-6.259 853 398 704 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -6.259 853 398 704 6(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
-6.259 853 398 704 6(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:

|-6.259 853 398 704 6| = 6.259 853 398 704 6


2. First, convert to binary (in base 2) the integer part: 6.
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;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

6(10) =

110(2)


4. Convert to binary (base 2) the fractional part: 0.259 853 398 704 6.

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) 0.259 853 398 704 6 × 2 = 0 + 0.519 706 797 409 2;
  • 2) 0.519 706 797 409 2 × 2 = 1 + 0.039 413 594 818 4;
  • 3) 0.039 413 594 818 4 × 2 = 0 + 0.078 827 189 636 8;
  • 4) 0.078 827 189 636 8 × 2 = 0 + 0.157 654 379 273 6;
  • 5) 0.157 654 379 273 6 × 2 = 0 + 0.315 308 758 547 2;
  • 6) 0.315 308 758 547 2 × 2 = 0 + 0.630 617 517 094 4;
  • 7) 0.630 617 517 094 4 × 2 = 1 + 0.261 235 034 188 8;
  • 8) 0.261 235 034 188 8 × 2 = 0 + 0.522 470 068 377 6;
  • 9) 0.522 470 068 377 6 × 2 = 1 + 0.044 940 136 755 2;
  • 10) 0.044 940 136 755 2 × 2 = 0 + 0.089 880 273 510 4;
  • 11) 0.089 880 273 510 4 × 2 = 0 + 0.179 760 547 020 8;
  • 12) 0.179 760 547 020 8 × 2 = 0 + 0.359 521 094 041 6;
  • 13) 0.359 521 094 041 6 × 2 = 0 + 0.719 042 188 083 2;
  • 14) 0.719 042 188 083 2 × 2 = 1 + 0.438 084 376 166 4;
  • 15) 0.438 084 376 166 4 × 2 = 0 + 0.876 168 752 332 8;
  • 16) 0.876 168 752 332 8 × 2 = 1 + 0.752 337 504 665 6;
  • 17) 0.752 337 504 665 6 × 2 = 1 + 0.504 675 009 331 2;
  • 18) 0.504 675 009 331 2 × 2 = 1 + 0.009 350 018 662 4;
  • 19) 0.009 350 018 662 4 × 2 = 0 + 0.018 700 037 324 8;
  • 20) 0.018 700 037 324 8 × 2 = 0 + 0.037 400 074 649 6;
  • 21) 0.037 400 074 649 6 × 2 = 0 + 0.074 800 149 299 2;
  • 22) 0.074 800 149 299 2 × 2 = 0 + 0.149 600 298 598 4;
  • 23) 0.149 600 298 598 4 × 2 = 0 + 0.299 200 597 196 8;
  • 24) 0.299 200 597 196 8 × 2 = 0 + 0.598 401 194 393 6;
  • 25) 0.598 401 194 393 6 × 2 = 1 + 0.196 802 388 787 2;
  • 26) 0.196 802 388 787 2 × 2 = 0 + 0.393 604 777 574 4;
  • 27) 0.393 604 777 574 4 × 2 = 0 + 0.787 209 555 148 8;
  • 28) 0.787 209 555 148 8 × 2 = 1 + 0.574 419 110 297 6;
  • 29) 0.574 419 110 297 6 × 2 = 1 + 0.148 838 220 595 2;
  • 30) 0.148 838 220 595 2 × 2 = 0 + 0.297 676 441 190 4;
  • 31) 0.297 676 441 190 4 × 2 = 0 + 0.595 352 882 380 8;
  • 32) 0.595 352 882 380 8 × 2 = 1 + 0.190 705 764 761 6;
  • 33) 0.190 705 764 761 6 × 2 = 0 + 0.381 411 529 523 2;
  • 34) 0.381 411 529 523 2 × 2 = 0 + 0.762 823 059 046 4;
  • 35) 0.762 823 059 046 4 × 2 = 1 + 0.525 646 118 092 8;
  • 36) 0.525 646 118 092 8 × 2 = 1 + 0.051 292 236 185 6;
  • 37) 0.051 292 236 185 6 × 2 = 0 + 0.102 584 472 371 2;
  • 38) 0.102 584 472 371 2 × 2 = 0 + 0.205 168 944 742 4;
  • 39) 0.205 168 944 742 4 × 2 = 0 + 0.410 337 889 484 8;
  • 40) 0.410 337 889 484 8 × 2 = 0 + 0.820 675 778 969 6;
  • 41) 0.820 675 778 969 6 × 2 = 1 + 0.641 351 557 939 2;
  • 42) 0.641 351 557 939 2 × 2 = 1 + 0.282 703 115 878 4;
  • 43) 0.282 703 115 878 4 × 2 = 0 + 0.565 406 231 756 8;
  • 44) 0.565 406 231 756 8 × 2 = 1 + 0.130 812 463 513 6;
  • 45) 0.130 812 463 513 6 × 2 = 0 + 0.261 624 927 027 2;
  • 46) 0.261 624 927 027 2 × 2 = 0 + 0.523 249 854 054 4;
  • 47) 0.523 249 854 054 4 × 2 = 1 + 0.046 499 708 108 8;
  • 48) 0.046 499 708 108 8 × 2 = 0 + 0.092 999 416 217 6;
  • 49) 0.092 999 416 217 6 × 2 = 0 + 0.185 998 832 435 2;
  • 50) 0.185 998 832 435 2 × 2 = 0 + 0.371 997 664 870 4;
  • 51) 0.371 997 664 870 4 × 2 = 0 + 0.743 995 329 740 8;
  • 52) 0.743 995 329 740 8 × 2 = 1 + 0.487 990 659 481 6;
  • 53) 0.487 990 659 481 6 × 2 = 0 + 0.975 981 318 963 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:


0.259 853 398 704 6(10) =

0.0100 0010 1000 0101 1100 0000 1001 1001 0011 0000 1101 0010 0001 0(2)

6. Positive number before normalization:

6.259 853 398 704 6(10) =

110.0100 0010 1000 0101 1100 0000 1001 1001 0011 0000 1101 0010 0001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


6.259 853 398 704 6(10) =

110.0100 0010 1000 0101 1100 0000 1001 1001 0011 0000 1101 0010 0001 0(2) =

110.0100 0010 1000 0101 1100 0000 1001 1001 0011 0000 1101 0010 0001 0(2) × 20 =

1.1001 0000 1010 0001 0111 0000 0010 0110 0100 1100 0011 0100 1000 010(2) × 22


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): 2

Mantissa (not normalized):
1.1001 0000 1010 0001 0111 0000 0010 0110 0100 1100 0011 0100 1000 010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1025(10) =

100 0000 0001(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1001 0000 1010 0001 0111 0000 0010 0110 0100 1100 0011 0100 1000 010 =

1001 0000 1010 0001 0111 0000 0010 0110 0100 1100 0011 0100 1000


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) =
100 0000 0001

Mantissa (52 bits) =
1001 0000 1010 0001 0111 0000 0010 0110 0100 1100 0011 0100 1000


Decimal number -6.259 853 398 704 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0001 - 1001 0000 1010 0001 0111 0000 0010 0110 0100 1100 0011 0100 1000

Share

» Convert decimal -6.259 853 398 710 2 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 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - 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