-0.000 035 666 844 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 844 4(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
-0.000 035 666 844 4(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:

|-0.000 035 666 844 4| = 0.000 035 666 844 4


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: 0.000 035 666 844 4.

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) 0.000 035 666 844 4 × 2 = 0 + 0.000 071 333 688 8;
  • 2) 0.000 071 333 688 8 × 2 = 0 + 0.000 142 667 377 6;
  • 3) 0.000 142 667 377 6 × 2 = 0 + 0.000 285 334 755 2;
  • 4) 0.000 285 334 755 2 × 2 = 0 + 0.000 570 669 510 4;
  • 5) 0.000 570 669 510 4 × 2 = 0 + 0.001 141 339 020 8;
  • 6) 0.001 141 339 020 8 × 2 = 0 + 0.002 282 678 041 6;
  • 7) 0.002 282 678 041 6 × 2 = 0 + 0.004 565 356 083 2;
  • 8) 0.004 565 356 083 2 × 2 = 0 + 0.009 130 712 166 4;
  • 9) 0.009 130 712 166 4 × 2 = 0 + 0.018 261 424 332 8;
  • 10) 0.018 261 424 332 8 × 2 = 0 + 0.036 522 848 665 6;
  • 11) 0.036 522 848 665 6 × 2 = 0 + 0.073 045 697 331 2;
  • 12) 0.073 045 697 331 2 × 2 = 0 + 0.146 091 394 662 4;
  • 13) 0.146 091 394 662 4 × 2 = 0 + 0.292 182 789 324 8;
  • 14) 0.292 182 789 324 8 × 2 = 0 + 0.584 365 578 649 6;
  • 15) 0.584 365 578 649 6 × 2 = 1 + 0.168 731 157 299 2;
  • 16) 0.168 731 157 299 2 × 2 = 0 + 0.337 462 314 598 4;
  • 17) 0.337 462 314 598 4 × 2 = 0 + 0.674 924 629 196 8;
  • 18) 0.674 924 629 196 8 × 2 = 1 + 0.349 849 258 393 6;
  • 19) 0.349 849 258 393 6 × 2 = 0 + 0.699 698 516 787 2;
  • 20) 0.699 698 516 787 2 × 2 = 1 + 0.399 397 033 574 4;
  • 21) 0.399 397 033 574 4 × 2 = 0 + 0.798 794 067 148 8;
  • 22) 0.798 794 067 148 8 × 2 = 1 + 0.597 588 134 297 6;
  • 23) 0.597 588 134 297 6 × 2 = 1 + 0.195 176 268 595 2;
  • 24) 0.195 176 268 595 2 × 2 = 0 + 0.390 352 537 190 4;
  • 25) 0.390 352 537 190 4 × 2 = 0 + 0.780 705 074 380 8;
  • 26) 0.780 705 074 380 8 × 2 = 1 + 0.561 410 148 761 6;
  • 27) 0.561 410 148 761 6 × 2 = 1 + 0.122 820 297 523 2;
  • 28) 0.122 820 297 523 2 × 2 = 0 + 0.245 640 595 046 4;
  • 29) 0.245 640 595 046 4 × 2 = 0 + 0.491 281 190 092 8;
  • 30) 0.491 281 190 092 8 × 2 = 0 + 0.982 562 380 185 6;
  • 31) 0.982 562 380 185 6 × 2 = 1 + 0.965 124 760 371 2;
  • 32) 0.965 124 760 371 2 × 2 = 1 + 0.930 249 520 742 4;
  • 33) 0.930 249 520 742 4 × 2 = 1 + 0.860 499 041 484 8;
  • 34) 0.860 499 041 484 8 × 2 = 1 + 0.720 998 082 969 6;
  • 35) 0.720 998 082 969 6 × 2 = 1 + 0.441 996 165 939 2;
  • 36) 0.441 996 165 939 2 × 2 = 0 + 0.883 992 331 878 4;
  • 37) 0.883 992 331 878 4 × 2 = 1 + 0.767 984 663 756 8;
  • 38) 0.767 984 663 756 8 × 2 = 1 + 0.535 969 327 513 6;
  • 39) 0.535 969 327 513 6 × 2 = 1 + 0.071 938 655 027 2;
  • 40) 0.071 938 655 027 2 × 2 = 0 + 0.143 877 310 054 4;
  • 41) 0.143 877 310 054 4 × 2 = 0 + 0.287 754 620 108 8;
  • 42) 0.287 754 620 108 8 × 2 = 0 + 0.575 509 240 217 6;
  • 43) 0.575 509 240 217 6 × 2 = 1 + 0.151 018 480 435 2;
  • 44) 0.151 018 480 435 2 × 2 = 0 + 0.302 036 960 870 4;
