1.745 459 324 169 999 827 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 827 59(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
1.745 459 324 169 999 827 59(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: 0.745 459 324 169 999 827 59.

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) 0.745 459 324 169 999 827 59 × 2 = 1 + 0.490 918 648 339 999 655 18;
  • 2) 0.490 918 648 339 999 655 18 × 2 = 0 + 0.981 837 296 679 999 310 36;
  • 3) 0.981 837 296 679 999 310 36 × 2 = 1 + 0.963 674 593 359 998 620 72;
  • 4) 0.963 674 593 359 998 620 72 × 2 = 1 + 0.927 349 186 719 997 241 44;
  • 5) 0.927 349 186 719 997 241 44 × 2 = 1 + 0.854 698 373 439 994 482 88;
  • 6) 0.854 698 373 439 994 482 88 × 2 = 1 + 0.709 396 746 879 988 965 76;
  • 7) 0.709 396 746 879 988 965 76 × 2 = 1 + 0.418 793 493 759 977 931 52;
  • 8) 0.418 793 493 759 977 931 52 × 2 = 0 + 0.837 586 987 519 955 863 04;
  • 9) 0.837 586 987 519 955 863 04 × 2 = 1 + 0.675 173 975 039 911 726 08;
  • 10) 0.675 173 975 039 911 726 08 × 2 = 1 + 0.350 347 950 079 823 452 16;
  • 11) 0.350 347 950 079 823 452 16 × 2 = 0 + 0.700 695 900 159 646 904 32;
  • 12) 0.700 695 900 159 646 904 32 × 2 = 1 + 0.401 391 800 319 293 808 64;
  • 13) 0.401 391 800 319 293 808 64 × 2 = 0 + 0.802 783 600 638 587 617 28;
  • 14) 0.802 783 600 638 587 617 28 × 2 = 1 + 0.605 567 201 277 175 234 56;
  • 15) 0.605 567 201 277 175 234 56 × 2 = 1 + 0.211 134 402 554 350 469 12;
  • 16) 0.211 134 402 554 350 469 12 × 2 = 0 + 0.422 268 805 108 700 938 24;
  • 17) 0.422 268 805 108 700 938 24 × 2 = 0 + 0.844 537 610 217 401 876 48;
  • 18) 0.844 537 610 217 401 876 48 × 2 = 1 + 0.689 075 220 434 803 752 96;
  • 19) 0.689 075 220 434 803 752 96 × 2 = 1 + 0.378 150 440 869 607 505 92;
  • 20) 0.378 150 440 869 607 505 92 × 2 = 0 + 0.756 300 881 739 215 011 84;
  • 21) 0.756 300 881 739 215 011 84 × 2 = 1 + 0.512 601 763 478 430 023 68;
  • 22) 0.512 601 763 478 430 023 68 × 2 = 1 + 0.025 203 526 956 860 047 36;
  • 23) 0.025 203 526 956 860 047 36 × 2 = 0 + 0.050 407 053 913 720 094 72;
  • 24) 0.050 407 053 913 720 094 72 × 2 = 0 + 0.100 814 107 827 440 189 44;
  • 25) 0.100 814 107 827 440 189 44 × 2 = 0 + 0.201 628 215 654 880 378 88;
  • 26) 0.201 628 215 654 880 378 88 × 2 = 0 + 0.403 256 431 309 760 757 76;
  • 27) 0.403 256 431 309 760 757 76 × 2 = 0 + 0.806 512 862 619 521 515 52;
  • 28) 0.806 512 862 619 521 515 52 × 2 = 1 + 0.613 025 725 239 043 031 04;
  • 29) 0.613 025 725 239 043 031 04 × 2 = 1 + 0.226 051 450 478 086 062 08;
  • 30) 0.226 051 450 478 086 062 08 × 2 = 0 + 0.452 102 900 956 172 124 16;
  • 31) 0.452 102 900 956 172 124 16 × 2 = 0 + 0.904 205 801 912 344 248 32;
  • 32) 0.904 205 801 912 344 248 32 × 2 = 1 + 0.808 411 603 824 688 496 64;
  • 33) 0.808 411 603 824 688 496 64 × 2 = 1 + 0.616 823 207 649 376 993 28;
  • 34) 0.616 823 207 649 376 993 28 × 2 = 1 + 0.233 646 415 298 753 986 56;
  • 35) 0.233 646 415 298 753 986 56 × 2 = 0 + 0.467 292 830 597 507 973 12;
  • 36) 0.467 292 830 597 507 973 12 × 2 = 0 + 0.934 585 661 195 015 946 24;
  • 37) 0.934 585 661 195 015 946 24 × 2 = 1 + 0.869 171 322 390 031 892 48;
  • 38) 0.869 171 322 390 031 892 48 × 2 = 1 + 0.738 342 644 780 063 784 96;
  • 39) 0.738 342 644 780 063 784 96 × 2 = 1 + 0.476 685 289 560 127 569 92;
  • 40) 0.476 685 289 560 127 569 92 × 2 = 0 + 0.953 370 579 120 255 139 84;
  • 41) 0.953 370 579 120 255 139 84 × 2 = 1 + 0.906 741 158 240 510 279 68;
  • 42) 0.906 741 158 240 510 279 68 × 2 = 1 + 0.813 482 316 481 020 559 36;
  • 43) 0.813 482 316 481 020 559 36 × 2 = 1 + 0.626 964 632 962 041 118 72;
  • 44) 0.626 964 632 962 041 118 72 × 2 = 1 + 0.253 929 265 924 082 237 44;
  • 45) 0.253 929 265 924 082 237 44 × 2 = 0 + 0.507 858 531 848 164 474 88;
  • 46) 0.507 858 531 848 164 474 88 × 2 = 1 + 0.015 717 063 696 328 949 76;
  • 47) 0.015 717 063 696 328 949 76 × 2 = 0 + 0.031 434 127 392 657 899 52;
  • 48) 0.031 434 127 392 657 899 52 × 2 = 0 + 0.062 868 254 785 315 799 04;
  • 49) 0.062 868 254 785 315 799 04 × 2 = 0 + 0.125 736 509 570 631 598 08;
  • 50) 0.125 736 509 570 631 598 08 × 2 = 0 + 0.251 473 019 141 263 196 16;
  • 51) 0.251 473 019 141 263 196 16 × 2 = 0 + 0.502 946 038 282 526 392 32;
  • 52) 0.502 946 038 282 526 392 32 × 2 = 1 + 0.005 892 076 565 052 784 64;
  • 53) 0.005 892 076 565 052 784 64 × 2 = 0 + 0.011 784 153 130 105 569 28;

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:


0.745 459 324 169 999 827 59(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 827 59(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:


1.745 459 324 169 999 827 59(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 1.745 459 324 169 999 827 59 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