1.004 999 999 999 783 732 995 695 118 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.004 999 999 999 783 732 995 695 118 7(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
1.004 999 999 999 783 732 995 695 118 7(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: 0.004 999 999 999 783 732 995 695 118 7.

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) 0.004 999 999 999 783 732 995 695 118 7 × 2 = 0 + 0.009 999 999 999 567 465 991 390 237 4;
  • 2) 0.009 999 999 999 567 465 991 390 237 4 × 2 = 0 + 0.019 999 999 999 134 931 982 780 474 8;
  • 3) 0.019 999 999 999 134 931 982 780 474 8 × 2 = 0 + 0.039 999 999 998 269 863 965 560 949 6;
  • 4) 0.039 999 999 998 269 863 965 560 949 6 × 2 = 0 + 0.079 999 999 996 539 727 931 121 899 2;
  • 5) 0.079 999 999 996 539 727 931 121 899 2 × 2 = 0 + 0.159 999 999 993 079 455 862 243 798 4;
  • 6) 0.159 999 999 993 079 455 862 243 798 4 × 2 = 0 + 0.319 999 999 986 158 911 724 487 596 8;
  • 7) 0.319 999 999 986 158 911 724 487 596 8 × 2 = 0 + 0.639 999 999 972 317 823 448 975 193 6;
  • 8) 0.639 999 999 972 317 823 448 975 193 6 × 2 = 1 + 0.279 999 999 944 635 646 897 950 387 2;
  • 9) 0.279 999 999 944 635 646 897 950 387 2 × 2 = 0 + 0.559 999 999 889 271 293 795 900 774 4;
  • 10) 0.559 999 999 889 271 293 795 900 774 4 × 2 = 1 + 0.119 999 999 778 542 587 591 801 548 8;
  • 11) 0.119 999 999 778 542 587 591 801 548 8 × 2 = 0 + 0.239 999 999 557 085 175 183 603 097 6;
  • 12) 0.239 999 999 557 085 175 183 603 097 6 × 2 = 0 + 0.479 999 999 114 170 350 367 206 195 2;
  • 13) 0.479 999 999 114 170 350 367 206 195 2 × 2 = 0 + 0.959 999 998 228 340 700 734 412 390 4;
  • 14) 0.959 999 998 228 340 700 734 412 390 4 × 2 = 1 + 0.919 999 996 456 681 401 468 824 780 8;
  • 15) 0.919 999 996 456 681 401 468 824 780 8 × 2 = 1 + 0.839 999 992 913 362 802 937 649 561 6;
  • 16) 0.839 999 992 913 362 802 937 649 561 6 × 2 = 1 + 0.679 999 985 826 725 605 875 299 123 2;
  • 17) 0.679 999 985 826 725 605 875 299 123 2 × 2 = 1 + 0.359 999 971 653 451 211 750 598 246 4;
  • 18) 0.359 999 971 653 451 211 750 598 246 4 × 2 = 0 + 0.719 999 943 306 902 423 501 196 492 8;
  • 19) 0.719 999 943 306 902 423 501 196 492 8 × 2 = 1 + 0.439 999 886 613 804 847 002 392 985 6;
  • 20) 0.439 999 886 613 804 847 002 392 985 6 × 2 = 0 + 0.879 999 773 227 609 694 004 785 971 2;
  • 21) 0.879 999 773 227 609 694 004 785 971 2 × 2 = 1 + 0.759 999 546 455 219 388 009 571 942 4;
  • 22) 0.759 999 546 455 219 388 009 571 942 4 × 2 = 1 + 0.519 999 092 910 438 776 019 143 884 8;
  • 23) 0.519 999 092 910 438 776 019 143 884 8 × 2 = 1 + 0.039 998 185 820 877 552 038 287 769 6;
  • 24) 0.039 998 185 820 877 552 038 287 769 6 × 2 = 0 + 0.079 996 371 641 755 104 076 575 539 2;
  • 25) 0.079 996 371 641 755 104 076 575 539 2 × 2 = 0 + 0.159 992 743 283 510 208 153 151 078 4;
  • 26) 0.159 992 743 283 510 208 153 151 078 4 × 2 = 0 + 0.319 985 486 567 020 416 306 302 156 8;
  • 27) 0.319 985 486 567 020 416 306 302 156 8 × 2 = 0 + 0.639 970 973 134 040 832 612 604 313 6;
  • 28) 0.639 970 973 134 040 832 612 604 313 6 × 2 = 1 + 0.279 941 946 268 081 665 225 208 627 2;
  • 29) 0.279 941 946 268 081 665 225 208 627 2 × 2 = 0 + 0.559 883 892 536 163 330 450 417 254 4;
  • 30) 0.559 883 892 536 163 330 450 417 254 4 × 2 = 1 + 0.119 767 785 072 326 660 900 834 508 8;
