Decimal to 64 Bit IEEE 754 Binary: Convert Number 4.440 892 098 500 6 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 4.440 892 098 500 6(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 4.
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;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

4(10) =

100(2)


3. Convert to binary (base 2) the fractional part: 0.440 892 098 500 6.

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) 0.440 892 098 500 6 × 2 = 0 + 0.881 784 197 001 2;
  • 2) 0.881 784 197 001 2 × 2 = 1 + 0.763 568 394 002 4;
  • 3) 0.763 568 394 002 4 × 2 = 1 + 0.527 136 788 004 8;
  • 4) 0.527 136 788 004 8 × 2 = 1 + 0.054 273 576 009 6;
  • 5) 0.054 273 576 009 6 × 2 = 0 + 0.108 547 152 019 2;
  • 6) 0.108 547 152 019 2 × 2 = 0 + 0.217 094 304 038 4;
  • 7) 0.217 094 304 038 4 × 2 = 0 + 0.434 188 608 076 8;
  • 8) 0.434 188 608 076 8 × 2 = 0 + 0.868 377 216 153 6;
  • 9) 0.868 377 216 153 6 × 2 = 1 + 0.736 754 432 307 2;
  • 10) 0.736 754 432 307 2 × 2 = 1 + 0.473 508 864 614 4;
  • 11) 0.473 508 864 614 4 × 2 = 0 + 0.947 017 729 228 8;
  • 12) 0.947 017 729 228 8 × 2 = 1 + 0.894 035 458 457 6;
  • 13) 0.894 035 458 457 6 × 2 = 1 + 0.788 070 916 915 2;
  • 14) 0.788 070 916 915 2 × 2 = 1 + 0.576 141 833 830 4;
  • 15) 0.576 141 833 830 4 × 2 = 1 + 0.152 283 667 660 8;
  • 16) 0.152 283 667 660 8 × 2 = 0 + 0.304 567 335 321 6;
  • 17) 0.304 567 335 321 6 × 2 = 0 + 0.609 134 670 643 2;
  • 18) 0.609 134 670 643 2 × 2 = 1 + 0.218 269 341 286 4;
  • 19) 0.218 269 341 286 4 × 2 = 0 + 0.436 538 682 572 8;
  • 20) 0.436 538 682 572 8 × 2 = 0 + 0.873 077 365 145 6;
  • 21) 0.873 077 365 145 6 × 2 = 1 + 0.746 154 730 291 2;
  • 22) 0.746 154 730 291 2 × 2 = 1 + 0.492 309 460 582 4;
  • 23) 0.492 309 460 582 4 × 2 = 0 + 0.984 618 921 164 8;
  • 24) 0.984 618 921 164 8 × 2 = 1 + 0.969 237 842 329 6;
  • 25) 0.969 237 842 329 6 × 2 = 1 + 0.938 475 684 659 2;
  • 26) 0.938 475 684 659 2 × 2 = 1 + 0.876 951 369 318 4;
  • 27) 0.876 951 369 318 4 × 2 = 1 + 0.753 902 738 636 8;
  • 28) 0.753 902 738 636 8 × 2 = 1 + 0.507 805 477 273 6;
  • 29) 0.507 805 477 273 6 × 2 = 1 + 0.015 610 954 547 2;
  • 30) 0.015 610 954 547 2 × 2 = 0 + 0.031 221 909 094 4;
  • 31) 0.031 221 909 094 4 × 2 = 0 + 0.062 443 818 188 8;
  • 32) 0.062 443 818 188 8 × 2 = 0 + 0.124 887 636 377 6;
  • 33) 0.124 887 636 377 6 × 2 = 0 + 0.249 775 272 755 2;
  • 34) 0.249 775 272 755 2 × 2 = 0 + 0.499 550 545 510 4;
  • 35) 0.499 550 545 510 4 × 2 = 0 + 0.999 101 091 020 8;
  • 36) 0.999 101 091 020 8 × 2 = 1 + 0.998 202 182 041 6;
  • 37) 0.998 202 182 041 6 × 2 = 1 + 0.996 404 364 083 2;
  • 38) 0.996 404 364 083 2 × 2 = 1 + 0.992 808 728 166 4;
  • 39) 0.992 808 728 166 4 × 2 = 1 + 0.985 617 456 332 8;
  • 40) 0.985 617 456 332 8 × 2 = 1 + 0.971 234 912 665 6;
  • 41) 0.971 234 912 665 6 × 2 = 1 + 0.942 469 825 331 2;
  • 42) 0.942 469 825 331 2 × 2 = 1 + 0.884 939 650 662 4;
  • 43) 0.884 939 650 662 4 × 2 = 1 + 0.769 879 301 324 8;
  • 44) 0.769 879 301 324 8 × 2 = 1 + 0.539 758 602 649 6;
  • 45) 0.539 758 602 649 6 × 2 = 1 + 0.079 517 205 299 2;
  • 46) 0.079 517 205 299 2 × 2 = 0 + 0.159 034 410 598 4;
  • 47) 0.159 034 410 598 4 × 2 = 0 + 0.318 068 821 196 8;
  • 48) 0.318 068 821 196 8 × 2 = 0 + 0.636 137 642 393 6;
  • 49) 0.636 137 642 393 6 × 2 = 1 + 0.272 275 284 787 2;
  • 50) 0.272 275 284 787 2 × 2 = 0 + 0.544 550 569 574 4;
  • 51) 0.544 550 569 574 4 × 2 = 1 + 0.089 101 139 148 8;
  • 52) 0.089 101 139 148 8 × 2 = 0 + 0.178 202 278 297 6;
  • 53) 0.178 202 278 297 6 × 2 = 0 + 0.356 404 556 595 2;

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:


0.440 892 098 500 6(10) =

0.0111 0000 1101 1110 0100 1101 1111 1000 0001 1111 1111 1000 1010 0(2)

5. Positive number before normalization:

4.440 892 098 500 6(10) =

100.0111 0000 1101 1110 0100 1101 1111 1000 0001 1111 1111 1000 1010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


4.440 892 098 500 6(10) =

100.0111 0000 1101 1110 0100 1101 1111 1000 0001 1111 1111 1000 1010 0(2) =

100.0111 0000 1101 1110 0100 1101 1111 1000 0001 1111 1111 1000 1010 0(2) × 20 =

1.0001 1100 0011 0111 1001 0011 0111 1110 0000 0111 1111 1110 0010 100(2) × 22


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): 2

Mantissa (not normalized):
1.0001 1100 0011 0111 1001 0011 0111 1110 0000 0111 1111 1110 0010 100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1025(10) =

100 0000 0001(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. 0001 1100 0011 0111 1001 0011 0111 1110 0000 0111 1111 1110 0010 100 =

0001 1100 0011 0111 1001 0011 0111 1110 0000 0111 1111 1110 0010


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) =
100 0000 0001

Mantissa (52 bits) =
0001 1100 0011 0111 1001 0011 0111 1110 0000 0111 1111 1110 0010


The base ten decimal number 4.440 892 098 500 6 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 0001 1100 0011 0111 1001 0011 0111 1110 0000 0111 1111 1110 0010

» Decimal number in base ten 4.440 892 098 491 8 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

» New: All the Calculations Performed by Our Visitors: Decimal Numbers Converted and Written as 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 03, 2025 [March]: Decimal Numbers Converted and Written as 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Monthly Calculations performed during the month of: March - by our visitors

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