1.745 459 324 169 999 826 281 694 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 694 69(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
1.745 459 324 169 999 826 281 694 69(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: 0.745 459 324 169 999 826 281 694 69.

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) 0.745 459 324 169 999 826 281 694 69 × 2 = 1 + 0.490 918 648 339 999 652 563 389 38;
  • 2) 0.490 918 648 339 999 652 563 389 38 × 2 = 0 + 0.981 837 296 679 999 305 126 778 76;
  • 3) 0.981 837 296 679 999 305 126 778 76 × 2 = 1 + 0.963 674 593 359 998 610 253 557 52;
  • 4) 0.963 674 593 359 998 610 253 557 52 × 2 = 1 + 0.927 349 186 719 997 220 507 115 04;
  • 5) 0.927 349 186 719 997 220 507 115 04 × 2 = 1 + 0.854 698 373 439 994 441 014 230 08;
  • 6) 0.854 698 373 439 994 441 014 230 08 × 2 = 1 + 0.709 396 746 879 988 882 028 460 16;
  • 7) 0.709 396 746 879 988 882 028 460 16 × 2 = 1 + 0.418 793 493 759 977 764 056 920 32;
  • 8) 0.418 793 493 759 977 764 056 920 32 × 2 = 0 + 0.837 586 987 519 955 528 113 840 64;
  • 9) 0.837 586 987 519 955 528 113 840 64 × 2 = 1 + 0.675 173 975 039 911 056 227 681 28;
  • 10) 0.675 173 975 039 911 056 227 681 28 × 2 = 1 + 0.350 347 950 079 822 112 455 362 56;
  • 11) 0.350 347 950 079 822 112 455 362 56 × 2 = 0 + 0.700 695 900 159 644 224 910 725 12;
  • 12) 0.700 695 900 159 644 224 910 725 12 × 2 = 1 + 0.401 391 800 319 288 449 821 450 24;
  • 13) 0.401 391 800 319 288 449 821 450 24 × 2 = 0 + 0.802 783 600 638 576 899 642 900 48;
  • 14) 0.802 783 600 638 576 899 642 900 48 × 2 = 1 + 0.605 567 201 277 153 799 285 800 96;
  • 15) 0.605 567 201 277 153 799 285 800 96 × 2 = 1 + 0.211 134 402 554 307 598 571 601 92;
  • 16) 0.211 134 402 554 307 598 571 601 92 × 2 = 0 + 0.422 268 805 108 615 197 143 203 84;
  • 17) 0.422 268 805 108 615 197 143 203 84 × 2 = 0 + 0.844 537 610 217 230 394 286 407 68;
  • 18) 0.844 537 610 217 230 394 286 407 68 × 2 = 1 + 0.689 075 220 434 460 788 572 815 36;
  • 19) 0.689 075 220 434 460 788 572 815 36 × 2 = 1 + 0.378 150 440 868 921 577 145 630 72;
  • 20) 0.378 150 440 868 921 577 145 630 72 × 2 = 0 + 0.756 300 881 737 843 154 291 261 44;
  • 21) 0.756 300 881 737 843 154 291 261 44 × 2 = 1 + 0.512 601 763 475 686 308 582 522 88;
  • 22) 0.512 601 763 475 686 308 582 522 88 × 2 = 1 + 0.025 203 526 951 372 617 165 045 76;
  • 23) 0.025 203 526 951 372 617 165 045 76 × 2 = 0 + 0.050 407 053 902 745 234 330 091 52;
  • 24) 0.050 407 053 902 745 234 330 091 52 × 2 = 0 + 0.100 814 107 805 490 468 660 183 04;
  • 25) 0.100 814 107 805 490 468 660 183 04 × 2 = 0 + 0.201 628 215 610 980 937 320 366 08;
  • 26) 0.201 628 215 610 980 937 320 366 08 × 2 = 0 + 0.403 256 431 221 961 874 640 732 16;
  • 27) 0.403 256 431 221 961 874 640 732 16 × 2 = 0 + 0.806 512 862 443 923 749 281 464 32;
  • 28) 0.806 512 862 443 923 749 281 464 32 × 2 = 1 + 0.613 025 724 887 847 498 562 928 64;
  • 29) 0.613 025 724 887 847 498 562 928 64 × 2 = 1 + 0.226 051 449 775 694 997 125 857 28;
  • 30) 0.226 051 449 775 694 997 125 857 28 × 2 = 0 + 0.452 102 899 551 389 994 251 714 56;
