1.307 100 402 889 773 249 626 159 591 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.307 100 402 889 773 249 626 159 591(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
1.307 100 402 889 773 249 626 159 591(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: 0.307 100 402 889 773 249 626 159 591.

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) 0.307 100 402 889 773 249 626 159 591 × 2 = 0 + 0.614 200 805 779 546 499 252 319 182;
  • 2) 0.614 200 805 779 546 499 252 319 182 × 2 = 1 + 0.228 401 611 559 092 998 504 638 364;
  • 3) 0.228 401 611 559 092 998 504 638 364 × 2 = 0 + 0.456 803 223 118 185 997 009 276 728;
  • 4) 0.456 803 223 118 185 997 009 276 728 × 2 = 0 + 0.913 606 446 236 371 994 018 553 456;
  • 5) 0.913 606 446 236 371 994 018 553 456 × 2 = 1 + 0.827 212 892 472 743 988 037 106 912;
  • 6) 0.827 212 892 472 743 988 037 106 912 × 2 = 1 + 0.654 425 784 945 487 976 074 213 824;
  • 7) 0.654 425 784 945 487 976 074 213 824 × 2 = 1 + 0.308 851 569 890 975 952 148 427 648;
  • 8) 0.308 851 569 890 975 952 148 427 648 × 2 = 0 + 0.617 703 139 781 951 904 296 855 296;
  • 9) 0.617 703 139 781 951 904 296 855 296 × 2 = 1 + 0.235 406 279 563 903 808 593 710 592;
  • 10) 0.235 406 279 563 903 808 593 710 592 × 2 = 0 + 0.470 812 559 127 807 617 187 421 184;
  • 11) 0.470 812 559 127 807 617 187 421 184 × 2 = 0 + 0.941 625 118 255 615 234 374 842 368;
  • 12) 0.941 625 118 255 615 234 374 842 368 × 2 = 1 + 0.883 250 236 511 230 468 749 684 736;
  • 13) 0.883 250 236 511 230 468 749 684 736 × 2 = 1 + 0.766 500 473 022 460 937 499 369 472;
  • 14) 0.766 500 473 022 460 937 499 369 472 × 2 = 1 + 0.533 000 946 044 921 874 998 738 944;
  • 15) 0.533 000 946 044 921 874 998 738 944 × 2 = 1 + 0.066 001 892 089 843 749 997 477 888;
  • 16) 0.066 001 892 089 843 749 997 477 888 × 2 = 0 + 0.132 003 784 179 687 499 994 955 776;
  • 17) 0.132 003 784 179 687 499 994 955 776 × 2 = 0 + 0.264 007 568 359 374 999 989 911 552;
  • 18) 0.264 007 568 359 374 999 989 911 552 × 2 = 0 + 0.528 015 136 718 749 999 979 823 104;
  • 19) 0.528 015 136 718 749 999 979 823 104 × 2 = 1 + 0.056 030 273 437 499 999 959 646 208;
  • 20) 0.056 030 273 437 499 999 959 646 208 × 2 = 0 + 0.112 060 546 874 999 999 919 292 416;
  • 21) 0.112 060 546 874 999 999 919 292 416 × 2 = 0 + 0.224 121 093 749 999 999 838 584 832;
  • 22) 0.224 121 093 749 999 999 838 584 832 × 2 = 0 + 0.448 242 187 499 999 999 677 169 664;
  • 23) 0.448 242 187 499 999 999 677 169 664 × 2 = 0 + 0.896 484 374 999 999 999 354 339 328;
  • 24) 0.896 484 374 999 999 999 354 339 328 × 2 = 1 + 0.792 968 749 999 999 998 708 678 656;
  • 25) 0.792 968 749 999 999 998 708 678 656 × 2 = 1 + 0.585 937 499 999 999 997 417 357 312;
  • 26) 0.585 937 499 999 999 997 417 357 312 × 2 = 1 + 0.171 874 999 999 999 994 834 714 624;
  • 27) 0.171 874 999 999 999 994 834 714 624 × 2 = 0 + 0.343 749 999 999 999 989 669 429 248;
  • 28) 0.343 749 999 999 999 989 669 429 248 × 2 = 0 + 0.687 499 999 999 999 979 338 858 496;
  • 29) 0.687 499 999 999 999 979 338 858 496 × 2 = 1 + 0.374 999 999 999 999 958 677 716 992;
  • 30) 0.374 999 999 999 999 958 677 716 992 × 2 = 0 + 0.749 999 999 999 999 917 355 433 984;
  • 31) 0.749 999 999 999 999 917 355 433 984 × 2 = 1 + 0.499 999 999 999 999 834 710 867 968;
