Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.365 481 236 489 619 22 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.365 481 236 489 619 22(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.365 481 236 489 619 22| = 0.365 481 236 489 619 22


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: 0.365 481 236 489 619 22.

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) 0.365 481 236 489 619 22 × 2 = 0 + 0.730 962 472 979 238 44;
  • 2) 0.730 962 472 979 238 44 × 2 = 1 + 0.461 924 945 958 476 88;
  • 3) 0.461 924 945 958 476 88 × 2 = 0 + 0.923 849 891 916 953 76;
  • 4) 0.923 849 891 916 953 76 × 2 = 1 + 0.847 699 783 833 907 52;
  • 5) 0.847 699 783 833 907 52 × 2 = 1 + 0.695 399 567 667 815 04;
  • 6) 0.695 399 567 667 815 04 × 2 = 1 + 0.390 799 135 335 630 08;
  • 7) 0.390 799 135 335 630 08 × 2 = 0 + 0.781 598 270 671 260 16;
  • 8) 0.781 598 270 671 260 16 × 2 = 1 + 0.563 196 541 342 520 32;
  • 9) 0.563 196 541 342 520 32 × 2 = 1 + 0.126 393 082 685 040 64;
  • 10) 0.126 393 082 685 040 64 × 2 = 0 + 0.252 786 165 370 081 28;
  • 11) 0.252 786 165 370 081 28 × 2 = 0 + 0.505 572 330 740 162 56;
  • 12) 0.505 572 330 740 162 56 × 2 = 1 + 0.011 144 661 480 325 12;
  • 13) 0.011 144 661 480 325 12 × 2 = 0 + 0.022 289 322 960 650 24;
  • 14) 0.022 289 322 960 650 24 × 2 = 0 + 0.044 578 645 921 300 48;
  • 15) 0.044 578 645 921 300 48 × 2 = 0 + 0.089 157 291 842 600 96;
  • 16) 0.089 157 291 842 600 96 × 2 = 0 + 0.178 314 583 685 201 92;
  • 17) 0.178 314 583 685 201 92 × 2 = 0 + 0.356 629 167 370 403 84;
  • 18) 0.356 629 167 370 403 84 × 2 = 0 + 0.713 258 334 740 807 68;
  • 19) 0.713 258 334 740 807 68 × 2 = 1 + 0.426 516 669 481 615 36;
  • 20) 0.426 516 669 481 615 36 × 2 = 0 + 0.853 033 338 963 230 72;
  • 21) 0.853 033 338 963 230 72 × 2 = 1 + 0.706 066 677 926 461 44;
  • 22) 0.706 066 677 926 461 44 × 2 = 1 + 0.412 133 355 852 922 88;
  • 23) 0.412 133 355 852 922 88 × 2 = 0 + 0.824 266 711 705 845 76;
  • 24) 0.824 266 711 705 845 76 × 2 = 1 + 0.648 533 423 411 691 52;
  • 25) 0.648 533 423 411 691 52 × 2 = 1 + 0.297 066 846 823 383 04;
  • 26) 0.297 066 846 823 383 04 × 2 = 0 + 0.594 133 693 646 766 08;
  • 27) 0.594 133 693 646 766 08 × 2 = 1 + 0.188 267 387 293 532 16;
  • 28) 0.188 267 387 293 532 16 × 2 = 0 + 0.376 534 774 587 064 32;
  • 29) 0.376 534 774 587 064 32 × 2 = 0 + 0.753 069 549 174 128 64;
  • 30) 0.753 069 549 174 128 64 × 2 = 1 + 0.506 139 098 348 257 28;
  • 31) 0.506 139 098 348 257 28 × 2 = 1 + 0.012 278 196 696 514 56;
  • 32) 0.012 278 196 696 514 56 × 2 = 0 + 0.024 556 393 393 029 12;
  • 33) 0.024 556 393 393 029 12 × 2 = 0 + 0.049 112 786 786 058 24;
  • 34) 0.049 112 786 786 058 24 × 2 = 0 + 0.098 225 573 572 116 48;
  • 35) 0.098 225 573 572 116 48 × 2 = 0 + 0.196 451 147 144 232 96;
  • 36) 0.196 451 147 144 232 96 × 2 = 0 + 0.392 902 294 288 465 92;
  • 37) 0.392 902 294 288 465 92 × 2 = 0 + 0.785 804 588 576 931 84;
  • 38) 0.785 804 588 576 931 84 × 2 = 1 + 0.571 609 177 153 863 68;
  • 39) 0.571 609 177 153 863 68 × 2 = 1 + 0.143 218 354 307 727 36;
  • 40) 0.143 218 354 307 727 36 × 2 = 0 + 0.286 436 708 615 454 72;
  • 41) 0.286 436 708 615 454 72 × 2 = 0 + 0.572 873 417 230 909 44;
  • 42) 0.572 873 417 230 909 44 × 2 = 1 + 0.145 746 834 461 818 88;
  • 43) 0.145 746 834 461 818 88 × 2 = 0 + 0.291 493 668 923 637 76;
  • 44) 0.291 493 668 923 637 76 × 2 = 0 + 0.582 987 337 847 275 52;
  • 45) 0.582 987 337 847 275 52 × 2 = 1 + 0.165 974 675 694 551 04;
  • 46) 0.165 974 675 694 551 04 × 2 = 0 + 0.331 949 351 389 102 08;
  • 47) 0.331 949 351 389 102 08 × 2 = 0 + 0.663 898 702 778 204 16;
  • 48) 0.663 898 702 778 204 16 × 2 = 1 + 0.327 797 405 556 408 32;
  • 49) 0.327 797 405 556 408 32 × 2 = 0 + 0.655 594 811 112 816 64;
  • 50) 0.655 594 811 112 816 64 × 2 = 1 + 0.311 189 622 225 633 28;
  • 51) 0.311 189 622 225 633 28 × 2 = 0 + 0.622 379 244 451 266 56;
  • 52) 0.622 379 244 451 266 56 × 2 = 1 + 0.244 758 488 902 533 12;
  • 53) 0.244 758 488 902 533 12 × 2 = 0 + 0.489 516 977 805 066 24;
  • 54) 0.489 516 977 805 066 24 × 2 = 0 + 0.979 033 955 610 132 48;