  • 31) 0.452 102 899 551 389 994 251 714 56 × 2 = 0 + 0.904 205 799 102 779 988 503 429 12;
  • 32) 0.904 205 799 102 779 988 503 429 12 × 2 = 1 + 0.808 411 598 205 559 977 006 858 24;
  • 33) 0.808 411 598 205 559 977 006 858 24 × 2 = 1 + 0.616 823 196 411 119 954 013 716 48;
  • 34) 0.616 823 196 411 119 954 013 716 48 × 2 = 1 + 0.233 646 392 822 239 908 027 432 96;
  • 35) 0.233 646 392 822 239 908 027 432 96 × 2 = 0 + 0.467 292 785 644 479 816 054 865 92;
  • 36) 0.467 292 785 644 479 816 054 865 92 × 2 = 0 + 0.934 585 571 288 959 632 109 731 84;
  • 37) 0.934 585 571 288 959 632 109 731 84 × 2 = 1 + 0.869 171 142 577 919 264 219 463 68;
  • 38) 0.869 171 142 577 919 264 219 463 68 × 2 = 1 + 0.738 342 285 155 838 528 438 927 36;
  • 39) 0.738 342 285 155 838 528 438 927 36 × 2 = 1 + 0.476 684 570 311 677 056 877 854 72;
  • 40) 0.476 684 570 311 677 056 877 854 72 × 2 = 0 + 0.953 369 140 623 354 113 755 709 44;
  • 41) 0.953 369 140 623 354 113 755 709 44 × 2 = 1 + 0.906 738 281 246 708 227 511 418 88;
  • 42) 0.906 738 281 246 708 227 511 418 88 × 2 = 1 + 0.813 476 562 493 416 455 022 837 76;
  • 43) 0.813 476 562 493 416 455 022 837 76 × 2 = 1 + 0.626 953 124 986 832 910 045 675 52;
  • 44) 0.626 953 124 986 832 910 045 675 52 × 2 = 1 + 0.253 906 249 973 665 820 091 351 04;
  • 45) 0.253 906 249 973 665 820 091 351 04 × 2 = 0 + 0.507 812 499 947 331 640 182 702 08;
  • 46) 0.507 812 499 947 331 640 182 702 08 × 2 = 1 + 0.015 624 999 894 663 280 365 404 16;
  • 47) 0.015 624 999 894 663 280 365 404 16 × 2 = 0 + 0.031 249 999 789 326 560 730 808 32;
  • 48) 0.031 249 999 789 326 560 730 808 32 × 2 = 0 + 0.062 499 999 578 653 121 461 616 64;
  • 49) 0.062 499 999 578 653 121 461 616 64 × 2 = 0 + 0.124 999 999 157 306 242 923 233 28;
  • 50) 0.124 999 999 157 306 242 923 233 28 × 2 = 0 + 0.249 999 998 314 612 485 846 466 56;
  • 51) 0.249 999 998 314 612 485 846 466 56 × 2 = 0 + 0.499 999 996 629 224 971 692 933 12;
  • 52) 0.499 999 996 629 224 971 692 933 12 × 2 = 0 + 0.999 999 993 258 449 943 385 866 24;
  • 53) 0.999 999 993 258 449 943 385 866 24 × 2 = 1 + 0.999 999 986 516 899 886 771 732 48;

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:


0.745 459 324 169 999 826 281 694 69(10) =


0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 694 69(10) =


1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 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:


1.745 459 324 169 999 826 281 694 69(10) =


1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) =


1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 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):
1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 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. 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1 =


1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


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 0000


Decimal number 1.745 459 324 169 999 826 281 694 69 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 0000


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