-0.145 067 813 487 901 050 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 57(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
-0.145 067 813 487 901 050 57(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. Start with the positive version of the number:

|-0.145 067 813 487 901 050 57| = 0.145 067 813 487 901 050 57


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

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.145 067 813 487 901 050 57.

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) 0.145 067 813 487 901 050 57 × 2 = 0 + 0.290 135 626 975 802 101 14;
  • 2) 0.290 135 626 975 802 101 14 × 2 = 0 + 0.580 271 253 951 604 202 28;
  • 3) 0.580 271 253 951 604 202 28 × 2 = 1 + 0.160 542 507 903 208 404 56;
  • 4) 0.160 542 507 903 208 404 56 × 2 = 0 + 0.321 085 015 806 416 809 12;
  • 5) 0.321 085 015 806 416 809 12 × 2 = 0 + 0.642 170 031 612 833 618 24;
  • 6) 0.642 170 031 612 833 618 24 × 2 = 1 + 0.284 340 063 225 667 236 48;
  • 7) 0.284 340 063 225 667 236 48 × 2 = 0 + 0.568 680 126 451 334 472 96;
  • 8) 0.568 680 126 451 334 472 96 × 2 = 1 + 0.137 360 252 902 668 945 92;
  • 9) 0.137 360 252 902 668 945 92 × 2 = 0 + 0.274 720 505 805 337 891 84;
  • 10) 0.274 720 505 805 337 891 84 × 2 = 0 + 0.549 441 011 610 675 783 68;
  • 11) 0.549 441 011 610 675 783 68 × 2 = 1 + 0.098 882 023 221 351 567 36;
  • 12) 0.098 882 023 221 351 567 36 × 2 = 0 + 0.197 764 046 442 703 134 72;
  • 13) 0.197 764 046 442 703 134 72 × 2 = 0 + 0.395 528 092 885 406 269 44;
  • 14) 0.395 528 092 885 406 269 44 × 2 = 0 + 0.791 056 185 770 812 538 88;
  • 15) 0.791 056 185 770 812 538 88 × 2 = 1 + 0.582 112 371 541 625 077 76;
  • 16) 0.582 112 371 541 625 077 76 × 2 = 1 + 0.164 224 743 083 250 155 52;
  • 17) 0.164 224 743 083 250 155 52 × 2 = 0 + 0.328 449 486 166 500 311 04;
  • 18) 0.328 449 486 166 500 311 04 × 2 = 0 + 0.656 898 972 333 000 622 08;
  • 19) 0.656 898 972 333 000 622 08 × 2 = 1 + 0.313 797 944 666 001 244 16;
  • 20) 0.313 797 944 666 001 244 16 × 2 = 0 + 0.627 595 889 332 002 488 32;
  • 21) 0.627 595 889 332 002 488 32 × 2 = 1 + 0.255 191 778 664 004 976 64;
  • 22) 0.255 191 778 664 004 976 64 × 2 = 0 + 0.510 383 557 328 009 953 28;
  • 23) 0.510 383 557 328 009 953 28 × 2 = 1 + 0.020 767 114 656 019 906 56;
  • 24) 0.020 767 114 656 019 906 56 × 2 = 0 + 0.041 534 229 312 039 813 12;
  • 25) 0.041 534 229 312 039 813 12 × 2 = 0 + 0.083 068 458 624 079 626 24;
  • 26) 0.083 068 458 624 079 626 24 × 2 = 0 + 0.166 136 917 248 159 252 48;
  • 27) 0.166 136 917 248 159 252 48 × 2 = 0 + 0.332 273 834 496 318 504 96;
  • 28) 0.332 273 834 496 318 504 96 × 2 = 0 + 0.664 547 668 992 637 009 92;
  • 29) 0.664 547 668 992 637 009 92 × 2 = 1 + 0.329 095 337 985 274 019 84;
  • 30) 0.329 095 337 985 274 019 84 × 2 = 0 + 0.658 190 675 970 548 039 68;
  • 31) 0.658 190 675 970 548 039 68 × 2 = 1 + 0.316 381 351 941 096 079 36;
  • 32) 0.316 381 351 941 096 079 36 × 2 = 0 + 0.632 762 703 882 192 158 72;
  • 33) 0.632 762 703 882 192 158 72 × 2 = 1 + 0.265 525 407 764 384 317 44;
  • 34) 0.265 525 407 764 384 317 44 × 2 = 0 + 0.531 050 815 528 768 634 88;
  • 35) 0.531 050 815 528 768 634 88 × 2 = 1 + 0.062 101 631 057 537 269 76;
  • 36) 0.062 101 631 057 537 269 76 × 2 = 0 + 0.124 203 262 115 074 539 52;
  • 37) 0.124 203 262 115 074 539 52 × 2 = 0 + 0.248 406 524 230 149 079 04;
  • 38) 0.248 406 524 230 149 079 04 × 2 = 0 + 0.496 813 048 460 298 158 08;
  • 39) 0.496 813 048 460 298 158 08 × 2 = 0 + 0.993 626 096 920 596 316 16;
  • 40) 0.993 626 096 920 596 316 16 × 2 = 1 + 0.987 252 193 841 192 632 32;
  • 41) 0.987 252 193 841 192 632 32 × 2 = 1 + 0.974 504 387 682 385 264 64;
  • 42) 0.974 504 387 682 385 264 64 × 2 = 1 + 0.949 008 775 364 770 529 28;
  • 43) 0.949 008 775 364 770 529 28 × 2 = 1 + 0.898 017 550 729 541 058 56;
  • 44) 0.898 017 550 729 541 058 56 × 2 = 1 + 0.796 035 101 459 082 117 12;
  • 45) 0.796 035 101 459 082 117 12 × 2 = 1 + 0.592 070 202 918 164 234 24;
  • 46) 0.592 070 202 918 164 234 24 × 2 = 1 + 0.184 140 405 836 328 468 48;
  • 47) 0.184 140 405 836 328 468 48 × 2 = 0 + 0.368 280 811 672 656 936 96;
  • 48) 0.368 280 811 672 656 936 96 × 2 = 0 + 0.736 561 623 345 313 873 92;
  • 49) 0.736 561 623 345 313 873 92 × 2 = 1 + 0.473 123 246 690 627 747 84;
  • 50) 0.473 123 246 690 627 747 84 × 2 = 0 + 0.946 246 493 381 255 495 68;
  • 51) 0.946 246 493 381 255 495 68 × 2 = 1 + 0.892 492 986 762 510 991 36;
  • 52) 0.892 492 986 762 510 991 36 × 2 = 1 + 0.784 985 973 525 021 982 72;
  • 53) 0.784 985 973 525 021 982 72 × 2 = 1 + 0.569 971 947 050 043 965 44;
  • 54) 0.569 971 947 050 043 965 44 × 2 = 1 + 0.139 943 894 100 087 930 88;
  • 55) 0.139 943 894 100 087 930 88 × 2 = 0 + 0.279 887 788 200 175 861 76;

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).


5. 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:


0.145 067 813 487 901 050 57(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2)

6. Positive number before normalization:

0.145 067 813 487 901 050 57(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.145 067 813 487 901 050 57(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2) × 20 =

1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110(2) × 2-3


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): -3

Mantissa (not normalized):
1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(10)


10. 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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1020(10) =

011 1111 1100(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110 =

0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1100

Mantissa (52 bits) =
0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


Decimal number -0.145 067 813 487 901 050 57 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110

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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