1.745 459 324 169 998 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 998 93(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
1.745 459 324 169 998 93(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: 0.745 459 324 169 998 93.

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) 0.745 459 324 169 998 93 × 2 = 1 + 0.490 918 648 339 997 86;
  • 2) 0.490 918 648 339 997 86 × 2 = 0 + 0.981 837 296 679 995 72;
  • 3) 0.981 837 296 679 995 72 × 2 = 1 + 0.963 674 593 359 991 44;
  • 4) 0.963 674 593 359 991 44 × 2 = 1 + 0.927 349 186 719 982 88;
  • 5) 0.927 349 186 719 982 88 × 2 = 1 + 0.854 698 373 439 965 76;
  • 6) 0.854 698 373 439 965 76 × 2 = 1 + 0.709 396 746 879 931 52;
  • 7) 0.709 396 746 879 931 52 × 2 = 1 + 0.418 793 493 759 863 04;
  • 8) 0.418 793 493 759 863 04 × 2 = 0 + 0.837 586 987 519 726 08;
  • 9) 0.837 586 987 519 726 08 × 2 = 1 + 0.675 173 975 039 452 16;
  • 10) 0.675 173 975 039 452 16 × 2 = 1 + 0.350 347 950 078 904 32;
  • 11) 0.350 347 950 078 904 32 × 2 = 0 + 0.700 695 900 157 808 64;
  • 12) 0.700 695 900 157 808 64 × 2 = 1 + 0.401 391 800 315 617 28;
  • 13) 0.401 391 800 315 617 28 × 2 = 0 + 0.802 783 600 631 234 56;
  • 14) 0.802 783 600 631 234 56 × 2 = 1 + 0.605 567 201 262 469 12;
  • 15) 0.605 567 201 262 469 12 × 2 = 1 + 0.211 134 402 524 938 24;
  • 16) 0.211 134 402 524 938 24 × 2 = 0 + 0.422 268 805 049 876 48;
  • 17) 0.422 268 805 049 876 48 × 2 = 0 + 0.844 537 610 099 752 96;
  • 18) 0.844 537 610 099 752 96 × 2 = 1 + 0.689 075 220 199 505 92;
  • 19) 0.689 075 220 199 505 92 × 2 = 1 + 0.378 150 440 399 011 84;
  • 20) 0.378 150 440 399 011 84 × 2 = 0 + 0.756 300 880 798 023 68;
  • 21) 0.756 300 880 798 023 68 × 2 = 1 + 0.512 601 761 596 047 36;
  • 22) 0.512 601 761 596 047 36 × 2 = 1 + 0.025 203 523 192 094 72;
  • 23) 0.025 203 523 192 094 72 × 2 = 0 + 0.050 407 046 384 189 44;
  • 24) 0.050 407 046 384 189 44 × 2 = 0 + 0.100 814 092 768 378 88;
  • 25) 0.100 814 092 768 378 88 × 2 = 0 + 0.201 628 185 536 757 76;
  • 26) 0.201 628 185 536 757 76 × 2 = 0 + 0.403 256 371 073 515 52;
  • 27) 0.403 256 371 073 515 52 × 2 = 0 + 0.806 512 742 147 031 04;
  • 28) 0.806 512 742 147 031 04 × 2 = 1 + 0.613 025 484 294 062 08;
  • 29) 0.613 025 484 294 062 08 × 2 = 1 + 0.226 050 968 588 124 16;
  • 30) 0.226 050 968 588 124 16 × 2 = 0 + 0.452 101 937 176 248 32;
  • 31) 0.452 101 937 176 248 32 × 2 = 0 + 0.904 203 874 352 496 64;
  • 32) 0.904 203 874 352 496 64 × 2 = 1 + 0.808 407 748 704 993 28;
  • 33) 0.808 407 748 704 993 28 × 2 = 1 + 0.616 815 497 409 986 56;
  • 34) 0.616 815 497 409 986 56 × 2 = 1 + 0.233 630 994 819 973 12;
  • 35) 0.233 630 994 819 973 12 × 2 = 0 + 0.467 261 989 639 946 24;
  • 36) 0.467 261 989 639 946 24 × 2 = 0 + 0.934 523 979 279 892 48;
  • 37) 0.934 523 979 279 892 48 × 2 = 1 + 0.869 047 958 559 784 96;
  • 38) 0.869 047 958 559 784 96 × 2 = 1 + 0.738 095 917 119 569 92;
  • 39) 0.738 095 917 119 569 92 × 2 = 1 + 0.476 191 834 239 139 84;
  • 40) 0.476 191 834 239 139 84 × 2 = 0 + 0.952 383 668 478 279 68;
  • 41) 0.952 383 668 478 279 68 × 2 = 1 + 0.904 767 336 956 559 36;
  • 42) 0.904 767 336 956 559 36 × 2 = 1 + 0.809 534 673 913 118 72;
  • 43) 0.809 534 673 913 118 72 × 2 = 1 + 0.619 069 347 826 237 44;
  • 44) 0.619 069 347 826 237 44 × 2 = 1 + 0.238 138 695 652 474 88;
  • 45) 0.238 138 695 652 474 88 × 2 = 0 + 0.476 277 391 304 949 76;
  • 46) 0.476 277 391 304 949 76 × 2 = 0 + 0.952 554 782 609 899 52;
  • 47) 0.952 554 782 609 899 52 × 2 = 1 + 0.905 109 565 219 799 04;
  • 48) 0.905 109 565 219 799 04 × 2 = 1 + 0.810 219 130 439 598 08;
  • 49) 0.810 219 130 439 598 08 × 2 = 1 + 0.620 438 260 879 196 16;
  • 50) 0.620 438 260 879 196 16 × 2 = 1 + 0.240 876 521 758 392 32;
  • 51) 0.240 876 521 758 392 32 × 2 = 0 + 0.481 753 043 516 784 64;
  • 52) 0.481 753 043 516 784 64 × 2 = 0 + 0.963 506 087 033 569 28;
  • 53) 0.963 506 087 033 569 28 × 2 = 1 + 0.927 012 174 067 138 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:


0.745 459 324 169 998 93(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0011 1100 1(2)

5. Positive number before normalization:

1.745 459 324 169 998 93(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0011 1100 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:


1.745 459 324 169 998 93(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0011 1100 1(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0011 1100 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 0011 1100 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 0011 1100 1 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0011 1100


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 0011 1100


Decimal number 1.745 459 324 169 998 93 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 0011 1100

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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