64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 134.555 555 555 555 555 556 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 134.555 555 555 555 555 556(10) converted and written in 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: 134.
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;
  • 134 ÷ 2 = 67 + 0;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 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.


134(10) =

1000 0110(2)


3. Convert to binary (base 2) the fractional part: 0.555 555 555 555 555 556.

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) 0.555 555 555 555 555 556 × 2 = 1 + 0.111 111 111 111 111 112;
  • 2) 0.111 111 111 111 111 112 × 2 = 0 + 0.222 222 222 222 222 224;
  • 3) 0.222 222 222 222 222 224 × 2 = 0 + 0.444 444 444 444 444 448;
  • 4) 0.444 444 444 444 444 448 × 2 = 0 + 0.888 888 888 888 888 896;
  • 5) 0.888 888 888 888 888 896 × 2 = 1 + 0.777 777 777 777 777 792;
  • 6) 0.777 777 777 777 777 792 × 2 = 1 + 0.555 555 555 555 555 584;
  • 7) 0.555 555 555 555 555 584 × 2 = 1 + 0.111 111 111 111 111 168;
  • 8) 0.111 111 111 111 111 168 × 2 = 0 + 0.222 222 222 222 222 336;
  • 9) 0.222 222 222 222 222 336 × 2 = 0 + 0.444 444 444 444 444 672;
  • 10) 0.444 444 444 444 444 672 × 2 = 0 + 0.888 888 888 888 889 344;
  • 11) 0.888 888 888 888 889 344 × 2 = 1 + 0.777 777 777 777 778 688;
  • 12) 0.777 777 777 777 778 688 × 2 = 1 + 0.555 555 555 555 557 376;
  • 13) 0.555 555 555 555 557 376 × 2 = 1 + 0.111 111 111 111 114 752;
  • 14) 0.111 111 111 111 114 752 × 2 = 0 + 0.222 222 222 222 229 504;
  • 15) 0.222 222 222 222 229 504 × 2 = 0 + 0.444 444 444 444 459 008;
  • 16) 0.444 444 444 444 459 008 × 2 = 0 + 0.888 888 888 888 918 016;
  • 17) 0.888 888 888 888 918 016 × 2 = 1 + 0.777 777 777 777 836 032;
  • 18) 0.777 777 777 777 836 032 × 2 = 1 + 0.555 555 555 555 672 064;
  • 19) 0.555 555 555 555 672 064 × 2 = 1 + 0.111 111 111 111 344 128;
  • 20) 0.111 111 111 111 344 128 × 2 = 0 + 0.222 222 222 222 688 256;
  • 21) 0.222 222 222 222 688 256 × 2 = 0 + 0.444 444 444 445 376 512;
  • 22) 0.444 444 444 445 376 512 × 2 = 0 + 0.888 888 888 890 753 024;
  • 23) 0.888 888 888 890 753 024 × 2 = 1 + 0.777 777 777 781 506 048;
  • 24) 0.777 777 777 781 506 048 × 2 = 1 + 0.555 555 555 563 012 096;
  • 25) 0.555 555 555 563 012 096 × 2 = 1 + 0.111 111 111 126 024 192;
  • 26) 0.111 111 111 126 024 192 × 2 = 0 + 0.222 222 222 252 048 384;
  • 27) 0.222 222 222 252 048 384 × 2 = 0 + 0.444 444 444 504 096 768;
  • 28) 0.444 444 444 504 096 768 × 2 = 0 + 0.888 888 889 008 193 536;
  • 29) 0.888 888 889 008 193 536 × 2 = 1 + 0.777 777 778 016 387 072;
  • 30) 0.777 777 778 016 387 072 × 2 = 1 + 0.555 555 556 032 774 144;
  • 31) 0.555 555 556 032 774 144 × 2 = 1 + 0.111 111 112 065 548 288;
  • 32) 0.111 111 112 065 548 288 × 2 = 0 + 0.222 222 224 131 096 576;
  • 33) 0.222 222 224 131 096 576 × 2 = 0 + 0.444 444 448 262 193 152;
  • 34) 0.444 444 448 262 193 152 × 2 = 0 + 0.888 888 896 524 386 304;
  • 35) 0.888 888 896 524 386 304 × 2 = 1 + 0.777 777 793 048 772 608;
  • 36) 0.777 777 793 048 772 608 × 2 = 1 + 0.555 555 586 097 545 216;
  • 37) 0.555 555 586 097 545 216 × 2 = 1 + 0.111 111 172 195 090 432;
  • 38) 0.111 111 172 195 090 432 × 2 = 0 + 0.222 222 344 390 180 864;
  • 39) 0.222 222 344 390 180 864 × 2 = 0 + 0.444 444 688 780 361 728;
  • 40) 0.444 444 688 780 361 728 × 2 = 0 + 0.888 889 377 560 723 456;
  • 41) 0.888 889 377 560 723 456 × 2 = 1 + 0.777 778 755 121 446 912;
  • 42) 0.777 778 755 121 446 912 × 2 = 1 + 0.555 557 510 242 893 824;
  • 43) 0.555 557 510 242 893 824 × 2 = 1 + 0.111 115 020 485 787 648;
  • 44) 0.111 115 020 485 787 648 × 2 = 0 + 0.222 230 040 971 575 296;
  • 45) 0.222 230 040 971 575 296 × 2 = 0 + 0.444 460 081 943 150 592;
  • 46) 0.444 460 081 943 150 592 × 2 = 0 + 0.888 920 163 886 301 184;
  • 47) 0.888 920 163 886 301 184 × 2 = 1 + 0.777 840 327 772 602 368;
  • 48) 0.777 840 327 772 602 368 × 2 = 1 + 0.555 680 655 545 204 736;
  • 49) 0.555 680 655 545 204 736 × 2 = 1 + 0.111 361 311 090 409 472;
  • 50) 0.111 361 311 090 409 472 × 2 = 0 + 0.222 722 622 180 818 944;
  • 51) 0.222 722 622 180 818 944 × 2 = 0 + 0.445 445 244 361 637 888;
  • 52) 0.445 445 244 361 637 888 × 2 = 0 + 0.890 890 488 723 275 776;
  • 53) 0.890 890 488 723 275 776 × 2 = 1 + 0.781 780 977 446 551 552;

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...)


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:


0.555 555 555 555 555 556(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2)


5. Positive number before normalization:

134.555 555 555 555 555 556(10) =

1000 0110.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


134.555 555 555 555 555 556(10) =

1000 0110.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) =

1000 0110.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) × 20 =

1.0000 1101 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001(2) × 27


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): 7

Mantissa (not normalized):
1.0000 1101 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

7 + 2(11-1) - 1 =

(7 + 1 023)(10) =

1 030(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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 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) =

1030(10) =

100 0000 0110(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. 0000 1101 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 =

0000 1101 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 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) =
100 0000 0110

Mantissa (52 bits) =
0000 1101 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100


The base ten decimal number 134.555 555 555 555 555 556 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0110 - 0000 1101 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 134.555 555 555 555 555 555 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 134.555 555 555 555 555 557 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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