-0.145 067 813 487 901 050 833 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 833(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
-0.145 067 813 487 901 050 833(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:

|-0.145 067 813 487 901 050 833| = 0.145 067 813 487 901 050 833


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: 0.145 067 813 487 901 050 833.

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) 0.145 067 813 487 901 050 833 × 2 = 0 + 0.290 135 626 975 802 101 666;
  • 2) 0.290 135 626 975 802 101 666 × 2 = 0 + 0.580 271 253 951 604 203 332;
  • 3) 0.580 271 253 951 604 203 332 × 2 = 1 + 0.160 542 507 903 208 406 664;
  • 4) 0.160 542 507 903 208 406 664 × 2 = 0 + 0.321 085 015 806 416 813 328;
  • 5) 0.321 085 015 806 416 813 328 × 2 = 0 + 0.642 170 031 612 833 626 656;
  • 6) 0.642 170 031 612 833 626 656 × 2 = 1 + 0.284 340 063 225 667 253 312;
  • 7) 0.284 340 063 225 667 253 312 × 2 = 0 + 0.568 680 126 451 334 506 624;
  • 8) 0.568 680 126 451 334 506 624 × 2 = 1 + 0.137 360 252 902 669 013 248;
  • 9) 0.137 360 252 902 669 013 248 × 2 = 0 + 0.274 720 505 805 338 026 496;
  • 10) 0.274 720 505 805 338 026 496 × 2 = 0 + 0.549 441 011 610 676 052 992;
  • 11) 0.549 441 011 610 676 052 992 × 2 = 1 + 0.098 882 023 221 352 105 984;
  • 12) 0.098 882 023 221 352 105 984 × 2 = 0 + 0.197 764 046 442 704 211 968;
  • 13) 0.197 764 046 442 704 211 968 × 2 = 0 + 0.395 528 092 885 408 423 936;
  • 14) 0.395 528 092 885 408 423 936 × 2 = 0 + 0.791 056 185 770 816 847 872;
  • 15) 0.791 056 185 770 816 847 872 × 2 = 1 + 0.582 112 371 541 633 695 744;
  • 16) 0.582 112 371 541 633 695 744 × 2 = 1 + 0.164 224 743 083 267 391 488;
  • 17) 0.164 224 743 083 267 391 488 × 2 = 0 + 0.328 449 486 166 534 782 976;
  • 18) 0.328 449 486 166 534 782 976 × 2 = 0 + 0.656 898 972 333 069 565 952;
  • 19) 0.656 898 972 333 069 565 952 × 2 = 1 + 0.313 797 944 666 139 131 904;
  • 20) 0.313 797 944 666 139 131 904 × 2 = 0 + 0.627 595 889 332 278 263 808;
  • 21) 0.627 595 889 332 278 263 808 × 2 = 1 + 0.255 191 778 664 556 527 616;
  • 22) 0.255 191 778 664 556 527 616 × 2 = 0 + 0.510 383 557 329 113 055 232;
  • 23) 0.510 383 557 329 113 055 232 × 2 = 1 + 0.020 767 114 658 226 110 464;
  • 24) 0.020 767 114 658 226 110 464 × 2 = 0 + 0.041 534 229 316 452 220 928;
  • 25) 0.041 534 229 316 452 220 928 × 2 = 0 + 0.083 068 458 632 904 441 856;
  • 26) 0.083 068 458 632 904 441 856 × 2 = 0 + 0.166 136 917 265 808 883 712;
  • 27) 0.166 136 917 265 808 883 712 × 2 = 0 + 0.332 273 834 531 617 767 424;
  • 28) 0.332 273 834 531 617 767 424 × 2 = 0 + 0.664 547 669 063 235 534 848;
  • 29) 0.664 547 669 063 235 534 848 × 2 = 1 + 0.329 095 338 126 471 069 696;
  • 30) 0.329 095 338 126 471 069 696 × 2 = 0 + 0.658 190 676 252 942 139 392;
  • 31) 0.658 190 676 252 942 139 392 × 2 = 1 + 0.316 381 352 505 884 278 784;
  • 32) 0.316 381 352 505 884 278 784 × 2 = 0 + 0.632 762 705 011 768 557 568;
  • 33) 0.632 762 705 011 768 557 568 × 2 = 1 + 0.265 525 410 023 537 115 136;
  • 34) 0.265 525 410 023 537 115 136 × 2 = 0 + 0.531 050 820 047 074 230 272;
  • 35) 0.531 050 820 047 074 230 272 × 2 = 1 + 0.062 101 640 094 148 460 544;
  • 36) 0.062 101 640 094 148 460 544 × 2 = 0 + 0.124 203 280 188 296 921 088;
  • 37) 0.124 203 280 188 296 921 088 × 2 = 0 + 0.248 406 560 376 593 842 176;
  • 38) 0.248 406 560 376 593 842 176 × 2 = 0 + 0.496 813 120 753 187 684 352;
  • 39) 0.496 813 120 753 187 684 352 × 2 = 0 + 0.993 626 241 506 375 368 704;
  • 40) 0.993 626 241 506 375 368 704 × 2 = 1 + 0.987 252 483 012 750 737 408;
  • 41) 0.987 252 483 012 750 737 408 × 2 = 1 + 0.974 504 966 025 501 474 816;
  • 42) 0.974 504 966 025 501 474 816 × 2 = 1 + 0.949 009 932 051 002 949 632;
  • 43) 0.949 009 932 051 002 949 632 × 2 = 1 + 0.898 019 864 102 005 899 264;
  • 44) 0.898 019 864 102 005 899 264 × 2 = 1 + 0.796 039 728 204 011 798 528;
  • 45) 0.796 039 728 204 011 798 528 × 2 = 1 + 0.592 079 456 408 023 597 056;
  • 46) 0.592 079 456 408 023 597 056 × 2 = 1 + 0.184 158 912 816 047 194 112;
  • 47) 0.184 158 912 816 047 194 112 × 2 = 0 + 0.368 317 825 632 094 388 224;
  • 48) 0.368 317 825 632 094 388 224 × 2 = 0 + 0.736 635 651 264 188 776 448;
  • 49) 0.736 635 651 264 188 776 448 × 2 = 1 + 0.473 271 302 528 377 552 896;
  • 50) 0.473 271 302 528 377 552 896 × 2 = 0 + 0.946 542 605 056 755 105 792;
  • 51) 0.946 542 605 056 755 105 792 × 2 = 1 + 0.893 085 210 113 510 211 584;
  • 52) 0.893 085 210 113 510 211 584 × 2 = 1 + 0.786 170 420 227 020 423 168;
  • 53) 0.786 170 420 227 020 423 168 × 2 = 1 + 0.572 340 840 454 040 846 336;
  • 54) 0.572 340 840 454 040 846 336 × 2 = 1 + 0.144 681 680 908 081 692 672;
  • 55) 0.144 681 680 908 081 692 672 × 2 = 0 + 0.289 363 361 816 163 385 344;

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:


0.145 067 813 487 901 050 833(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 833(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:


0.145 067 813 487 901 050 833(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 -0.145 067 813 487 901 050 833 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 -0.145 067 813 487 901 050 894 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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