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

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

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


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: 0.145 067 813 487 901 050 813.

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) 0.145 067 813 487 901 050 813 × 2 = 0 + 0.290 135 626 975 802 101 626;
  • 2) 0.290 135 626 975 802 101 626 × 2 = 0 + 0.580 271 253 951 604 203 252;
  • 3) 0.580 271 253 951 604 203 252 × 2 = 1 + 0.160 542 507 903 208 406 504;
  • 4) 0.160 542 507 903 208 406 504 × 2 = 0 + 0.321 085 015 806 416 813 008;
  • 5) 0.321 085 015 806 416 813 008 × 2 = 0 + 0.642 170 031 612 833 626 016;
  • 6) 0.642 170 031 612 833 626 016 × 2 = 1 + 0.284 340 063 225 667 252 032;
  • 7) 0.284 340 063 225 667 252 032 × 2 = 0 + 0.568 680 126 451 334 504 064;
  • 8) 0.568 680 126 451 334 504 064 × 2 = 1 + 0.137 360 252 902 669 008 128;
  • 9) 0.137 360 252 902 669 008 128 × 2 = 0 + 0.274 720 505 805 338 016 256;
  • 10) 0.274 720 505 805 338 016 256 × 2 = 0 + 0.549 441 011 610 676 032 512;
  • 11) 0.549 441 011 610 676 032 512 × 2 = 1 + 0.098 882 023 221 352 065 024;
  • 12) 0.098 882 023 221 352 065 024 × 2 = 0 + 0.197 764 046 442 704 130 048;
  • 13) 0.197 764 046 442 704 130 048 × 2 = 0 + 0.395 528 092 885 408 260 096;
  • 14) 0.395 528 092 885 408 260 096 × 2 = 0 + 0.791 056 185 770 816 520 192;
  • 15) 0.791 056 185 770 816 520 192 × 2 = 1 + 0.582 112 371 541 633 040 384;
  • 16) 0.582 112 371 541 633 040 384 × 2 = 1 + 0.164 224 743 083 266 080 768;
  • 17) 0.164 224 743 083 266 080 768 × 2 = 0 + 0.328 449 486 166 532 161 536;
  • 18) 0.328 449 486 166 532 161 536 × 2 = 0 + 0.656 898 972 333 064 323 072;
  • 19) 0.656 898 972 333 064 323 072 × 2 = 1 + 0.313 797 944 666 128 646 144;
  • 20) 0.313 797 944 666 128 646 144 × 2 = 0 + 0.627 595 889 332 257 292 288;
  • 21) 0.627 595 889 332 257 292 288 × 2 = 1 + 0.255 191 778 664 514 584 576;
  • 22) 0.255 191 778 664 514 584 576 × 2 = 0 + 0.510 383 557 329 029 169 152;
  • 23) 0.510 383 557 329 029 169 152 × 2 = 1 + 0.020 767 114 658 058 338 304;
  • 24) 0.020 767 114 658 058 338 304 × 2 = 0 + 0.041 534 229 316 116 676 608;
  • 25) 0.041 534 229 316 116 676 608 × 2 = 0 + 0.083 068 458 632 233 353 216;
  • 26) 0.083 068 458 632 233 353 216 × 2 = 0 + 0.166 136 917 264 466 706 432;
  • 27) 0.166 136 917 264 466 706 432 × 2 = 0 + 0.332 273 834 528 933 412 864;
  • 28) 0.332 273 834 528 933 412 864 × 2 = 0 + 0.664 547 669 057 866 825 728;
  • 29) 0.664 547 669 057 866 825 728 × 2 = 1 + 0.329 095 338 115 733 651 456;
  • 30) 0.329 095 338 115 733 651 456 × 2 = 0 + 0.658 190 676 231 467 302 912;
  • 31) 0.658 190 676 231 467 302 912 × 2 = 1 + 0.316 381 352 462 934 605 824;
  • 32) 0.316 381 352 462 934 605 824 × 2 = 0 + 0.632 762 704 925 869 211 648;
  • 33) 0.632 762 704 925 869 211 648 × 2 = 1 + 0.265 525 409 851 738 423 296;
  • 34) 0.265 525 409 851 738 423 296 × 2 = 0 + 0.531 050 819 703 476 846 592;
  • 35) 0.531 050 819 703 476 846 592 × 2 = 1 + 0.062 101 639 406 953 693 184;
  • 36) 0.062 101 639 406 953 693 184 × 2 = 0 + 0.124 203 278 813 907 386 368;
  • 37) 0.124 203 278 813 907 386 368 × 2 = 0 + 0.248 406 557 627 814 772 736;
  • 38) 0.248 406 557 627 814 772 736 × 2 = 0 + 0.496 813 115 255 629 545 472;
  • 39) 0.496 813 115 255 629 545 472 × 2 = 0 + 0.993 626 230 511 259 090 944;
  • 40) 0.993 626 230 511 259 090 944 × 2 = 1 + 0.987 252 461 022 518 181 888;
  • 41) 0.987 252 461 022 518 181 888 × 2 = 1 + 0.974 504 922 045 036 363 776;
  • 42) 0.974 504 922 045 036 363 776 × 2 = 1 + 0.949 009 844 090 072 727 552;
  • 43) 0.949 009 844 090 072 727 552 × 2 = 1 + 0.898 019 688 180 145 455 104;
  • 44) 0.898 019 688 180 145 455 104 × 2 = 1 + 0.796 039 376 360 290 910 208;
  • 45) 0.796 039 376 360 290 910 208 × 2 = 1 + 0.592 078 752 720 581 820 416;
  • 46) 0.592 078 752 720 581 820 416 × 2 = 1 + 0.184 157 505 441 163 640 832;
  • 47) 0.184 157 505 441 163 640 832 × 2 = 0 + 0.368 315 010 882 327 281 664;
  • 48) 0.368 315 010 882 327 281 664 × 2 = 0 + 0.736 630 021 764 654 563 328;
  • 49) 0.736 630 021 764 654 563 328 × 2 = 1 + 0.473 260 043 529 309 126 656;
  • 50) 0.473 260 043 529 309 126 656 × 2 = 0 + 0.946 520 087 058 618 253 312;
  • 51) 0.946 520 087 058 618 253 312 × 2 = 1 + 0.893 040 174 117 236 506 624;
  • 52) 0.893 040 174 117 236 506 624 × 2 = 1 + 0.786 080 348 234 473 013 248;
  • 53) 0.786 080 348 234 473 013 248 × 2 = 1 + 0.572 160 696 468 946 026 496;
  • 54) 0.572 160 696 468 946 026 496 × 2 = 1 + 0.144 321 392 937 892 052 992;
  • 55) 0.144 321 392 937 892 052 992 × 2 = 0 + 0.288 642 785 875 784 105 984;

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:


0.145 067 813 487 901 050 813(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 813(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:


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

Share

» Convert decimal -0.145 067 813 487 901 050 893 to 64 bit double precision IEEE 754 binary floating point representation standard

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

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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