  • 32) 0.499 999 999 999 999 834 710 867 968 × 2 = 0 + 0.999 999 999 999 999 669 421 735 936;
  • 33) 0.999 999 999 999 999 669 421 735 936 × 2 = 1 + 0.999 999 999 999 999 338 843 471 872;
  • 34) 0.999 999 999 999 999 338 843 471 872 × 2 = 1 + 0.999 999 999 999 998 677 686 943 744;
  • 35) 0.999 999 999 999 998 677 686 943 744 × 2 = 1 + 0.999 999 999 999 997 355 373 887 488;
  • 36) 0.999 999 999 999 997 355 373 887 488 × 2 = 1 + 0.999 999 999 999 994 710 747 774 976;
  • 37) 0.999 999 999 999 994 710 747 774 976 × 2 = 1 + 0.999 999 999 999 989 421 495 549 952;
  • 38) 0.999 999 999 999 989 421 495 549 952 × 2 = 1 + 0.999 999 999 999 978 842 991 099 904;
  • 39) 0.999 999 999 999 978 842 991 099 904 × 2 = 1 + 0.999 999 999 999 957 685 982 199 808;
  • 40) 0.999 999 999 999 957 685 982 199 808 × 2 = 1 + 0.999 999 999 999 915 371 964 399 616;
  • 41) 0.999 999 999 999 915 371 964 399 616 × 2 = 1 + 0.999 999 999 999 830 743 928 799 232;
  • 42) 0.999 999 999 999 830 743 928 799 232 × 2 = 1 + 0.999 999 999 999 661 487 857 598 464;
  • 43) 0.999 999 999 999 661 487 857 598 464 × 2 = 1 + 0.999 999 999 999 322 975 715 196 928;
  • 44) 0.999 999 999 999 322 975 715 196 928 × 2 = 1 + 0.999 999 999 998 645 951 430 393 856;
  • 45) 0.999 999 999 998 645 951 430 393 856 × 2 = 1 + 0.999 999 999 997 291 902 860 787 712;
  • 46) 0.999 999 999 997 291 902 860 787 712 × 2 = 1 + 0.999 999 999 994 583 805 721 575 424;
  • 47) 0.999 999 999 994 583 805 721 575 424 × 2 = 1 + 0.999 999 999 989 167 611 443 150 848;
  • 48) 0.999 999 999 989 167 611 443 150 848 × 2 = 1 + 0.999 999 999 978 335 222 886 301 696;
  • 49) 0.999 999 999 978 335 222 886 301 696 × 2 = 1 + 0.999 999 999 956 670 445 772 603 392;
  • 50) 0.999 999 999 956 670 445 772 603 392 × 2 = 1 + 0.999 999 999 913 340 891 545 206 784;
  • 51) 0.999 999 999 913 340 891 545 206 784 × 2 = 1 + 0.999 999 999 826 681 783 090 413 568;
  • 52) 0.999 999 999 826 681 783 090 413 568 × 2 = 1 + 0.999 999 999 653 363 566 180 827 136;
  • 53) 0.999 999 999 653 363 566 180 827 136 × 2 = 1 + 0.999 999 999 306 727 132 361 654 272;

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:


0.307 100 402 889 773 249 626 159 591(10) =

0.0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111 1(2)

5. Positive number before normalization:

1.307 100 402 889 773 249 626 159 591(10) =

1.0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111 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:


1.307 100 402 889 773 249 626 159 591(10) =

1.0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111 1(2) =

1.0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111 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):
1.0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111 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. 0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111 1 =

0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111


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) =
0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111


Decimal number 1.307 100 402 889 773 249 626 159 591 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0100 1110 1001 1110 0010 0001 1100 1010 1111 1111 1111 1111 1111

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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