-0.145 067 813 487 901 050 577 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 577(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
-0.145 067 813 487 901 050 577(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:

|-0.145 067 813 487 901 050 577| = 0.145 067 813 487 901 050 577


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: 0.145 067 813 487 901 050 577.

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) 0.145 067 813 487 901 050 577 × 2 = 0 + 0.290 135 626 975 802 101 154;
  • 2) 0.290 135 626 975 802 101 154 × 2 = 0 + 0.580 271 253 951 604 202 308;
  • 3) 0.580 271 253 951 604 202 308 × 2 = 1 + 0.160 542 507 903 208 404 616;
  • 4) 0.160 542 507 903 208 404 616 × 2 = 0 + 0.321 085 015 806 416 809 232;
  • 5) 0.321 085 015 806 416 809 232 × 2 = 0 + 0.642 170 031 612 833 618 464;
  • 6) 0.642 170 031 612 833 618 464 × 2 = 1 + 0.284 340 063 225 667 236 928;
  • 7) 0.284 340 063 225 667 236 928 × 2 = 0 + 0.568 680 126 451 334 473 856;
  • 8) 0.568 680 126 451 334 473 856 × 2 = 1 + 0.137 360 252 902 668 947 712;
  • 9) 0.137 360 252 902 668 947 712 × 2 = 0 + 0.274 720 505 805 337 895 424;
  • 10) 0.274 720 505 805 337 895 424 × 2 = 0 + 0.549 441 011 610 675 790 848;
  • 11) 0.549 441 011 610 675 790 848 × 2 = 1 + 0.098 882 023 221 351 581 696;
  • 12) 0.098 882 023 221 351 581 696 × 2 = 0 + 0.197 764 046 442 703 163 392;
  • 13) 0.197 764 046 442 703 163 392 × 2 = 0 + 0.395 528 092 885 406 326 784;
  • 14) 0.395 528 092 885 406 326 784 × 2 = 0 + 0.791 056 185 770 812 653 568;
  • 15) 0.791 056 185 770 812 653 568 × 2 = 1 + 0.582 112 371 541 625 307 136;
  • 16) 0.582 112 371 541 625 307 136 × 2 = 1 + 0.164 224 743 083 250 614 272;
  • 17) 0.164 224 743 083 250 614 272 × 2 = 0 + 0.328 449 486 166 501 228 544;
  • 18) 0.328 449 486 166 501 228 544 × 2 = 0 + 0.656 898 972 333 002 457 088;
  • 19) 0.656 898 972 333 002 457 088 × 2 = 1 + 0.313 797 944 666 004 914 176;
  • 20) 0.313 797 944 666 004 914 176 × 2 = 0 + 0.627 595 889 332 009 828 352;
  • 21) 0.627 595 889 332 009 828 352 × 2 = 1 + 0.255 191 778 664 019 656 704;
  • 22) 0.255 191 778 664 019 656 704 × 2 = 0 + 0.510 383 557 328 039 313 408;
  • 23) 0.510 383 557 328 039 313 408 × 2 = 1 + 0.020 767 114 656 078 626 816;
  • 24) 0.020 767 114 656 078 626 816 × 2 = 0 + 0.041 534 229 312 157 253 632;
  • 25) 0.041 534 229 312 157 253 632 × 2 = 0 + 0.083 068 458 624 314 507 264;
  • 26) 0.083 068 458 624 314 507 264 × 2 = 0 + 0.166 136 917 248 629 014 528;
  • 27) 0.166 136 917 248 629 014 528 × 2 = 0 + 0.332 273 834 497 258 029 056;
  • 28) 0.332 273 834 497 258 029 056 × 2 = 0 + 0.664 547 668 994 516 058 112;
  • 29) 0.664 547 668 994 516 058 112 × 2 = 1 + 0.329 095 337 989 032 116 224;
  • 30) 0.329 095 337 989 032 116 224 × 2 = 0 + 0.658 190 675 978 064 232 448;
  • 31) 0.658 190 675 978 064 232 448 × 2 = 1 + 0.316 381 351 956 128 464 896;
  • 32) 0.316 381 351 956 128 464 896 × 2 = 0 + 0.632 762 703 912 256 929 792;
  • 33) 0.632 762 703 912 256 929 792 × 2 = 1 + 0.265 525 407 824 513 859 584;
  • 34) 0.265 525 407 824 513 859 584 × 2 = 0 + 0.531 050 815 649 027 719 168;
  • 35) 0.531 050 815 649 027 719 168 × 2 = 1 + 0.062 101 631 298 055 438 336;
  • 36) 0.062 101 631 298 055 438 336 × 2 = 0 + 0.124 203 262 596 110 876 672;
  • 37) 0.124 203 262 596 110 876 672 × 2 = 0 + 0.248 406 525 192 221 753 344;
  • 38) 0.248 406 525 192 221 753 344 × 2 = 0 + 0.496 813 050 384 443 506 688;
  • 39) 0.496 813 050 384 443 506 688 × 2 = 0 + 0.993 626 100 768 887 013 376;
  • 40) 0.993 626 100 768 887 013 376 × 2 = 1 + 0.987 252 201 537 774 026 752;
  • 41) 0.987 252 201 537 774 026 752 × 2 = 1 + 0.974 504 403 075 548 053 504;
  • 42) 0.974 504 403 075 548 053 504 × 2 = 1 + 0.949 008 806 151 096 107 008;
  • 43) 0.949 008 806 151 096 107 008 × 2 = 1 + 0.898 017 612 302 192 214 016;
  • 44) 0.898 017 612 302 192 214 016 × 2 = 1 + 0.796 035 224 604 384 428 032;
  • 45) 0.796 035 224 604 384 428 032 × 2 = 1 + 0.592 070 449 208 768 856 064;
  • 46) 0.592 070 449 208 768 856 064 × 2 = 1 + 0.184 140 898 417 537 712 128;
  • 47) 0.184 140 898 417 537 712 128 × 2 = 0 + 0.368 281 796 835 075 424 256;
  • 48) 0.368 281 796 835 075 424 256 × 2 = 0 + 0.736 563 593 670 150 848 512;
  • 49) 0.736 563 593 670 150 848 512 × 2 = 1 + 0.473 127 187 340 301 697 024;
  • 50) 0.473 127 187 340 301 697 024 × 2 = 0 + 0.946 254 374 680 603 394 048;
  • 51) 0.946 254 374 680 603 394 048 × 2 = 1 + 0.892 508 749 361 206 788 096;
  • 52) 0.892 508 749 361 206 788 096 × 2 = 1 + 0.785 017 498 722 413 576 192;
  • 53) 0.785 017 498 722 413 576 192 × 2 = 1 + 0.570 034 997 444 827 152 384;
  • 54) 0.570 034 997 444 827 152 384 × 2 = 1 + 0.140 069 994 889 654 304 768;
  • 55) 0.140 069 994 889 654 304 768 × 2 = 0 + 0.280 139 989 779 308 609 536;

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:


0.145 067 813 487 901 050 577(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 577(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:


0.145 067 813 487 901 050 577(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 -0.145 067 813 487 901 050 577 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