0.785 430 024 824 850 447 974 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 430 024 824 850 447 974(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
0.785 430 024 824 850 447 974(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: 0.785 430 024 824 850 447 974.

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) 0.785 430 024 824 850 447 974 × 2 = 1 + 0.570 860 049 649 700 895 948;
  • 2) 0.570 860 049 649 700 895 948 × 2 = 1 + 0.141 720 099 299 401 791 896;
  • 3) 0.141 720 099 299 401 791 896 × 2 = 0 + 0.283 440 198 598 803 583 792;
  • 4) 0.283 440 198 598 803 583 792 × 2 = 0 + 0.566 880 397 197 607 167 584;
  • 5) 0.566 880 397 197 607 167 584 × 2 = 1 + 0.133 760 794 395 214 335 168;
  • 6) 0.133 760 794 395 214 335 168 × 2 = 0 + 0.267 521 588 790 428 670 336;
  • 7) 0.267 521 588 790 428 670 336 × 2 = 0 + 0.535 043 177 580 857 340 672;
  • 8) 0.535 043 177 580 857 340 672 × 2 = 1 + 0.070 086 355 161 714 681 344;
  • 9) 0.070 086 355 161 714 681 344 × 2 = 0 + 0.140 172 710 323 429 362 688;
  • 10) 0.140 172 710 323 429 362 688 × 2 = 0 + 0.280 345 420 646 858 725 376;
  • 11) 0.280 345 420 646 858 725 376 × 2 = 0 + 0.560 690 841 293 717 450 752;
  • 12) 0.560 690 841 293 717 450 752 × 2 = 1 + 0.121 381 682 587 434 901 504;
  • 13) 0.121 381 682 587 434 901 504 × 2 = 0 + 0.242 763 365 174 869 803 008;
  • 14) 0.242 763 365 174 869 803 008 × 2 = 0 + 0.485 526 730 349 739 606 016;
  • 15) 0.485 526 730 349 739 606 016 × 2 = 0 + 0.971 053 460 699 479 212 032;
  • 16) 0.971 053 460 699 479 212 032 × 2 = 1 + 0.942 106 921 398 958 424 064;
  • 17) 0.942 106 921 398 958 424 064 × 2 = 1 + 0.884 213 842 797 916 848 128;
  • 18) 0.884 213 842 797 916 848 128 × 2 = 1 + 0.768 427 685 595 833 696 256;
  • 19) 0.768 427 685 595 833 696 256 × 2 = 1 + 0.536 855 371 191 667 392 512;
  • 20) 0.536 855 371 191 667 392 512 × 2 = 1 + 0.073 710 742 383 334 785 024;
  • 21) 0.073 710 742 383 334 785 024 × 2 = 0 + 0.147 421 484 766 669 570 048;
  • 22) 0.147 421 484 766 669 570 048 × 2 = 0 + 0.294 842 969 533 339 140 096;
  • 23) 0.294 842 969 533 339 140 096 × 2 = 0 + 0.589 685 939 066 678 280 192;
  • 24) 0.589 685 939 066 678 280 192 × 2 = 1 + 0.179 371 878 133 356 560 384;
  • 25) 0.179 371 878 133 356 560 384 × 2 = 0 + 0.358 743 756 266 713 120 768;
  • 26) 0.358 743 756 266 713 120 768 × 2 = 0 + 0.717 487 512 533 426 241 536;
  • 27) 0.717 487 512 533 426 241 536 × 2 = 1 + 0.434 975 025 066 852 483 072;
  • 28) 0.434 975 025 066 852 483 072 × 2 = 0 + 0.869 950 050 133 704 966 144;
  • 29) 0.869 950 050 133 704 966 144 × 2 = 1 + 0.739 900 100 267 409 932 288;
  • 30) 0.739 900 100 267 409 932 288 × 2 = 1 + 0.479 800 200 534 819 864 576;
  • 31) 0.479 800 200 534 819 864 576 × 2 = 0 + 0.959 600 401 069 639 729 152;
  • 32) 0.959 600 401 069 639 729 152 × 2 = 1 + 0.919 200 802 139 279 458 304;
  • 33) 0.919 200 802 139 279 458 304 × 2 = 1 + 0.838 401 604 278 558 916 608;
  • 34) 0.838 401 604 278 558 916 608 × 2 = 1 + 0.676 803 208 557 117 833 216;
  • 35) 0.676 803 208 557 117 833 216 × 2 = 1 + 0.353 606 417 114 235 666 432;
  • 36) 0.353 606 417 114 235 666 432 × 2 = 0 + 0.707 212 834 228 471 332 864;
  • 37) 0.707 212 834 228 471 332 864 × 2 = 1 + 0.414 425 668 456 942 665 728;
  • 38) 0.414 425 668 456 942 665 728 × 2 = 0 + 0.828 851 336 913 885 331 456;
  • 39) 0.828 851 336 913 885 331 456 × 2 = 1 + 0.657 702 673 827 770 662 912;
  • 40) 0.657 702 673 827 770 662 912 × 2 = 1 + 0.315 405 347 655 541 325 824;
  • 41) 0.315 405 347 655 541 325 824 × 2 = 0 + 0.630 810 695 311 082 651 648;
  • 42) 0.630 810 695 311 082 651 648 × 2 = 1 + 0.261 621 390 622 165 303 296;
  • 43) 0.261 621 390 622 165 303 296 × 2 = 0 + 0.523 242 781 244 330 606 592;
  • 44) 0.523 242 781 244 330 606 592 × 2 = 1 + 0.046 485 562 488 661 213 184;
  • 45) 0.046 485 562 488 661 213 184 × 2 = 0 + 0.092 971 124 977 322 426 368;
  • 46) 0.092 971 124 977 322 426 368 × 2 = 0 + 0.185 942 249 954 644 852 736;
  • 47) 0.185 942 249 954 644 852 736 × 2 = 0 + 0.371 884 499 909 289 705 472;
  • 48) 0.371 884 499 909 289 705 472 × 2 = 0 + 0.743 768 999 818 579 410 944;
  • 49) 0.743 768 999 818 579 410 944 × 2 = 1 + 0.487 537 999 637 158 821 888;
  • 50) 0.487 537 999 637 158 821 888 × 2 = 0 + 0.975 075 999 274 317 643 776;
  • 51) 0.975 075 999 274 317 643 776 × 2 = 1 + 0.950 151 998 548 635 287 552;
  • 52) 0.950 151 998 548 635 287 552 × 2 = 1 + 0.900 303 997 097 270 575 104;
  • 53) 0.900 303 997 097 270 575 104 × 2 = 1 + 0.800 607 994 194 541 150 208;

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:


0.785 430 024 824 850 447 974(10) =

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

5. Positive number before normalization:

0.785 430 024 824 850 447 974(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:


0.785 430 024 824 850 447 974(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 0.785 430 024 824 850 447 974 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