1.745 459 324 169 999 826 281 696 184 39 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 184 39(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
1.745 459 324 169 999 826 281 696 184 39(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: 0.745 459 324 169 999 826 281 696 184 39.

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) 0.745 459 324 169 999 826 281 696 184 39 × 2 = 1 + 0.490 918 648 339 999 652 563 392 368 78;
  • 2) 0.490 918 648 339 999 652 563 392 368 78 × 2 = 0 + 0.981 837 296 679 999 305 126 784 737 56;
  • 3) 0.981 837 296 679 999 305 126 784 737 56 × 2 = 1 + 0.963 674 593 359 998 610 253 569 475 12;
  • 4) 0.963 674 593 359 998 610 253 569 475 12 × 2 = 1 + 0.927 349 186 719 997 220 507 138 950 24;
  • 5) 0.927 349 186 719 997 220 507 138 950 24 × 2 = 1 + 0.854 698 373 439 994 441 014 277 900 48;
  • 6) 0.854 698 373 439 994 441 014 277 900 48 × 2 = 1 + 0.709 396 746 879 988 882 028 555 800 96;
  • 7) 0.709 396 746 879 988 882 028 555 800 96 × 2 = 1 + 0.418 793 493 759 977 764 057 111 601 92;
  • 8) 0.418 793 493 759 977 764 057 111 601 92 × 2 = 0 + 0.837 586 987 519 955 528 114 223 203 84;
  • 9) 0.837 586 987 519 955 528 114 223 203 84 × 2 = 1 + 0.675 173 975 039 911 056 228 446 407 68;
  • 10) 0.675 173 975 039 911 056 228 446 407 68 × 2 = 1 + 0.350 347 950 079 822 112 456 892 815 36;
  • 11) 0.350 347 950 079 822 112 456 892 815 36 × 2 = 0 + 0.700 695 900 159 644 224 913 785 630 72;
  • 12) 0.700 695 900 159 644 224 913 785 630 72 × 2 = 1 + 0.401 391 800 319 288 449 827 571 261 44;
  • 13) 0.401 391 800 319 288 449 827 571 261 44 × 2 = 0 + 0.802 783 600 638 576 899 655 142 522 88;
  • 14) 0.802 783 600 638 576 899 655 142 522 88 × 2 = 1 + 0.605 567 201 277 153 799 310 285 045 76;
  • 15) 0.605 567 201 277 153 799 310 285 045 76 × 2 = 1 + 0.211 134 402 554 307 598 620 570 091 52;
  • 16) 0.211 134 402 554 307 598 620 570 091 52 × 2 = 0 + 0.422 268 805 108 615 197 241 140 183 04;
  • 17) 0.422 268 805 108 615 197 241 140 183 04 × 2 = 0 + 0.844 537 610 217 230 394 482 280 366 08;
  • 18) 0.844 537 610 217 230 394 482 280 366 08 × 2 = 1 + 0.689 075 220 434 460 788 964 560 732 16;
  • 19) 0.689 075 220 434 460 788 964 560 732 16 × 2 = 1 + 0.378 150 440 868 921 577 929 121 464 32;
  • 20) 0.378 150 440 868 921 577 929 121 464 32 × 2 = 0 + 0.756 300 881 737 843 155 858 242 928 64;
  • 21) 0.756 300 881 737 843 155 858 242 928 64 × 2 = 1 + 0.512 601 763 475 686 311 716 485 857 28;
  • 22) 0.512 601 763 475 686 311 716 485 857 28 × 2 = 1 + 0.025 203 526 951 372 623 432 971 714 56;
  • 23) 0.025 203 526 951 372 623 432 971 714 56 × 2 = 0 + 0.050 407 053 902 745 246 865 943 429 12;
  • 24) 0.050 407 053 902 745 246 865 943 429 12 × 2 = 0 + 0.100 814 107 805 490 493 731 886 858 24;
  • 25) 0.100 814 107 805 490 493 731 886 858 24 × 2 = 0 + 0.201 628 215 610 980 987 463 773 716 48;
  • 26) 0.201 628 215 610 980 987 463 773 716 48 × 2 = 0 + 0.403 256 431 221 961 974 927 547 432 96;
  • 27) 0.403 256 431 221 961 974 927 547 432 96 × 2 = 0 + 0.806 512 862 443 923 949 855 094 865 92;