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:


0.365 481 236 489 619 22(10) =

0.0101 1101 1001 0000 0010 1101 1010 0110 0000 0110 0100 1001 0101 00(2)

6. Positive number before normalization:

0.365 481 236 489 619 22(10) =

0.0101 1101 1001 0000 0010 1101 1010 0110 0000 0110 0100 1001 0101 00(2)

7. Normalize the binary representation of the number.

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


0.365 481 236 489 619 22(10) =

0.0101 1101 1001 0000 0010 1101 1010 0110 0000 0110 0100 1001 0101 00(2) =

0.0101 1101 1001 0000 0010 1101 1010 0110 0000 0110 0100 1001 0101 00(2) × 20 =

1.0111 0110 0100 0000 1011 0110 1001 1000 0001 1001 0010 0101 0100(2) × 2-2


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):
1.0111 0110 0100 0000 1011 0110 1001 1000 0001 1001 0010 0101 0100


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) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1021(10) =

011 1111 1101(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. 0111 0110 0100 0000 1011 0110 1001 1000 0001 1001 0010 0101 0100 =

0111 0110 0100 0000 1011 0110 1001 1000 0001 1001 0010 0101 0100


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 1101

Mantissa (52 bits) =
0111 0110 0100 0000 1011 0110 1001 1000 0001 1001 0010 0101 0100


The base ten decimal number -0.365 481 236 489 619 22 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0111 0110 0100 0000 1011 0110 1001 1000 0001 1001 0010 0101 0100

» Decimal number in base ten -0.365 481 236 489 618 41 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

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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