  • 45) 0.302 036 960 870 4 × 2 = 0 + 0.604 073 921 740 8;
  • 46) 0.604 073 921 740 8 × 2 = 1 + 0.208 147 843 481 6;
  • 47) 0.208 147 843 481 6 × 2 = 0 + 0.416 295 686 963 2;
  • 48) 0.416 295 686 963 2 × 2 = 0 + 0.832 591 373 926 4;
  • 49) 0.832 591 373 926 4 × 2 = 1 + 0.665 182 747 852 8;
  • 50) 0.665 182 747 852 8 × 2 = 1 + 0.330 365 495 705 6;
  • 51) 0.330 365 495 705 6 × 2 = 0 + 0.660 730 991 411 2;
  • 52) 0.660 730 991 411 2 × 2 = 1 + 0.321 461 982 822 4;
  • 53) 0.321 461 982 822 4 × 2 = 0 + 0.642 923 965 644 8;
  • 54) 0.642 923 965 644 8 × 2 = 1 + 0.285 847 931 289 6;
  • 55) 0.285 847 931 289 6 × 2 = 0 + 0.571 695 862 579 2;
  • 56) 0.571 695 862 579 2 × 2 = 1 + 0.143 391 725 158 4;
  • 57) 0.143 391 725 158 4 × 2 = 0 + 0.286 783 450 316 8;
  • 58) 0.286 783 450 316 8 × 2 = 0 + 0.573 566 900 633 6;
  • 59) 0.573 566 900 633 6 × 2 = 1 + 0.147 133 801 267 2;
  • 60) 0.147 133 801 267 2 × 2 = 0 + 0.294 267 602 534 4;
  • 61) 0.294 267 602 534 4 × 2 = 0 + 0.588 535 205 068 8;
  • 62) 0.588 535 205 068 8 × 2 = 1 + 0.177 070 410 137 6;
  • 63) 0.177 070 410 137 6 × 2 = 0 + 0.354 140 820 275 2;
  • 64) 0.354 140 820 275 2 × 2 = 0 + 0.708 281 640 550 4;
  • 65) 0.708 281 640 550 4 × 2 = 1 + 0.416 563 281 100 8;
  • 66) 0.416 563 281 100 8 × 2 = 0 + 0.833 126 562 201 6;
  • 67) 0.833 126 562 201 6 × 2 = 1 + 0.666 253 124 403 2;

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:


0.000 035 666 844 4(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1110 0010 0100 1101 0101 0010 0100 101(2)

6. Positive number before normalization:

0.000 035 666 844 4(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1110 0010 0100 1101 0101 0010 0100 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 035 666 844 4(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1110 0010 0100 1101 0101 0010 0100 101(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1110 0010 0100 1101 0101 0010 0100 101(2) × 20 =

1.0010 1011 0011 0001 1111 0111 0001 0010 0110 1010 1001 0010 0101(2) × 2-15


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): -15

Mantissa (not normalized):
1.0010 1011 0011 0001 1111 0111 0001 0010 0110 1010 1001 0010 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 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) =

1008(10) =

011 1111 0000(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 1011 0011 0001 1111 0111 0001 0010 0110 1010 1001 0010 0101 =

0010 1011 0011 0001 1111 0111 0001 0010 0110 1010 1001 0010 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 0000

Mantissa (52 bits) =
0010 1011 0011 0001 1111 0111 0001 0010 0110 1010 1001 0010 0101


Decimal number -0.000 035 666 844 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0111 0001 0010 0110 1010 1001 0010 0101

Share

» Convert decimal -0.000 035 666 844 3 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 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions & 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