0.785 430 024 824 850 447 697 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 430 024 824 850 447 697(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
0.785 430 024 824 850 447 697(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: 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.785 430 024 824 850 447 697.

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) 0.785 430 024 824 850 447 697 × 2 = 1 + 0.570 860 049 649 700 895 394;
  • 2) 0.570 860 049 649 700 895 394 × 2 = 1 + 0.141 720 099 299 401 790 788;
  • 3) 0.141 720 099 299 401 790 788 × 2 = 0 + 0.283 440 198 598 803 581 576;
  • 4) 0.283 440 198 598 803 581 576 × 2 = 0 + 0.566 880 397 197 607 163 152;
  • 5) 0.566 880 397 197 607 163 152 × 2 = 1 + 0.133 760 794 395 214 326 304;
  • 6) 0.133 760 794 395 214 326 304 × 2 = 0 + 0.267 521 588 790 428 652 608;
  • 7) 0.267 521 588 790 428 652 608 × 2 = 0 + 0.535 043 177 580 857 305 216;
  • 8) 0.535 043 177 580 857 305 216 × 2 = 1 + 0.070 086 355 161 714 610 432;
  • 9) 0.070 086 355 161 714 610 432 × 2 = 0 + 0.140 172 710 323 429 220 864;
  • 10) 0.140 172 710 323 429 220 864 × 2 = 0 + 0.280 345 420 646 858 441 728;
  • 11) 0.280 345 420 646 858 441 728 × 2 = 0 + 0.560 690 841 293 716 883 456;
  • 12) 0.560 690 841 293 716 883 456 × 2 = 1 + 0.121 381 682 587 433 766 912;
  • 13) 0.121 381 682 587 433 766 912 × 2 = 0 + 0.242 763 365 174 867 533 824;
  • 14) 0.242 763 365 174 867 533 824 × 2 = 0 + 0.485 526 730 349 735 067 648;
  • 15) 0.485 526 730 349 735 067 648 × 2 = 0 + 0.971 053 460 699 470 135 296;
  • 16) 0.971 053 460 699 470 135 296 × 2 = 1 + 0.942 106 921 398 940 270 592;
  • 17) 0.942 106 921 398 940 270 592 × 2 = 1 + 0.884 213 842 797 880 541 184;
  • 18) 0.884 213 842 797 880 541 184 × 2 = 1 + 0.768 427 685 595 761 082 368;
  • 19) 0.768 427 685 595 761 082 368 × 2 = 1 + 0.536 855 371 191 522 164 736;
  • 20) 0.536 855 371 191 522 164 736 × 2 = 1 + 0.073 710 742 383 044 329 472;
  • 21) 0.073 710 742 383 044 329 472 × 2 = 0 + 0.147 421 484 766 088 658 944;
  • 22) 0.147 421 484 766 088 658 944 × 2 = 0 + 0.294 842 969 532 177 317 888;
  • 23) 0.294 842 969 532 177 317 888 × 2 = 0 + 0.589 685 939 064 354 635 776;
  • 24) 0.589 685 939 064 354 635 776 × 2 = 1 + 0.179 371 878 128 709 271 552;
  • 25) 0.179 371 878 128 709 271 552 × 2 = 0 + 0.358 743 756 257 418 543 104;
  • 26) 0.358 743 756 257 418 543 104 × 2 = 0 + 0.717 487 512 514 837 086 208;
  • 27) 0.717 487 512 514 837 086 208 × 2 = 1 + 0.434 975 025 029 674 172 416;
  • 28) 0.434 975 025 029 674 172 416 × 2 = 0 + 0.869 950 050 059 348 344 832;
  • 29) 0.869 950 050 059 348 344 832 × 2 = 1 + 0.739 900 100 118 696 689 664;
  • 30) 0.739 900 100 118 696 689 664 × 2 = 1 + 0.479 800 200 237 393 379 328;
  • 31) 0.479 800 200 237 393 379 328 × 2 = 0 + 0.959 600 400 474 786 758 656;
  • 32) 0.959 600 400 474 786 758 656 × 2 = 1 + 0.919 200 800 949 573 517 312;
  • 33) 0.919 200 800 949 573 517 312 × 2 = 1 + 0.838 401 601 899 147 034 624;
  • 34) 0.838 401 601 899 147 034 624 × 2 = 1 + 0.676 803 203 798 294 069 248;
  • 35) 0.676 803 203 798 294 069 248 × 2 = 1 + 0.353 606 407 596 588 138 496;
  • 36) 0.353 606 407 596 588 138 496 × 2 = 0 + 0.707 212 815 193 176 276 992;
  • 37) 0.707 212 815 193 176 276 992 × 2 = 1 + 0.414 425 630 386 352 553 984;
  • 38) 0.414 425 630 386 352 553 984 × 2 = 0 + 0.828 851 260 772 705 107 968;
  • 39) 0.828 851 260 772 705 107 968 × 2 = 1 + 0.657 702 521 545 410 215 936;
  • 40) 0.657 702 521 545 410 215 936 × 2 = 1 + 0.315 405 043 090 820 431 872;
  • 41) 0.315 405 043 090 820 431 872 × 2 = 0 + 0.630 810 086 181 640 863 744;
  • 42) 0.630 810 086 181 640 863 744 × 2 = 1 + 0.261 620 172 363 281 727 488;
  • 43) 0.261 620 172 363 281 727 488 × 2 = 0 + 0.523 240 344 726 563 454 976;
  • 44) 0.523 240 344 726 563 454 976 × 2 = 1 + 0.046 480 689 453 126 909 952;
  • 45) 0.046 480 689 453 126 909 952 × 2 = 0 + 0.092 961 378 906 253 819 904;
  • 46) 0.092 961 378 906 253 819 904 × 2 = 0 + 0.185 922 757 812 507 639 808;
  • 47) 0.185 922 757 812 507 639 808 × 2 = 0 + 0.371 845 515 625 015 279 616;
  • 48) 0.371 845 515 625 015 279 616 × 2 = 0 + 0.743 691 031 250 030 559 232;
  • 49) 0.743 691 031 250 030 559 232 × 2 = 1 + 0.487 382 062 500 061 118 464;
  • 50) 0.487 382 062 500 061 118 464 × 2 = 0 + 0.974 764 125 000 122 236 928;
  • 51) 0.974 764 125 000 122 236 928 × 2 = 1 + 0.949 528 250 000 244 473 856;
  • 52) 0.949 528 250 000 244 473 856 × 2 = 1 + 0.899 056 500 000 488 947 712;
  • 53) 0.899 056 500 000 488 947 712 × 2 = 1 + 0.798 113 000 000 977 895 424;

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:


0.785 430 024 824 850 447 697(10) =

0.1100 1001 0001 0001 1111 0001 0010 1101 1110 1011 0101 0000 1011 1(2)

5. Positive number before normalization:

0.785 430 024 824 850 447 697(10) =

0.1100 1001 0001 0001 1111 0001 0010 1101 1110 1011 0101 0000 1011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.785 430 024 824 850 447 697(10) =

0.1100 1001 0001 0001 1111 0001 0010 1101 1110 1011 0101 0000 1011 1(2) =

0.1100 1001 0001 0001 1111 0001 0010 1101 1110 1011 0101 0000 1011 1(2) × 20 =

1.1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111(2) × 2-1


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): -1

Mantissa (not normalized):
1.1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 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) =

1022(10) =

011 1111 1110(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111 =

1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111


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 1110

Mantissa (52 bits) =
1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111


Decimal number 0.785 430 024 824 850 447 697 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111

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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