1.745 459 324 169 999 826 281 700 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 700 42(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
1.745 459 324 169 999 826 281 700 42(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: 0.745 459 324 169 999 826 281 700 42.

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) 0.745 459 324 169 999 826 281 700 42 × 2 = 1 + 0.490 918 648 339 999 652 563 400 84;
  • 2) 0.490 918 648 339 999 652 563 400 84 × 2 = 0 + 0.981 837 296 679 999 305 126 801 68;
  • 3) 0.981 837 296 679 999 305 126 801 68 × 2 = 1 + 0.963 674 593 359 998 610 253 603 36;
  • 4) 0.963 674 593 359 998 610 253 603 36 × 2 = 1 + 0.927 349 186 719 997 220 507 206 72;
  • 5) 0.927 349 186 719 997 220 507 206 72 × 2 = 1 + 0.854 698 373 439 994 441 014 413 44;
  • 6) 0.854 698 373 439 994 441 014 413 44 × 2 = 1 + 0.709 396 746 879 988 882 028 826 88;
  • 7) 0.709 396 746 879 988 882 028 826 88 × 2 = 1 + 0.418 793 493 759 977 764 057 653 76;
  • 8) 0.418 793 493 759 977 764 057 653 76 × 2 = 0 + 0.837 586 987 519 955 528 115 307 52;
  • 9) 0.837 586 987 519 955 528 115 307 52 × 2 = 1 + 0.675 173 975 039 911 056 230 615 04;
  • 10) 0.675 173 975 039 911 056 230 615 04 × 2 = 1 + 0.350 347 950 079 822 112 461 230 08;
  • 11) 0.350 347 950 079 822 112 461 230 08 × 2 = 0 + 0.700 695 900 159 644 224 922 460 16;
  • 12) 0.700 695 900 159 644 224 922 460 16 × 2 = 1 + 0.401 391 800 319 288 449 844 920 32;
  • 13) 0.401 391 800 319 288 449 844 920 32 × 2 = 0 + 0.802 783 600 638 576 899 689 840 64;
  • 14) 0.802 783 600 638 576 899 689 840 64 × 2 = 1 + 0.605 567 201 277 153 799 379 681 28;
  • 15) 0.605 567 201 277 153 799 379 681 28 × 2 = 1 + 0.211 134 402 554 307 598 759 362 56;
  • 16) 0.211 134 402 554 307 598 759 362 56 × 2 = 0 + 0.422 268 805 108 615 197 518 725 12;
  • 17) 0.422 268 805 108 615 197 518 725 12 × 2 = 0 + 0.844 537 610 217 230 395 037 450 24;
  • 18) 0.844 537 610 217 230 395 037 450 24 × 2 = 1 + 0.689 075 220 434 460 790 074 900 48;
  • 19) 0.689 075 220 434 460 790 074 900 48 × 2 = 1 + 0.378 150 440 868 921 580 149 800 96;
  • 20) 0.378 150 440 868 921 580 149 800 96 × 2 = 0 + 0.756 300 881 737 843 160 299 601 92;
  • 21) 0.756 300 881 737 843 160 299 601 92 × 2 = 1 + 0.512 601 763 475 686 320 599 203 84;
  • 22) 0.512 601 763 475 686 320 599 203 84 × 2 = 1 + 0.025 203 526 951 372 641 198 407 68;
  • 23) 0.025 203 526 951 372 641 198 407 68 × 2 = 0 + 0.050 407 053 902 745 282 396 815 36;
  • 24) 0.050 407 053 902 745 282 396 815 36 × 2 = 0 + 0.100 814 107 805 490 564 793 630 72;
  • 25) 0.100 814 107 805 490 564 793 630 72 × 2 = 0 + 0.201 628 215 610 981 129 587 261 44;
  • 26) 0.201 628 215 610 981 129 587 261 44 × 2 = 0 + 0.403 256 431 221 962 259 174 522 88;
  • 27) 0.403 256 431 221 962 259 174 522 88 × 2 = 0 + 0.806 512 862 443 924 518 349 045 76;
