1.745 459 324 169 999 826 281 746 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 746(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
1.745 459 324 169 999 826 281 746(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: 0.745 459 324 169 999 826 281 746.

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) 0.745 459 324 169 999 826 281 746 × 2 = 1 + 0.490 918 648 339 999 652 563 492;
  • 2) 0.490 918 648 339 999 652 563 492 × 2 = 0 + 0.981 837 296 679 999 305 126 984;
  • 3) 0.981 837 296 679 999 305 126 984 × 2 = 1 + 0.963 674 593 359 998 610 253 968;
  • 4) 0.963 674 593 359 998 610 253 968 × 2 = 1 + 0.927 349 186 719 997 220 507 936;
  • 5) 0.927 349 186 719 997 220 507 936 × 2 = 1 + 0.854 698 373 439 994 441 015 872;
  • 6) 0.854 698 373 439 994 441 015 872 × 2 = 1 + 0.709 396 746 879 988 882 031 744;
  • 7) 0.709 396 746 879 988 882 031 744 × 2 = 1 + 0.418 793 493 759 977 764 063 488;
  • 8) 0.418 793 493 759 977 764 063 488 × 2 = 0 + 0.837 586 987 519 955 528 126 976;
  • 9) 0.837 586 987 519 955 528 126 976 × 2 = 1 + 0.675 173 975 039 911 056 253 952;
  • 10) 0.675 173 975 039 911 056 253 952 × 2 = 1 + 0.350 347 950 079 822 112 507 904;
  • 11) 0.350 347 950 079 822 112 507 904 × 2 = 0 + 0.700 695 900 159 644 225 015 808;
  • 12) 0.700 695 900 159 644 225 015 808 × 2 = 1 + 0.401 391 800 319 288 450 031 616;
  • 13) 0.401 391 800 319 288 450 031 616 × 2 = 0 + 0.802 783 600 638 576 900 063 232;
  • 14) 0.802 783 600 638 576 900 063 232 × 2 = 1 + 0.605 567 201 277 153 800 126 464;
  • 15) 0.605 567 201 277 153 800 126 464 × 2 = 1 + 0.211 134 402 554 307 600 252 928;
  • 16) 0.211 134 402 554 307 600 252 928 × 2 = 0 + 0.422 268 805 108 615 200 505 856;
  • 17) 0.422 268 805 108 615 200 505 856 × 2 = 0 + 0.844 537 610 217 230 401 011 712;
  • 18) 0.844 537 610 217 230 401 011 712 × 2 = 1 + 0.689 075 220 434 460 802 023 424;
  • 19) 0.689 075 220 434 460 802 023 424 × 2 = 1 + 0.378 150 440 868 921 604 046 848;
  • 20) 0.378 150 440 868 921 604 046 848 × 2 = 0 + 0.756 300 881 737 843 208 093 696;
  • 21) 0.756 300 881 737 843 208 093 696 × 2 = 1 + 0.512 601 763 475 686 416 187 392;
  • 22) 0.512 601 763 475 686 416 187 392 × 2 = 1 + 0.025 203 526 951 372 832 374 784;
  • 23) 0.025 203 526 951 372 832 374 784 × 2 = 0 + 0.050 407 053 902 745 664 749 568;
  • 24) 0.050 407 053 902 745 664 749 568 × 2 = 0 + 0.100 814 107 805 491 329 499 136;
  • 25) 0.100 814 107 805 491 329 499 136 × 2 = 0 + 0.201 628 215 610 982 658 998 272;
  • 26) 0.201 628 215 610 982 658 998 272 × 2 = 0 + 0.403 256 431 221 965 317 996 544;
  • 27) 0.403 256 431 221 965 317 996 544 × 2 = 0 + 0.806 512 862 443 930 635 993 088;
  • 28) 0.806 512 862 443 930 635 993 088 × 2 = 1 + 0.613 025 724 887 861 271 986 176;
  • 29) 0.613 025 724 887 861 271 986 176 × 2 = 1 + 0.226 051 449 775 722 543 972 352;
  • 30) 0.226 051 449 775 722 543 972 352 × 2 = 0 + 0.452 102 899 551 445 087 944 704;
  • 31) 0.452 102 899 551 445 087 944 704 × 2 = 0 + 0.904 205 799 102 890 175 889 408;
  • 32) 0.904 205 799 102 890 175 889 408 × 2 = 1 + 0.808 411 598 205 780 351 778 816;
  • 33) 0.808 411 598 205 780 351 778 816 × 2 = 1 + 0.616 823 196 411 560 703 557 632;
  • 34) 0.616 823 196 411 560 703 557 632 × 2 = 1 + 0.233 646 392 823 121 407 115 264;
  • 35) 0.233 646 392 823 121 407 115 264 × 2 = 0 + 0.467 292 785 646 242 814 230 528;
  • 36) 0.467 292 785 646 242 814 230 528 × 2 = 0 + 0.934 585 571 292 485 628 461 056;
  • 37) 0.934 585 571 292 485 628 461 056 × 2 = 1 + 0.869 171 142 584 971 256 922 112;
  • 38) 0.869 171 142 584 971 256 922 112 × 2 = 1 + 0.738 342 285 169 942 513 844 224;
  • 39) 0.738 342 285 169 942 513 844 224 × 2 = 1 + 0.476 684 570 339 885 027 688 448;
  • 40) 0.476 684 570 339 885 027 688 448 × 2 = 0 + 0.953 369 140 679 770 055 376 896;
  • 41) 0.953 369 140 679 770 055 376 896 × 2 = 1 + 0.906 738 281 359 540 110 753 792;
  • 42) 0.906 738 281 359 540 110 753 792 × 2 = 1 + 0.813 476 562 719 080 221 507 584;
  • 43) 0.813 476 562 719 080 221 507 584 × 2 = 1 + 0.626 953 125 438 160 443 015 168;
  • 44) 0.626 953 125 438 160 443 015 168 × 2 = 1 + 0.253 906 250 876 320 886 030 336;
  • 45) 0.253 906 250 876 320 886 030 336 × 2 = 0 + 0.507 812 501 752 641 772 060 672;
  • 46) 0.507 812 501 752 641 772 060 672 × 2 = 1 + 0.015 625 003 505 283 544 121 344;
  • 47) 0.015 625 003 505 283 544 121 344 × 2 = 0 + 0.031 250 007 010 567 088 242 688;
  • 48) 0.031 250 007 010 567 088 242 688 × 2 = 0 + 0.062 500 014 021 134 176 485 376;
  • 49) 0.062 500 014 021 134 176 485 376 × 2 = 0 + 0.125 000 028 042 268 352 970 752;
  • 50) 0.125 000 028 042 268 352 970 752 × 2 = 0 + 0.250 000 056 084 536 705 941 504;
  • 51) 0.250 000 056 084 536 705 941 504 × 2 = 0 + 0.500 000 112 169 073 411 883 008;
  • 52) 0.500 000 112 169 073 411 883 008 × 2 = 1 + 0.000 000 224 338 146 823 766 016;
  • 53) 0.000 000 224 338 146 823 766 016 × 2 = 0 + 0.000 000 448 676 293 647 532 032;

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:


0.745 459 324 169 999 826 281 746(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 746(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(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:


1.745 459 324 169 999 826 281 746(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(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):
1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0


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. 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001


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) =
1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001


Decimal number 1.745 459 324 169 999 826 281 746 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001

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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