1.745 459 324 169 999 823 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 823 68(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
1.745 459 324 169 999 823 68(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: 0.745 459 324 169 999 823 68.

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) 0.745 459 324 169 999 823 68 × 2 = 1 + 0.490 918 648 339 999 647 36;
  • 2) 0.490 918 648 339 999 647 36 × 2 = 0 + 0.981 837 296 679 999 294 72;
  • 3) 0.981 837 296 679 999 294 72 × 2 = 1 + 0.963 674 593 359 998 589 44;
  • 4) 0.963 674 593 359 998 589 44 × 2 = 1 + 0.927 349 186 719 997 178 88;
  • 5) 0.927 349 186 719 997 178 88 × 2 = 1 + 0.854 698 373 439 994 357 76;
  • 6) 0.854 698 373 439 994 357 76 × 2 = 1 + 0.709 396 746 879 988 715 52;
  • 7) 0.709 396 746 879 988 715 52 × 2 = 1 + 0.418 793 493 759 977 431 04;
  • 8) 0.418 793 493 759 977 431 04 × 2 = 0 + 0.837 586 987 519 954 862 08;
  • 9) 0.837 586 987 519 954 862 08 × 2 = 1 + 0.675 173 975 039 909 724 16;
  • 10) 0.675 173 975 039 909 724 16 × 2 = 1 + 0.350 347 950 079 819 448 32;
  • 11) 0.350 347 950 079 819 448 32 × 2 = 0 + 0.700 695 900 159 638 896 64;
  • 12) 0.700 695 900 159 638 896 64 × 2 = 1 + 0.401 391 800 319 277 793 28;
  • 13) 0.401 391 800 319 277 793 28 × 2 = 0 + 0.802 783 600 638 555 586 56;
  • 14) 0.802 783 600 638 555 586 56 × 2 = 1 + 0.605 567 201 277 111 173 12;
  • 15) 0.605 567 201 277 111 173 12 × 2 = 1 + 0.211 134 402 554 222 346 24;
  • 16) 0.211 134 402 554 222 346 24 × 2 = 0 + 0.422 268 805 108 444 692 48;
  • 17) 0.422 268 805 108 444 692 48 × 2 = 0 + 0.844 537 610 216 889 384 96;
  • 18) 0.844 537 610 216 889 384 96 × 2 = 1 + 0.689 075 220 433 778 769 92;
  • 19) 0.689 075 220 433 778 769 92 × 2 = 1 + 0.378 150 440 867 557 539 84;
  • 20) 0.378 150 440 867 557 539 84 × 2 = 0 + 0.756 300 881 735 115 079 68;
  • 21) 0.756 300 881 735 115 079 68 × 2 = 1 + 0.512 601 763 470 230 159 36;
  • 22) 0.512 601 763 470 230 159 36 × 2 = 1 + 0.025 203 526 940 460 318 72;
  • 23) 0.025 203 526 940 460 318 72 × 2 = 0 + 0.050 407 053 880 920 637 44;
  • 24) 0.050 407 053 880 920 637 44 × 2 = 0 + 0.100 814 107 761 841 274 88;
  • 25) 0.100 814 107 761 841 274 88 × 2 = 0 + 0.201 628 215 523 682 549 76;
  • 26) 0.201 628 215 523 682 549 76 × 2 = 0 + 0.403 256 431 047 365 099 52;
  • 27) 0.403 256 431 047 365 099 52 × 2 = 0 + 0.806 512 862 094 730 199 04;
  • 28) 0.806 512 862 094 730 199 04 × 2 = 1 + 0.613 025 724 189 460 398 08;
  • 29) 0.613 025 724 189 460 398 08 × 2 = 1 + 0.226 051 448 378 920 796 16;
  • 30) 0.226 051 448 378 920 796 16 × 2 = 0 + 0.452 102 896 757 841 592 32;
  • 31) 0.452 102 896 757 841 592 32 × 2 = 0 + 0.904 205 793 515 683 184 64;
  • 32) 0.904 205 793 515 683 184 64 × 2 = 1 + 0.808 411 587 031 366 369 28;
  • 33) 0.808 411 587 031 366 369 28 × 2 = 1 + 0.616 823 174 062 732 738 56;
  • 34) 0.616 823 174 062 732 738 56 × 2 = 1 + 0.233 646 348 125 465 477 12;
  • 35) 0.233 646 348 125 465 477 12 × 2 = 0 + 0.467 292 696 250 930 954 24;
  • 36) 0.467 292 696 250 930 954 24 × 2 = 0 + 0.934 585 392 501 861 908 48;
  • 37) 0.934 585 392 501 861 908 48 × 2 = 1 + 0.869 170 785 003 723 816 96;
  • 38) 0.869 170 785 003 723 816 96 × 2 = 1 + 0.738 341 570 007 447 633 92;
  • 39) 0.738 341 570 007 447 633 92 × 2 = 1 + 0.476 683 140 014 895 267 84;
  • 40) 0.476 683 140 014 895 267 84 × 2 = 0 + 0.953 366 280 029 790 535 68;
  • 41) 0.953 366 280 029 790 535 68 × 2 = 1 + 0.906 732 560 059 581 071 36;
  • 42) 0.906 732 560 059 581 071 36 × 2 = 1 + 0.813 465 120 119 162 142 72;
  • 43) 0.813 465 120 119 162 142 72 × 2 = 1 + 0.626 930 240 238 324 285 44;
  • 44) 0.626 930 240 238 324 285 44 × 2 = 1 + 0.253 860 480 476 648 570 88;
  • 45) 0.253 860 480 476 648 570 88 × 2 = 0 + 0.507 720 960 953 297 141 76;
  • 46) 0.507 720 960 953 297 141 76 × 2 = 1 + 0.015 441 921 906 594 283 52;
  • 47) 0.015 441 921 906 594 283 52 × 2 = 0 + 0.030 883 843 813 188 567 04;
  • 48) 0.030 883 843 813 188 567 04 × 2 = 0 + 0.061 767 687 626 377 134 08;
  • 49) 0.061 767 687 626 377 134 08 × 2 = 0 + 0.123 535 375 252 754 268 16;
  • 50) 0.123 535 375 252 754 268 16 × 2 = 0 + 0.247 070 750 505 508 536 32;
  • 51) 0.247 070 750 505 508 536 32 × 2 = 0 + 0.494 141 501 011 017 072 64;
  • 52) 0.494 141 501 011 017 072 64 × 2 = 0 + 0.988 283 002 022 034 145 28;
  • 53) 0.988 283 002 022 034 145 28 × 2 = 1 + 0.976 566 004 044 068 290 56;

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:


0.745 459 324 169 999 823 68(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 823 68(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:


1.745 459 324 169 999 823 68(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 1.745 459 324 169 999 823 68 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