1.570 796 326 794 896 23 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.570 796 326 794 896 23(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
1.570 796 326 794 896 23(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.570 796 326 794 896 23.

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) 0.570 796 326 794 896 23 × 2 = 1 + 0.141 592 653 589 792 46;
  • 2) 0.141 592 653 589 792 46 × 2 = 0 + 0.283 185 307 179 584 92;
  • 3) 0.283 185 307 179 584 92 × 2 = 0 + 0.566 370 614 359 169 84;
  • 4) 0.566 370 614 359 169 84 × 2 = 1 + 0.132 741 228 718 339 68;
  • 5) 0.132 741 228 718 339 68 × 2 = 0 + 0.265 482 457 436 679 36;
  • 6) 0.265 482 457 436 679 36 × 2 = 0 + 0.530 964 914 873 358 72;
  • 7) 0.530 964 914 873 358 72 × 2 = 1 + 0.061 929 829 746 717 44;
  • 8) 0.061 929 829 746 717 44 × 2 = 0 + 0.123 859 659 493 434 88;
  • 9) 0.123 859 659 493 434 88 × 2 = 0 + 0.247 719 318 986 869 76;
  • 10) 0.247 719 318 986 869 76 × 2 = 0 + 0.495 438 637 973 739 52;
  • 11) 0.495 438 637 973 739 52 × 2 = 0 + 0.990 877 275 947 479 04;
  • 12) 0.990 877 275 947 479 04 × 2 = 1 + 0.981 754 551 894 958 08;
  • 13) 0.981 754 551 894 958 08 × 2 = 1 + 0.963 509 103 789 916 16;
  • 14) 0.963 509 103 789 916 16 × 2 = 1 + 0.927 018 207 579 832 32;
  • 15) 0.927 018 207 579 832 32 × 2 = 1 + 0.854 036 415 159 664 64;
  • 16) 0.854 036 415 159 664 64 × 2 = 1 + 0.708 072 830 319 329 28;
  • 17) 0.708 072 830 319 329 28 × 2 = 1 + 0.416 145 660 638 658 56;
  • 18) 0.416 145 660 638 658 56 × 2 = 0 + 0.832 291 321 277 317 12;
  • 19) 0.832 291 321 277 317 12 × 2 = 1 + 0.664 582 642 554 634 24;
  • 20) 0.664 582 642 554 634 24 × 2 = 1 + 0.329 165 285 109 268 48;
  • 21) 0.329 165 285 109 268 48 × 2 = 0 + 0.658 330 570 218 536 96;
  • 22) 0.658 330 570 218 536 96 × 2 = 1 + 0.316 661 140 437 073 92;
  • 23) 0.316 661 140 437 073 92 × 2 = 0 + 0.633 322 280 874 147 84;
  • 24) 0.633 322 280 874 147 84 × 2 = 1 + 0.266 644 561 748 295 68;
  • 25) 0.266 644 561 748 295 68 × 2 = 0 + 0.533 289 123 496 591 36;
  • 26) 0.533 289 123 496 591 36 × 2 = 1 + 0.066 578 246 993 182 72;
  • 27) 0.066 578 246 993 182 72 × 2 = 0 + 0.133 156 493 986 365 44;
  • 28) 0.133 156 493 986 365 44 × 2 = 0 + 0.266 312 987 972 730 88;
  • 29) 0.266 312 987 972 730 88 × 2 = 0 + 0.532 625 975 945 461 76;
  • 30) 0.532 625 975 945 461 76 × 2 = 1 + 0.065 251 951 890 923 52;
  • 31) 0.065 251 951 890 923 52 × 2 = 0 + 0.130 503 903 781 847 04;
  • 32) 0.130 503 903 781 847 04 × 2 = 0 + 0.261 007 807 563 694 08;
  • 33) 0.261 007 807 563 694 08 × 2 = 0 + 0.522 015 615 127 388 16;
  • 34) 0.522 015 615 127 388 16 × 2 = 1 + 0.044 031 230 254 776 32;
  • 35) 0.044 031 230 254 776 32 × 2 = 0 + 0.088 062 460 509 552 64;
  • 36) 0.088 062 460 509 552 64 × 2 = 0 + 0.176 124 921 019 105 28;
  • 37) 0.176 124 921 019 105 28 × 2 = 0 + 0.352 249 842 038 210 56;
  • 38) 0.352 249 842 038 210 56 × 2 = 0 + 0.704 499 684 076 421 12;
  • 39) 0.704 499 684 076 421 12 × 2 = 1 + 0.408 999 368 152 842 24;
  • 40) 0.408 999 368 152 842 24 × 2 = 0 + 0.817 998 736 305 684 48;
  • 41) 0.817 998 736 305 684 48 × 2 = 1 + 0.635 997 472 611 368 96;
  • 42) 0.635 997 472 611 368 96 × 2 = 1 + 0.271 994 945 222 737 92;
  • 43) 0.271 994 945 222 737 92 × 2 = 0 + 0.543 989 890 445 475 84;
  • 44) 0.543 989 890 445 475 84 × 2 = 1 + 0.087 979 780 890 951 68;
  • 45) 0.087 979 780 890 951 68 × 2 = 0 + 0.175 959 561 781 903 36;
  • 46) 0.175 959 561 781 903 36 × 2 = 0 + 0.351 919 123 563 806 72;
  • 47) 0.351 919 123 563 806 72 × 2 = 0 + 0.703 838 247 127 613 44;
  • 48) 0.703 838 247 127 613 44 × 2 = 1 + 0.407 676 494 255 226 88;
  • 49) 0.407 676 494 255 226 88 × 2 = 0 + 0.815 352 988 510 453 76;
  • 50) 0.815 352 988 510 453 76 × 2 = 1 + 0.630 705 977 020 907 52;
  • 51) 0.630 705 977 020 907 52 × 2 = 1 + 0.261 411 954 041 815 04;
  • 52) 0.261 411 954 041 815 04 × 2 = 0 + 0.522 823 908 083 630 08;
  • 53) 0.522 823 908 083 630 08 × 2 = 1 + 0.045 647 816 167 260 16;

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).


4. 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:


0.570 796 326 794 896 23(10) =

0.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 1(2)

5. Positive number before normalization:

1.570 796 326 794 896 23(10) =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.570 796 326 794 896 23(10) =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 1(2) =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 1(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(10)


9. 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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1023(10) =

011 1111 1111(2)


11. 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 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 1 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1111

Mantissa (52 bits) =
1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110


Decimal number 1.570 796 326 794 896 23 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110

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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