 = 0 + 0.886 578 192 384;
  • 33) 0.886 578 192 384 × 2 = 1 + 0.773 156 384 768;
  • 34) 0.773 156 384 768 × 2 = 1 + 0.546 312 769 536;
  • 35) 0.546 312 769 536 × 2 = 1 + 0.092 625 539 072;
  • 36) 0.092 625 539 072 × 2 = 0 + 0.185 251 078 144;
  • 37) 0.185 251 078 144 × 2 = 0 + 0.370 502 156 288;
  • 38) 0.370 502 156 288 × 2 = 0 + 0.741 004 312 576;
  • 39) 0.741 004 312 576 × 2 = 1 + 0.482 008 625 152;
  • 40) 0.482 008 625 152 × 2 = 0 + 0.964 017 250 304;
  • 41) 0.964 017 250 304 × 2 = 1 + 0.928 034 500 608;
  • 42) 0.928 034 500 608 × 2 = 1 + 0.856 069 001 216;
  • 43) 0.856 069 001 216 × 2 = 1 + 0.712 138 002 432;
  • 44) 0.712 138 002 432 × 2 = 1 + 0.424 276 004 864;
  • 45) 0.424 276 004 864 × 2 = 0 + 0.848 552 009 728;
  • 46) 0.848 552 009 728 × 2 = 1 + 0.697 104 019 456;
  • 47) 0.697 104 019 456 × 2 = 1 + 0.394 208 038 912;
  • 48) 0.394 208 038 912 × 2 = 0 + 0.788 416 077 824;
  • 49) 0.788 416 077 824 × 2 = 1 + 0.576 832 155 648;
  • 50) 0.576 832 155 648 × 2 = 1 + 0.153 664 311 296;
  • 51) 0.153 664 311 296 × 2 = 0 + 0.307 328 622 592;
  • 52) 0.307 328 622 592 × 2 = 0 + 0.614 657 245 184;
  • 53) 0.614 657 245 184 × 2 = 1 + 0.229 314 490 368;
  • 54) 0.229 314 490 368 × 2 = 0 + 0.458 628 980 736;
  • 55) 0.458 628 980 736 × 2 = 0 + 0.917 257 961 472;
  • 56) 0.917 257 961 472 × 2 = 1 + 0.834 515 922 944;
  • 57) 0.834 515 922 944 × 2 = 1 + 0.669 031 845 888;
  • 58) 0.669 031 845 888 × 2 = 1 + 0.338 063 691 776;
  • 59) 0.338 063 691 776 × 2 = 0 + 0.676 127 383 552;
  • 60) 0.676 127 383 552 × 2 = 1 + 0.352 254 767 104;
  • 61) 0.352 254 767 104 × 2 = 0 + 0.704 509 534 208;
  • 62) 0.704 509 534 208 × 2 = 1 + 0.409 019 068 416;
  • 63) 0.409 019 068 416 × 2 = 0 + 0.818 038 136 832;
  • 64) 0.818 038 136 832 × 2 = 1 + 0.636 076 273 664;

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:


0.000 282 006 079(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1110 0010 1111 0110 1100 1001 1101 0101(2)

6. Positive number before normalization:

0.000 282 006 079(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1110 0010 1111 0110 1100 1001 1101 0101(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:


0.000 282 006 079(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1110 0010 1111 0110 1100 1001 1101 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1110 0010 1111 0110 1100 1001 1101 0101(2) × 20 =

1.0010 0111 1011 0100 0110 1110 0010 1111 0110 1100 1001 1101 0101(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 1110 0010 1111 0110 1100 1001 1101 0101


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 1110 0010 1111 0110 1100 1001 1101 0101 =

0010 0111 1011 0100 0110 1110 0010 1111 0110 1100 1001 1101 0101


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 1110 0010 1111 0110 1100 1001 1101 0101


Decimal number -0.000 282 006 079 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1110 0010 1111 0110 1100 1001 1101 0101

Share

» Convert decimal -0.000 282 006 091 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