1.732 050 808 198 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.732 050 808 198(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
1.732 050 808 198(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: 0.732 050 808 198.

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) 0.732 050 808 198 × 2 = 1 + 0.464 101 616 396;
  • 2) 0.464 101 616 396 × 2 = 0 + 0.928 203 232 792;
  • 3) 0.928 203 232 792 × 2 = 1 + 0.856 406 465 584;
  • 4) 0.856 406 465 584 × 2 = 1 + 0.712 812 931 168;
  • 5) 0.712 812 931 168 × 2 = 1 + 0.425 625 862 336;
  • 6) 0.425 625 862 336 × 2 = 0 + 0.851 251 724 672;
  • 7) 0.851 251 724 672 × 2 = 1 + 0.702 503 449 344;
  • 8) 0.702 503 449 344 × 2 = 1 + 0.405 006 898 688;
  • 9) 0.405 006 898 688 × 2 = 0 + 0.810 013 797 376;
  • 10) 0.810 013 797 376 × 2 = 1 + 0.620 027 594 752;
  • 11) 0.620 027 594 752 × 2 = 1 + 0.240 055 189 504;
  • 12) 0.240 055 189 504 × 2 = 0 + 0.480 110 379 008;
  • 13) 0.480 110 379 008 × 2 = 0 + 0.960 220 758 016;
  • 14) 0.960 220 758 016 × 2 = 1 + 0.920 441 516 032;
  • 15) 0.920 441 516 032 × 2 = 1 + 0.840 883 032 064;
  • 16) 0.840 883 032 064 × 2 = 1 + 0.681 766 064 128;
  • 17) 0.681 766 064 128 × 2 = 1 + 0.363 532 128 256;
  • 18) 0.363 532 128 256 × 2 = 0 + 0.727 064 256 512;
  • 19) 0.727 064 256 512 × 2 = 1 + 0.454 128 513 024;
  • 20) 0.454 128 513 024 × 2 = 0 + 0.908 257 026 048;
  • 21) 0.908 257 026 048 × 2 = 1 + 0.816 514 052 096;
  • 22) 0.816 514 052 096 × 2 = 1 + 0.633 028 104 192;
  • 23) 0.633 028 104 192 × 2 = 1 + 0.266 056 208 384;
  • 24) 0.266 056 208 384 × 2 = 0 + 0.532 112 416 768;
  • 25) 0.532 112 416 768 × 2 = 1 + 0.064 224 833 536;
  • 26) 0.064 224 833 536 × 2 = 0 + 0.128 449 667 072;
  • 27) 0.128 449 667 072 × 2 = 0 + 0.256 899 334 144;
  • 28) 0.256 899 334 144 × 2 = 0 + 0.513 798 668 288;
  • 29) 0.513 798 668 288 × 2 = 1 + 0.027 597 336 576;
  • 30) 0.027 597 336 576 × 2 = 0 + 0.055 194 673 152;
  • 31) 0.055 194 673 152 × 2 = 0 + 0.110 389 346 304;
  • 32) 0.110 389 346 304 × 2 = 0 + 0.220 778 692 608;
  • 33) 0.220 778 692 608 × 2 = 0 + 0.441 557 385 216;
  • 34) 0.441 557 385 216 × 2 = 0 + 0.883 114 770 432;
  • 35) 0.883 114 770 432 × 2 = 1 + 0.766 229 540 864;
  • 36) 0.766 229 540 864 × 2 = 1 + 0.532 459 081 728;
  • 37) 0.532 459 081 728 × 2 = 1 + 0.064 918 163 456;
  • 38) 0.064 918 163 456 × 2 = 0 + 0.129 836 326 912;
  • 39) 0.129 836 326 912 × 2 = 0 + 0.259 672 653 824;
  • 40) 0.259 672 653 824 × 2 = 0 + 0.519 345 307 648;
  • 41) 0.519 345 307 648 × 2 = 1 + 0.038 690 615 296;
  • 42) 0.038 690 615 296 × 2 = 0 + 0.077 381 230 592;
  • 43) 0.077 381 230 592 × 2 = 0 + 0.154 762 461 184;
  • 44) 0.154 762 461 184 × 2 = 0 + 0.309 524 922 368;
  • 45) 0.309 524 922 368 × 2 = 0 + 0.619 049 844 736;
  • 46) 0.619 049 844 736 × 2 = 1 + 0.238 099 689 472;
  • 47) 0.238 099 689 472 × 2 = 0 + 0.476 199 378 944;
  • 48) 0.476 199 378 944 × 2 = 0 + 0.952 398 757 888;
  • 49) 0.952 398 757 888 × 2 = 1 + 0.904 797 515 776;
  • 50) 0.904 797 515 776 × 2 = 1 + 0.809 595 031 552;
  • 51) 0.809 595 031 552 × 2 = 1 + 0.619 190 063 104;
  • 52) 0.619 190 063 104 × 2 = 1 + 0.238 380 126 208;
  • 53) 0.238 380 126 208 × 2 = 0 + 0.476 760 252 416;

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:


0.732 050 808 198(10) =

0.1011 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111 0(2)

5. Positive number before normalization:

1.732 050 808 198(10) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111 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:


1.732 050 808 198(10) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111 0(2) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111 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 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111 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 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111 0 =

1011 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111


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 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111


Decimal number 1.732 050 808 198 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1011 0110 0111 1010 1110 1000 1000 0011 1000 1000 0100 1111

» Convert decimal 1.732 050 808 227 to 64 bit double precision IEEE 754 binary floating point representation standard

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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