1.745 459 324 169 999 826 281 744 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 744(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
1.745 459 324 169 999 826 281 744(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: 0.745 459 324 169 999 826 281 744.

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) 0.745 459 324 169 999 826 281 744 × 2 = 1 + 0.490 918 648 339 999 652 563 488;
  • 2) 0.490 918 648 339 999 652 563 488 × 2 = 0 + 0.981 837 296 679 999 305 126 976;
  • 3) 0.981 837 296 679 999 305 126 976 × 2 = 1 + 0.963 674 593 359 998 610 253 952;
  • 4) 0.963 674 593 359 998 610 253 952 × 2 = 1 + 0.927 349 186 719 997 220 507 904;
  • 5) 0.927 349 186 719 997 220 507 904 × 2 = 1 + 0.854 698 373 439 994 441 015 808;
  • 6) 0.854 698 373 439 994 441 015 808 × 2 = 1 + 0.709 396 746 879 988 882 031 616;
  • 7) 0.709 396 746 879 988 882 031 616 × 2 = 1 + 0.418 793 493 759 977 764 063 232;
  • 8) 0.418 793 493 759 977 764 063 232 × 2 = 0 + 0.837 586 987 519 955 528 126 464;
  • 9) 0.837 586 987 519 955 528 126 464 × 2 = 1 + 0.675 173 975 039 911 056 252 928;
  • 10) 0.675 173 975 039 911 056 252 928 × 2 = 1 + 0.350 347 950 079 822 112 505 856;
  • 11) 0.350 347 950 079 822 112 505 856 × 2 = 0 + 0.700 695 900 159 644 225 011 712;
  • 12) 0.700 695 900 159 644 225 011 712 × 2 = 1 + 0.401 391 800 319 288 450 023 424;
  • 13) 0.401 391 800 319 288 450 023 424 × 2 = 0 + 0.802 783 600 638 576 900 046 848;
  • 14) 0.802 783 600 638 576 900 046 848 × 2 = 1 + 0.605 567 201 277 153 800 093 696;
  • 15) 0.605 567 201 277 153 800 093 696 × 2 = 1 + 0.211 134 402 554 307 600 187 392;
  • 16) 0.211 134 402 554 307 600 187 392 × 2 = 0 + 0.422 268 805 108 615 200 374 784;
  • 17) 0.422 268 805 108 615 200 374 784 × 2 = 0 + 0.844 537 610 217 230 400 749 568;
  • 18) 0.844 537 610 217 230 400 749 568 × 2 = 1 + 0.689 075 220 434 460 801 499 136;
  • 19) 0.689 075 220 434 460 801 499 136 × 2 = 1 + 0.378 150 440 868 921 602 998 272;
  • 20) 0.378 150 440 868 921 602 998 272 × 2 = 0 + 0.756 300 881 737 843 205 996 544;
  • 21) 0.756 300 881 737 843 205 996 544 × 2 = 1 + 0.512 601 763 475 686 411 993 088;
  • 22) 0.512 601 763 475 686 411 993 088 × 2 = 1 + 0.025 203 526 951 372 823 986 176;
  • 23) 0.025 203 526 951 372 823 986 176 × 2 = 0 + 0.050 407 053 902 745 647 972 352;
  • 24) 0.050 407 053 902 745 647 972 352 × 2 = 0 + 0.100 814 107 805 491 295 944 704;
  • 25) 0.100 814 107 805 491 295 944 704 × 2 = 0 + 0.201 628 215 610 982 591 889 408;
  • 26) 0.201 628 215 610 982 591 889 408 × 2 = 0 + 0.403 256 431 221 965 183 778 816;
  • 27) 0.403 256 431 221 965 183 778 816 × 2 = 0 + 0.806 512 862 443 930 367 557 632;
  • 28) 0.806 512 862 443 930 367 557 632 × 2 = 1 + 0.613 025 724 887 860 735 115 264;
  • 29) 0.613 025 724 887 860 735 115 264 × 2 = 1 + 0.226 051 449 775 721 470 230 528;
  • 30) 0.226 051 449 775 721 470 230 528 × 2 = 0 + 0.452 102 899 551 442 940 461 056;
  • 31) 0.452 102 899 551 442 940 461 056 × 2 = 0 + 0.904 205 799 102 885 880 922 112;
  • 32) 0.904 205 799 102 885 880 922 112 × 2 = 1 + 0.808 411 598 205 771 761 844 224;
  • 33) 0.808 411 598 205 771 761 844 224 × 2 = 1 + 0.616 823 196 411 543 523 688 448;
  • 34) 0.616 823 196 411 543 523 688 448 × 2 = 1 + 0.233 646 392 823 087 047 376 896;
  • 35) 0.233 646 392 823 087 047 376 896 × 2 = 0 + 0.467 292 785 646 174 094 753 792;
  • 36) 0.467 292 785 646 174 094 753 792 × 2 = 0 + 0.934 585 571 292 348 189 507 584;
  • 37) 0.934 585 571 292 348 189 507 584 × 2 = 1 + 0.869 171 142 584 696 379 015 168;
  • 38) 0.869 171 142 584 696 379 015 168 × 2 = 1 + 0.738 342 285 169 392 758 030 336;
  • 39) 0.738 342 285 169 392 758 030 336 × 2 = 1 + 0.476 684 570 338 785 516 060 672;
  • 40) 0.476 684 570 338 785 516 060 672 × 2 = 0 + 0.953 369 140 677 571 032 121 344;
  • 41) 0.953 369 140 677 571 032 121 344 × 2 = 1 + 0.906 738 281 355 142 064 242 688;
  • 42) 0.906 738 281 355 142 064 242 688 × 2 = 1 + 0.813 476 562 710 284 128 485 376;
  • 43) 0.813 476 562 710 284 128 485 376 × 2 = 1 + 0.626 953 125 420 568 256 970 752;
  • 44) 0.626 953 125 420 568 256 970 752 × 2 = 1 + 0.253 906 250 841 136 513 941 504;
  • 45) 0.253 906 250 841 136 513 941 504 × 2 = 0 + 0.507 812 501 682 273 027 883 008;
  • 46) 0.507 812 501 682 273 027 883 008 × 2 = 1 + 0.015 625 003 364 546 055 766 016;
  • 47) 0.015 625 003 364 546 055 766 016 × 2 = 0 + 0.031 250 006 729 092 111 532 032;
  • 48) 0.031 250 006 729 092 111 532 032 × 2 = 0 + 0.062 500 013 458 184 223 064 064;
  • 49) 0.062 500 013 458 184 223 064 064 × 2 = 0 + 0.125 000 026 916 368 446 128 128;
  • 50) 0.125 000 026 916 368 446 128 128 × 2 = 0 + 0.250 000 053 832 736 892 256 256;
  • 51) 0.250 000 053 832 736 892 256 256 × 2 = 0 + 0.500 000 107 665 473 784 512 512;
  • 52) 0.500 000 107 665 473 784 512 512 × 2 = 1 + 0.000 000 215 330 947 569 025 024;
  • 53) 0.000 000 215 330 947 569 025 024 × 2 = 0 + 0.000 000 430 661 895 138 050 048;

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:


0.745 459 324 169 999 826 281 744(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 744(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:


1.745 459 324 169 999 826 281 744(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 1.745 459 324 169 999 826 281 744 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