  • 28) 0.806 512 862 443 923 949 855 094 865 92 × 2 = 1 + 0.613 025 724 887 847 899 710 189 731 84;
  • 29) 0.613 025 724 887 847 899 710 189 731 84 × 2 = 1 + 0.226 051 449 775 695 799 420 379 463 68;
  • 30) 0.226 051 449 775 695 799 420 379 463 68 × 2 = 0 + 0.452 102 899 551 391 598 840 758 927 36;
  • 31) 0.452 102 899 551 391 598 840 758 927 36 × 2 = 0 + 0.904 205 799 102 783 197 681 517 854 72;
  • 32) 0.904 205 799 102 783 197 681 517 854 72 × 2 = 1 + 0.808 411 598 205 566 395 363 035 709 44;
  • 33) 0.808 411 598 205 566 395 363 035 709 44 × 2 = 1 + 0.616 823 196 411 132 790 726 071 418 88;
  • 34) 0.616 823 196 411 132 790 726 071 418 88 × 2 = 1 + 0.233 646 392 822 265 581 452 142 837 76;
  • 35) 0.233 646 392 822 265 581 452 142 837 76 × 2 = 0 + 0.467 292 785 644 531 162 904 285 675 52;
  • 36) 0.467 292 785 644 531 162 904 285 675 52 × 2 = 0 + 0.934 585 571 289 062 325 808 571 351 04;
  • 37) 0.934 585 571 289 062 325 808 571 351 04 × 2 = 1 + 0.869 171 142 578 124 651 617 142 702 08;
  • 38) 0.869 171 142 578 124 651 617 142 702 08 × 2 = 1 + 0.738 342 285 156 249 303 234 285 404 16;
  • 39) 0.738 342 285 156 249 303 234 285 404 16 × 2 = 1 + 0.476 684 570 312 498 606 468 570 808 32;
  • 40) 0.476 684 570 312 498 606 468 570 808 32 × 2 = 0 + 0.953 369 140 624 997 212 937 141 616 64;
  • 41) 0.953 369 140 624 997 212 937 141 616 64 × 2 = 1 + 0.906 738 281 249 994 425 874 283 233 28;
  • 42) 0.906 738 281 249 994 425 874 283 233 28 × 2 = 1 + 0.813 476 562 499 988 851 748 566 466 56;
  • 43) 0.813 476 562 499 988 851 748 566 466 56 × 2 = 1 + 0.626 953 124 999 977 703 497 132 933 12;
  • 44) 0.626 953 124 999 977 703 497 132 933 12 × 2 = 1 + 0.253 906 249 999 955 406 994 265 866 24;
  • 45) 0.253 906 249 999 955 406 994 265 866 24 × 2 = 0 + 0.507 812 499 999 910 813 988 531 732 48;
  • 46) 0.507 812 499 999 910 813 988 531 732 48 × 2 = 1 + 0.015 624 999 999 821 627 977 063 464 96;
  • 47) 0.015 624 999 999 821 627 977 063 464 96 × 2 = 0 + 0.031 249 999 999 643 255 954 126 929 92;
  • 48) 0.031 249 999 999 643 255 954 126 929 92 × 2 = 0 + 0.062 499 999 999 286 511 908 253 859 84;
  • 49) 0.062 499 999 999 286 511 908 253 859 84 × 2 = 0 + 0.124 999 999 998 573 023 816 507 719 68;
  • 50) 0.124 999 999 998 573 023 816 507 719 68 × 2 = 0 + 0.249 999 999 997 146 047 633 015 439 36;
  • 51) 0.249 999 999 997 146 047 633 015 439 36 × 2 = 0 + 0.499 999 999 994 292 095 266 030 878 72;
  • 52) 0.499 999 999 994 292 095 266 030 878 72 × 2 = 0 + 0.999 999 999 988 584 190 532 061 757 44;
  • 53) 0.999 999 999 988 584 190 532 061 757 44 × 2 = 1 + 0.999 999 999 977 168 381 064 123 514 88;

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:


0.745 459 324 169 999 826 281 696 184 39(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 184 39(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:


1.745 459 324 169 999 826 281 696 184 39(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 1.745 459 324 169 999 826 281 696 184 39 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

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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