1.745 459 324 169 999 828 86 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 828 86(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
1.745 459 324 169 999 828 86(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: 0.745 459 324 169 999 828 86.

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) 0.745 459 324 169 999 828 86 × 2 = 1 + 0.490 918 648 339 999 657 72;
  • 2) 0.490 918 648 339 999 657 72 × 2 = 0 + 0.981 837 296 679 999 315 44;
  • 3) 0.981 837 296 679 999 315 44 × 2 = 1 + 0.963 674 593 359 998 630 88;
  • 4) 0.963 674 593 359 998 630 88 × 2 = 1 + 0.927 349 186 719 997 261 76;
  • 5) 0.927 349 186 719 997 261 76 × 2 = 1 + 0.854 698 373 439 994 523 52;
  • 6) 0.854 698 373 439 994 523 52 × 2 = 1 + 0.709 396 746 879 989 047 04;
  • 7) 0.709 396 746 879 989 047 04 × 2 = 1 + 0.418 793 493 759 978 094 08;
  • 8) 0.418 793 493 759 978 094 08 × 2 = 0 + 0.837 586 987 519 956 188 16;
  • 9) 0.837 586 987 519 956 188 16 × 2 = 1 + 0.675 173 975 039 912 376 32;
  • 10) 0.675 173 975 039 912 376 32 × 2 = 1 + 0.350 347 950 079 824 752 64;
  • 11) 0.350 347 950 079 824 752 64 × 2 = 0 + 0.700 695 900 159 649 505 28;
  • 12) 0.700 695 900 159 649 505 28 × 2 = 1 + 0.401 391 800 319 299 010 56;
  • 13) 0.401 391 800 319 299 010 56 × 2 = 0 + 0.802 783 600 638 598 021 12;
  • 14) 0.802 783 600 638 598 021 12 × 2 = 1 + 0.605 567 201 277 196 042 24;
  • 15) 0.605 567 201 277 196 042 24 × 2 = 1 + 0.211 134 402 554 392 084 48;
  • 16) 0.211 134 402 554 392 084 48 × 2 = 0 + 0.422 268 805 108 784 168 96;
  • 17) 0.422 268 805 108 784 168 96 × 2 = 0 + 0.844 537 610 217 568 337 92;
  • 18) 0.844 537 610 217 568 337 92 × 2 = 1 + 0.689 075 220 435 136 675 84;
  • 19) 0.689 075 220 435 136 675 84 × 2 = 1 + 0.378 150 440 870 273 351 68;
  • 20) 0.378 150 440 870 273 351 68 × 2 = 0 + 0.756 300 881 740 546 703 36;
  • 21) 0.756 300 881 740 546 703 36 × 2 = 1 + 0.512 601 763 481 093 406 72;
  • 22) 0.512 601 763 481 093 406 72 × 2 = 1 + 0.025 203 526 962 186 813 44;
  • 23) 0.025 203 526 962 186 813 44 × 2 = 0 + 0.050 407 053 924 373 626 88;
  • 24) 0.050 407 053 924 373 626 88 × 2 = 0 + 0.100 814 107 848 747 253 76;
  • 25) 0.100 814 107 848 747 253 76 × 2 = 0 + 0.201 628 215 697 494 507 52;
  • 26) 0.201 628 215 697 494 507 52 × 2 = 0 + 0.403 256 431 394 989 015 04;
  • 27) 0.403 256 431 394 989 015 04 × 2 = 0 + 0.806 512 862 789 978 030 08;
  • 28) 0.806 512 862 789 978 030 08 × 2 = 1 + 0.613 025 725 579 956 060 16;
  • 29) 0.613 025 725 579 956 060 16 × 2 = 1 + 0.226 051 451 159 912 120 32;
  • 30) 0.226 051 451 159 912 120 32 × 2 = 0 + 0.452 102 902 319 824 240 64;
  • 31) 0.452 102 902 319 824 240 64 × 2 = 0 + 0.904 205 804 639 648 481 28;
  • 32) 0.904 205 804 639 648 481 28 × 2 = 1 + 0.808 411 609 279 296 962 56;
  • 33) 0.808 411 609 279 296 962 56 × 2 = 1 + 0.616 823 218 558 593 925 12;
  • 34) 0.616 823 218 558 593 925 12 × 2 = 1 + 0.233 646 437 117 187 850 24;
  • 35) 0.233 646 437 117 187 850 24 × 2 = 0 + 0.467 292 874 234 375 700 48;
  • 36) 0.467 292 874 234 375 700 48 × 2 = 0 + 0.934 585 748 468 751 400 96;
  • 37) 0.934 585 748 468 751 400 96 × 2 = 1 + 0.869 171 496 937 502 801 92;
  • 38) 0.869 171 496 937 502 801 92 × 2 = 1 + 0.738 342 993 875 005 603 84;
  • 39) 0.738 342 993 875 005 603 84 × 2 = 1 + 0.476 685 987 750 011 207 68;
  • 40) 0.476 685 987 750 011 207 68 × 2 = 0 + 0.953 371 975 500 022 415 36;
  • 41) 0.953 371 975 500 022 415 36 × 2 = 1 + 0.906 743 951 000 044 830 72;
  • 42) 0.906 743 951 000 044 830 72 × 2 = 1 + 0.813 487 902 000 089 661 44;
  • 43) 0.813 487 902 000 089 661 44 × 2 = 1 + 0.626 975 804 000 179 322 88;
  • 44) 0.626 975 804 000 179 322 88 × 2 = 1 + 0.253 951 608 000 358 645 76;
  • 45) 0.253 951 608 000 358 645 76 × 2 = 0 + 0.507 903 216 000 717 291 52;
  • 46) 0.507 903 216 000 717 291 52 × 2 = 1 + 0.015 806 432 001 434 583 04;
  • 47) 0.015 806 432 001 434 583 04 × 2 = 0 + 0.031 612 864 002 869 166 08;
  • 48) 0.031 612 864 002 869 166 08 × 2 = 0 + 0.063 225 728 005 738 332 16;
  • 49) 0.063 225 728 005 738 332 16 × 2 = 0 + 0.126 451 456 011 476 664 32;
  • 50) 0.126 451 456 011 476 664 32 × 2 = 0 + 0.252 902 912 022 953 328 64;
  • 51) 0.252 902 912 022 953 328 64 × 2 = 0 + 0.505 805 824 045 906 657 28;
  • 52) 0.505 805 824 045 906 657 28 × 2 = 1 + 0.011 611 648 091 813 314 56;
  • 53) 0.011 611 648 091 813 314 56 × 2 = 0 + 0.023 223 296 183 626 629 12;

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:


0.745 459 324 169 999 828 86(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 828 86(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:


1.745 459 324 169 999 828 86(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 1.745 459 324 169 999 828 86 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