  • 31) 0.119 767 785 072 326 660 900 834 508 8 × 2 = 0 + 0.239 535 570 144 653 321 801 669 017 6;
  • 32) 0.239 535 570 144 653 321 801 669 017 6 × 2 = 0 + 0.479 071 140 289 306 643 603 338 035 2;
  • 33) 0.479 071 140 289 306 643 603 338 035 2 × 2 = 0 + 0.958 142 280 578 613 287 206 676 070 4;
  • 34) 0.958 142 280 578 613 287 206 676 070 4 × 2 = 1 + 0.916 284 561 157 226 574 413 352 140 8;
  • 35) 0.916 284 561 157 226 574 413 352 140 8 × 2 = 1 + 0.832 569 122 314 453 148 826 704 281 6;
  • 36) 0.832 569 122 314 453 148 826 704 281 6 × 2 = 1 + 0.665 138 244 628 906 297 653 408 563 2;
  • 37) 0.665 138 244 628 906 297 653 408 563 2 × 2 = 1 + 0.330 276 489 257 812 595 306 817 126 4;
  • 38) 0.330 276 489 257 812 595 306 817 126 4 × 2 = 0 + 0.660 552 978 515 625 190 613 634 252 8;
  • 39) 0.660 552 978 515 625 190 613 634 252 8 × 2 = 1 + 0.321 105 957 031 250 381 227 268 505 6;
  • 40) 0.321 105 957 031 250 381 227 268 505 6 × 2 = 0 + 0.642 211 914 062 500 762 454 537 011 2;
  • 41) 0.642 211 914 062 500 762 454 537 011 2 × 2 = 1 + 0.284 423 828 125 001 524 909 074 022 4;
  • 42) 0.284 423 828 125 001 524 909 074 022 4 × 2 = 0 + 0.568 847 656 250 003 049 818 148 044 8;
  • 43) 0.568 847 656 250 003 049 818 148 044 8 × 2 = 1 + 0.137 695 312 500 006 099 636 296 089 6;
  • 44) 0.137 695 312 500 006 099 636 296 089 6 × 2 = 0 + 0.275 390 625 000 012 199 272 592 179 2;
  • 45) 0.275 390 625 000 012 199 272 592 179 2 × 2 = 0 + 0.550 781 250 000 024 398 545 184 358 4;
  • 46) 0.550 781 250 000 024 398 545 184 358 4 × 2 = 1 + 0.101 562 500 000 048 797 090 368 716 8;
  • 47) 0.101 562 500 000 048 797 090 368 716 8 × 2 = 0 + 0.203 125 000 000 097 594 180 737 433 6;
  • 48) 0.203 125 000 000 097 594 180 737 433 6 × 2 = 0 + 0.406 250 000 000 195 188 361 474 867 2;
  • 49) 0.406 250 000 000 195 188 361 474 867 2 × 2 = 0 + 0.812 500 000 000 390 376 722 949 734 4;
  • 50) 0.812 500 000 000 390 376 722 949 734 4 × 2 = 1 + 0.625 000 000 000 780 753 445 899 468 8;
  • 51) 0.625 000 000 000 780 753 445 899 468 8 × 2 = 1 + 0.250 000 000 001 561 506 891 798 937 6;
  • 52) 0.250 000 000 001 561 506 891 798 937 6 × 2 = 0 + 0.500 000 000 003 123 013 783 597 875 2;
  • 53) 0.500 000 000 003 123 013 783 597 875 2 × 2 = 1 + 0.000 000 000 006 246 027 567 195 750 4;

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:


0.004 999 999 999 783 732 995 695 118 7(10) =

0.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1(2)

5. Positive number before normalization:

1.004 999 999 999 783 732 995 695 118 7(10) =

1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 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:


1.004 999 999 999 783 732 995 695 118 7(10) =

1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1(2) =

1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 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):
1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 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. 0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1 =

0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 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) =
0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110


Decimal number 1.004 999 999 999 783 732 995 695 118 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 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