  • 28) 0.806 512 862 443 924 518 349 045 76 × 2 = 1 + 0.613 025 724 887 849 036 698 091 52;
  • 29) 0.613 025 724 887 849 036 698 091 52 × 2 = 1 + 0.226 051 449 775 698 073 396 183 04;
  • 30) 0.226 051 449 775 698 073 396 183 04 × 2 = 0 + 0.452 102 899 551 396 146 792 366 08;
  • 31) 0.452 102 899 551 396 146 792 366 08 × 2 = 0 + 0.904 205 799 102 792 293 584 732 16;
  • 32) 0.904 205 799 102 792 293 584 732 16 × 2 = 1 + 0.808 411 598 205 584 587 169 464 32;
  • 33) 0.808 411 598 205 584 587 169 464 32 × 2 = 1 + 0.616 823 196 411 169 174 338 928 64;
  • 34) 0.616 823 196 411 169 174 338 928 64 × 2 = 1 + 0.233 646 392 822 338 348 677 857 28;
  • 35) 0.233 646 392 822 338 348 677 857 28 × 2 = 0 + 0.467 292 785 644 676 697 355 714 56;
  • 36) 0.467 292 785 644 676 697 355 714 56 × 2 = 0 + 0.934 585 571 289 353 394 711 429 12;
  • 37) 0.934 585 571 289 353 394 711 429 12 × 2 = 1 + 0.869 171 142 578 706 789 422 858 24;
  • 38) 0.869 171 142 578 706 789 422 858 24 × 2 = 1 + 0.738 342 285 157 413 578 845 716 48;
  • 39) 0.738 342 285 157 413 578 845 716 48 × 2 = 1 + 0.476 684 570 314 827 157 691 432 96;
  • 40) 0.476 684 570 314 827 157 691 432 96 × 2 = 0 + 0.953 369 140 629 654 315 382 865 92;
  • 41) 0.953 369 140 629 654 315 382 865 92 × 2 = 1 + 0.906 738 281 259 308 630 765 731 84;
  • 42) 0.906 738 281 259 308 630 765 731 84 × 2 = 1 + 0.813 476 562 518 617 261 531 463 68;
  • 43) 0.813 476 562 518 617 261 531 463 68 × 2 = 1 + 0.626 953 125 037 234 523 062 927 36;
  • 44) 0.626 953 125 037 234 523 062 927 36 × 2 = 1 + 0.253 906 250 074 469 046 125 854 72;
  • 45) 0.253 906 250 074 469 046 125 854 72 × 2 = 0 + 0.507 812 500 148 938 092 251 709 44;
  • 46) 0.507 812 500 148 938 092 251 709 44 × 2 = 1 + 0.015 625 000 297 876 184 503 418 88;
  • 47) 0.015 625 000 297 876 184 503 418 88 × 2 = 0 + 0.031 250 000 595 752 369 006 837 76;
  • 48) 0.031 250 000 595 752 369 006 837 76 × 2 = 0 + 0.062 500 001 191 504 738 013 675 52;
  • 49) 0.062 500 001 191 504 738 013 675 52 × 2 = 0 + 0.125 000 002 383 009 476 027 351 04;
  • 50) 0.125 000 002 383 009 476 027 351 04 × 2 = 0 + 0.250 000 004 766 018 952 054 702 08;
  • 51) 0.250 000 004 766 018 952 054 702 08 × 2 = 0 + 0.500 000 009 532 037 904 109 404 16;
  • 52) 0.500 000 009 532 037 904 109 404 16 × 2 = 1 + 0.000 000 019 064 075 808 218 808 32;
  • 53) 0.000 000 019 064 075 808 218 808 32 × 2 = 0 + 0.000 000 038 128 151 616 437 616 64;

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:


0.745 459 324 169 999 826 281 700 42(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 700 42(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:


1.745 459 324 169 999 826 281 700 42(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 1.745 459 324 169 999 826 281 700 42 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