-0.000 282 006 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 59(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 59(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 59| = 0.000 282 006 59


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 59.

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 59 × 2 = 0 + 0.000 564 013 18;
  • 2) 0.000 564 013 18 × 2 = 0 + 0.001 128 026 36;
  • 3) 0.001 128 026 36 × 2 = 0 + 0.002 256 052 72;
  • 4) 0.002 256 052 72 × 2 = 0 + 0.004 512 105 44;
  • 5) 0.004 512 105 44 × 2 = 0 + 0.009 024 210 88;
  • 6) 0.009 024 210 88 × 2 = 0 + 0.018 048 421 76;
  • 7) 0.018 048 421 76 × 2 = 0 + 0.036 096 843 52;
  • 8) 0.036 096 843 52 × 2 = 0 + 0.072 193 687 04;
  • 9) 0.072 193 687 04 × 2 = 0 + 0.144 387 374 08;
  • 10) 0.144 387 374 08 × 2 = 0 + 0.288 774 748 16;
  • 11) 0.288 774 748 16 × 2 = 0 + 0.577 549 496 32;
  • 12) 0.577 549 496 32 × 2 = 1 + 0.155 098 992 64;
  • 13) 0.155 098 992 64 × 2 = 0 + 0.310 197 985 28;
  • 14) 0.310 197 985 28 × 2 = 0 + 0.620 395 970 56;
  • 15) 0.620 395 970 56 × 2 = 1 + 0.240 791 941 12;
  • 16) 0.240 791 941 12 × 2 = 0 + 0.481 583 882 24;
  • 17) 0.481 583 882 24 × 2 = 0 + 0.963 167 764 48;
  • 18) 0.963 167 764 48 × 2 = 1 + 0.926 335 528 96;
  • 19) 0.926 335 528 96 × 2 = 1 + 0.852 671 057 92;
  • 20) 0.852 671 057 92 × 2 = 1 + 0.705 342 115 84;
  • 21) 0.705 342 115 84 × 2 = 1 + 0.410 684 231 68;
  • 22) 0.410 684 231 68 × 2 = 0 + 0.821 368 463 36;
  • 23) 0.821 368 463 36 × 2 = 1 + 0.642 736 926 72;
  • 24) 0.642 736 926 72 × 2 = 1 + 0.285 473 853 44;
  • 25) 0.285 473 853 44 × 2 = 0 + 0.570 947 706 88;
  • 26) 0.570 947 706 88 × 2 = 1 + 0.141 895 413 76;
  • 27) 0.141 895 413 76 × 2 = 0 + 0.283 790 827 52;
  • 28) 0.283 790 827 52 × 2 = 0 + 0.567 581 655 04;
  • 29) 0.567 581 655 04 × 2 = 1 + 0.135 163 310 08;
  • 30) 0.135 163 310 08 × 2 = 0 + 0.270 326 620 16;
  • 31) 0.270 326 620 16 × 2 = 0 + 0.540 653 240 32;
  • 32) 0.540 653 240 32 × 2 = 1 + 0.081 306 480 64;
  • 33) 0.081 306 480 64 × 2 = 0 + 0.162 612 961 28;
  • 34) 0.162 612 961 28 × 2 = 0 + 0.325 225 922 56;
  • 35) 0.325 225 922 56 × 2 = 0 + 0.650 451 845 12;
  • 36) 0.650 451 845 12 × 2 = 1 + 0.300 903 690 24;
  • 37) 0.300 903 690 24 × 2 = 0 + 0.601 807 380 48;
  • 38) 0.601 807 380 48 × 2 = 1 + 0.203 614 760 96;
  • 39) 0.203 614 760 96 × 2 = 0 + 0.407 229 521 92;
  • 40) 0.407 229 521 92 × 2 = 0 + 0.814 459 043 84;
  • 41) 0.814 459 043 84 × 2 = 1 + 0.628 918 087 68;
  • 42) 0.628 918 087 68 × 2 = 1 + 0.257 836 175 36;
  • 43) 0.257 836 175 36 × 2 = 0 + 0.515 672 350 72;
  • 44) 0.515 672 350 72 × 2 = 1 + 0.031 344 701 44;
  • 45) 0.031 344 701 44 × 2 = 0 + 0.062 689 402 88;
  • 46) 0.062 689 402 88 × 2 = 0 + 0.125 378 805 76;
  • 47) 0.125 378 805 76 × 2 = 0 + 0.250 757 611 52;
  • 48) 0.250 757 611 52 × 2 = 0 + 0.501 515 223 04;
  • 49) 0.501 515 223 04 × 2 = 1 + 0.003 030 446 08;
  • 50) 0.003 030 446 08 × 2 = 0 + 0.006 060 892 16;
  • 51) 0.006 060 892 16 × 2 = 0 + 0.012 121 784 32;
  • 52) 0.012 121 784 32 × 2 = 0 + 0.024 243 568 64;
  • 53) 0.024 243 568 64 × 2 = 0 + 0.048 487 137 28;
  • 54) 0.048 487 137 28 × 2 = 0 + 0.096 974 274 56;
  • 55) 0.096 974 274 56 × 2 = 0 + 0.193 948 549 12;
  • 56) 0.193 948 549 12 × 2 = 0 + 0.387 897 098 24;
  • 57) 0.387 897 098 24 × 2 = 0 + 0.775 794 196 48;
  • 58) 0.775 794 196 48 × 2 = 1 + 0.551 588 392 96;
  • 59) 0.551 588 392 96 × 2 = 1 + 0.103 176 785 92;
  • 60) 0.103 176 785 92 × 2 = 0 + 0.206 353 571 84;
  • 61) 0.206 353 571 84 × 2 = 0 + 0.412 707 143 68;
  • 62) 0.412 707 143 68 × 2 = 0 + 0.825 414 287 36;
  • 63) 0.825 414 287 36 × 2 = 1 + 0.650 828 574 72;
  • 64) 0.650 828 574 72 × 2 = 1 + 0.301 657 149 44;

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 59(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 0011(2)

6. Positive number before normalization:

0.000 282 006 59(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 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:


0.000 282 006 59(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 0011(2) × 20 =

1.0010 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 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 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 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 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 0011 =

0010 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 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 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 0011


Decimal number -0.000 282 006 59 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1001 0001 0100 1101 0000 1000 0000 0110 0011

Share

» Convert decimal -0.000 282 006 64 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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