-0.145 067 813 487 901 050 782 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 782(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
-0.145 067 813 487 901 050 782(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:

|-0.145 067 813 487 901 050 782| = 0.145 067 813 487 901 050 782


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: 0.145 067 813 487 901 050 782.

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) 0.145 067 813 487 901 050 782 × 2 = 0 + 0.290 135 626 975 802 101 564;
  • 2) 0.290 135 626 975 802 101 564 × 2 = 0 + 0.580 271 253 951 604 203 128;
  • 3) 0.580 271 253 951 604 203 128 × 2 = 1 + 0.160 542 507 903 208 406 256;
  • 4) 0.160 542 507 903 208 406 256 × 2 = 0 + 0.321 085 015 806 416 812 512;
  • 5) 0.321 085 015 806 416 812 512 × 2 = 0 + 0.642 170 031 612 833 625 024;
  • 6) 0.642 170 031 612 833 625 024 × 2 = 1 + 0.284 340 063 225 667 250 048;
  • 7) 0.284 340 063 225 667 250 048 × 2 = 0 + 0.568 680 126 451 334 500 096;
  • 8) 0.568 680 126 451 334 500 096 × 2 = 1 + 0.137 360 252 902 669 000 192;
  • 9) 0.137 360 252 902 669 000 192 × 2 = 0 + 0.274 720 505 805 338 000 384;
  • 10) 0.274 720 505 805 338 000 384 × 2 = 0 + 0.549 441 011 610 676 000 768;
  • 11) 0.549 441 011 610 676 000 768 × 2 = 1 + 0.098 882 023 221 352 001 536;
  • 12) 0.098 882 023 221 352 001 536 × 2 = 0 + 0.197 764 046 442 704 003 072;
  • 13) 0.197 764 046 442 704 003 072 × 2 = 0 + 0.395 528 092 885 408 006 144;
  • 14) 0.395 528 092 885 408 006 144 × 2 = 0 + 0.791 056 185 770 816 012 288;
  • 15) 0.791 056 185 770 816 012 288 × 2 = 1 + 0.582 112 371 541 632 024 576;
  • 16) 0.582 112 371 541 632 024 576 × 2 = 1 + 0.164 224 743 083 264 049 152;
  • 17) 0.164 224 743 083 264 049 152 × 2 = 0 + 0.328 449 486 166 528 098 304;
  • 18) 0.328 449 486 166 528 098 304 × 2 = 0 + 0.656 898 972 333 056 196 608;
  • 19) 0.656 898 972 333 056 196 608 × 2 = 1 + 0.313 797 944 666 112 393 216;
  • 20) 0.313 797 944 666 112 393 216 × 2 = 0 + 0.627 595 889 332 224 786 432;
  • 21) 0.627 595 889 332 224 786 432 × 2 = 1 + 0.255 191 778 664 449 572 864;
  • 22) 0.255 191 778 664 449 572 864 × 2 = 0 + 0.510 383 557 328 899 145 728;
  • 23) 0.510 383 557 328 899 145 728 × 2 = 1 + 0.020 767 114 657 798 291 456;
  • 24) 0.020 767 114 657 798 291 456 × 2 = 0 + 0.041 534 229 315 596 582 912;
  • 25) 0.041 534 229 315 596 582 912 × 2 = 0 + 0.083 068 458 631 193 165 824;
  • 26) 0.083 068 458 631 193 165 824 × 2 = 0 + 0.166 136 917 262 386 331 648;
  • 27) 0.166 136 917 262 386 331 648 × 2 = 0 + 0.332 273 834 524 772 663 296;
  • 28) 0.332 273 834 524 772 663 296 × 2 = 0 + 0.664 547 669 049 545 326 592;
  • 29) 0.664 547 669 049 545 326 592 × 2 = 1 + 0.329 095 338 099 090 653 184;
  • 30) 0.329 095 338 099 090 653 184 × 2 = 0 + 0.658 190 676 198 181 306 368;
  • 31) 0.658 190 676 198 181 306 368 × 2 = 1 + 0.316 381 352 396 362 612 736;
  • 32) 0.316 381 352 396 362 612 736 × 2 = 0 + 0.632 762 704 792 725 225 472;
  • 33) 0.632 762 704 792 725 225 472 × 2 = 1 + 0.265 525 409 585 450 450 944;
  • 34) 0.265 525 409 585 450 450 944 × 2 = 0 + 0.531 050 819 170 900 901 888;
  • 35) 0.531 050 819 170 900 901 888 × 2 = 1 + 0.062 101 638 341 801 803 776;
  • 36) 0.062 101 638 341 801 803 776 × 2 = 0 + 0.124 203 276 683 603 607 552;
  • 37) 0.124 203 276 683 603 607 552 × 2 = 0 + 0.248 406 553 367 207 215 104;
  • 38) 0.248 406 553 367 207 215 104 × 2 = 0 + 0.496 813 106 734 414 430 208;
  • 39) 0.496 813 106 734 414 430 208 × 2 = 0 + 0.993 626 213 468 828 860 416;
  • 40) 0.993 626 213 468 828 860 416 × 2 = 1 + 0.987 252 426 937 657 720 832;
  • 41) 0.987 252 426 937 657 720 832 × 2 = 1 + 0.974 504 853 875 315 441 664;
  • 42) 0.974 504 853 875 315 441 664 × 2 = 1 + 0.949 009 707 750 630 883 328;
  • 43) 0.949 009 707 750 630 883 328 × 2 = 1 + 0.898 019 415 501 261 766 656;
  • 44) 0.898 019 415 501 261 766 656 × 2 = 1 + 0.796 038 831 002 523 533 312;
  • 45) 0.796 038 831 002 523 533 312 × 2 = 1 + 0.592 077 662 005 047 066 624;
  • 46) 0.592 077 662 005 047 066 624 × 2 = 1 + 0.184 155 324 010 094 133 248;
  • 47) 0.184 155 324 010 094 133 248 × 2 = 0 + 0.368 310 648 020 188 266 496;
  • 48) 0.368 310 648 020 188 266 496 × 2 = 0 + 0.736 621 296 040 376 532 992;
  • 49) 0.736 621 296 040 376 532 992 × 2 = 1 + 0.473 242 592 080 753 065 984;
  • 50) 0.473 242 592 080 753 065 984 × 2 = 0 + 0.946 485 184 161 506 131 968;
  • 51) 0.946 485 184 161 506 131 968 × 2 = 1 + 0.892 970 368 323 012 263 936;
  • 52) 0.892 970 368 323 012 263 936 × 2 = 1 + 0.785 940 736 646 024 527 872;
  • 53) 0.785 940 736 646 024 527 872 × 2 = 1 + 0.571 881 473 292 049 055 744;
  • 54) 0.571 881 473 292 049 055 744 × 2 = 1 + 0.143 762 946 584 098 111 488;
  • 55) 0.143 762 946 584 098 111 488 × 2 = 0 + 0.287 525 893 168 196 222 976;

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:


0.145 067 813 487 901 050 782(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 782(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:


0.145 067 813 487 901 050 782(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 -0.145 067 813 487 901 050 782 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