-0.000 282 006 143 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 143(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
-0.000 282 006 143(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:

|-0.000 282 006 143| = 0.000 282 006 143


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: 0.000 282 006 143.

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) 0.000 282 006 143 × 2 = 0 + 0.000 564 012 286;
  • 2) 0.000 564 012 286 × 2 = 0 + 0.001 128 024 572;
  • 3) 0.001 128 024 572 × 2 = 0 + 0.002 256 049 144;
  • 4) 0.002 256 049 144 × 2 = 0 + 0.004 512 098 288;
  • 5) 0.004 512 098 288 × 2 = 0 + 0.009 024 196 576;
  • 6) 0.009 024 196 576 × 2 = 0 + 0.018 048 393 152;
  • 7) 0.018 048 393 152 × 2 = 0 + 0.036 096 786 304;
  • 8) 0.036 096 786 304 × 2 = 0 + 0.072 193 572 608;
  • 9) 0.072 193 572 608 × 2 = 0 + 0.144 387 145 216;
  • 10) 0.144 387 145 216 × 2 = 0 + 0.288 774 290 432;
  • 11) 0.288 774 290 432 × 2 = 0 + 0.577 548 580 864;
  • 12) 0.577 548 580 864 × 2 = 1 + 0.155 097 161 728;
  • 13) 0.155 097 161 728 × 2 = 0 + 0.310 194 323 456;
  • 14) 0.310 194 323 456 × 2 = 0 + 0.620 388 646 912;
  • 15) 0.620 388 646 912 × 2 = 1 + 0.240 777 293 824;
  • 16) 0.240 777 293 824 × 2 = 0 + 0.481 554 587 648;
  • 17) 0.481 554 587 648 × 2 = 0 + 0.963 109 175 296;
  • 18) 0.963 109 175 296 × 2 = 1 + 0.926 218 350 592;
  • 19) 0.926 218 350 592 × 2 = 1 + 0.852 436 701 184;
  • 20) 0.852 436 701 184 × 2 = 1 + 0.704 873 402 368;
  • 21) 0.704 873 402 368 × 2 = 1 + 0.409 746 804 736;
  • 22) 0.409 746 804 736 × 2 = 0 + 0.819 493 609 472;
  • 23) 0.819 493 609 472 × 2 = 1 + 0.638 987 218 944;
  • 24) 0.638 987 218 944 × 2 = 1 + 0.277 974 437 888;
  • 25) 0.277 974 437 888 × 2 = 0 + 0.555 948 875 776;
  • 26) 0.555 948 875 776 × 2 = 1 + 0.111 897 751 552;
  • 27) 0.111 897 751 552 × 2 = 0 + 0.223 795 503 104;
  • 28) 0.223 795 503 104 × 2 = 0 + 0.447 591 006 208;
  • 29) 0.447 591 006 208 × 2 = 0 + 0.895 182 012 416;
  • 30) 0.895 182 012 416 × 2 = 1 + 0.790 364 024 832;
  • 31) 0.790 364 024 832 × 2 = 1 + 0.580 728 049 664;
  • 32) 0.580 728 049 664 × 2 = 1 + 0.161 456 099 328;
  • 33) 0.161 456 099 328 × 2 = 0 + 0.322 912 198 656;
  • 34) 0.322 912 198 656 × 2 = 0 + 0.645 824 397 312;
  • 35) 0.645 824 397 312 × 2 = 1 + 0.291 648 794 624;
  • 36) 0.291 648 794 624 × 2 = 0 + 0.583 297 589 248;
  • 37) 0.583 297 589 248 × 2 = 1 + 0.166 595 178 496;
  • 38) 0.166 595 178 496 × 2 = 0 + 0.333 190 356 992;
  • 39) 0.333 190 356 992 × 2 = 0 + 0.666 380 713 984;
  • 40) 0.666 380 713 984 × 2 = 1 + 0.332 761 427 968;
  • 41) 0.332 761 427 968 × 2 = 0 + 0.665 522 855 936;
  • 42) 0.665 522 855 936 × 2 = 1 + 0.331 045 711 872;
  • 43) 0.331 045 711 872 × 2 = 0 + 0.662 091 423 744;
  • 44) 0.662 091 423 744 × 2 = 1 + 0.324 182 847 488;
  • 45) 0.324 182 847 488 × 2 = 0 + 0.648 365 694 976;
  • 46) 0.648 365 694 976 × 2 = 1 + 0.296 731 389 952;
  • 47) 0.296 731 389 952 × 2 = 0 + 0.593 462 779 904;
  • 48) 0.593 462 779 904 × 2 = 1 + 0.186 925 559 808;
  • 49) 0.186 925 559 808 × 2 = 0 + 0.373 851 119 616;
  • 50) 0.373 851 119 616 × 2 = 0 + 0.747 702 239 232;
  • 51) 0.747 702 239 232 × 2 = 1 + 0.495 404 478 464;
  • 52) 0.495 404 478 464 × 2 = 0 + 0.990 808 956 928;
  • 53) 0.990 808 956 928 × 2 = 1 + 0.981 617 913 856;
  • 54) 0.981 617 913 856 × 2 = 1 + 0.963 235 827 712;
  • 55) 0.963 235 827 712 × 2 = 1 + 0.926 471 655 424;
  • 56) 0.926 471 655 424 × 2 = 1 + 0.852 943 310 848;
  • 57) 0.852 943 310 848 × 2 = 1 + 0.705 886 621 696;
  • 58) 0.705 886 621 696 × 2 = 1 + 0.411 773 243 392;
  • 59) 0.411 773 243 392 × 2 = 0 + 0.823 546 486 784;
  • 60) 0.823 546 486 784 × 2 = 1 + 0.647 092 973 568;
  • 61) 0.647 092 973 568 × 2 = 1 + 0.294 185 947 136;
  • 62) 0.294 185 947 136 × 2 = 0 + 0.588 371 894 272;
  • 63) 0.588 371 894 272 × 2 = 1 + 0.176 743 788 544;
  • 64) 0.176 743 788 544 × 2 = 0 + 0.353 487 577 088;

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:


0.000 282 006 143(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 1001 0101 0101 0010 1111 1101 1010(2)

6. Positive number before normalization:

0.000 282 006 143(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 1001 0101 0101 0010 1111 1101 1010(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:


0.000 282 006 143(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 1001 0101 0101 0010 1111 1101 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 1001 0101 0101 0010 1111 1101 1010(2) × 20 =

1.0010 0111 1011 0100 0111 0010 1001 0101 0101 0010 1111 1101 1010(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 0010 1001 0101 0101 0010 1111 1101 1010


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 0010 1001 0101 0101 0010 1111 1101 1010 =

0010 0111 1011 0100 0111 0010 1001 0101 0101 0010 1111 1101 1010


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 0010 1001 0101 0101 0010 1111 1101 1010


Decimal number -0.000 282 006 143 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 0010 1001 0101 0101 0010 1111 1101 1010

Share

» Convert decimal -0.000 282 006 083 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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