-0.016 738 891 601 562 496 530 552 972 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 552 972(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
-0.016 738 891 601 562 496 530 552 972(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:

|-0.016 738 891 601 562 496 530 552 972| = 0.016 738 891 601 562 496 530 552 972


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: 0.016 738 891 601 562 496 530 552 972.

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) 0.016 738 891 601 562 496 530 552 972 × 2 = 0 + 0.033 477 783 203 124 993 061 105 944;
  • 2) 0.033 477 783 203 124 993 061 105 944 × 2 = 0 + 0.066 955 566 406 249 986 122 211 888;
  • 3) 0.066 955 566 406 249 986 122 211 888 × 2 = 0 + 0.133 911 132 812 499 972 244 423 776;
  • 4) 0.133 911 132 812 499 972 244 423 776 × 2 = 0 + 0.267 822 265 624 999 944 488 847 552;
  • 5) 0.267 822 265 624 999 944 488 847 552 × 2 = 0 + 0.535 644 531 249 999 888 977 695 104;
  • 6) 0.535 644 531 249 999 888 977 695 104 × 2 = 1 + 0.071 289 062 499 999 777 955 390 208;
  • 7) 0.071 289 062 499 999 777 955 390 208 × 2 = 0 + 0.142 578 124 999 999 555 910 780 416;
  • 8) 0.142 578 124 999 999 555 910 780 416 × 2 = 0 + 0.285 156 249 999 999 111 821 560 832;
  • 9) 0.285 156 249 999 999 111 821 560 832 × 2 = 0 + 0.570 312 499 999 998 223 643 121 664;
  • 10) 0.570 312 499 999 998 223 643 121 664 × 2 = 1 + 0.140 624 999 999 996 447 286 243 328;
  • 11) 0.140 624 999 999 996 447 286 243 328 × 2 = 0 + 0.281 249 999 999 992 894 572 486 656;
  • 12) 0.281 249 999 999 992 894 572 486 656 × 2 = 0 + 0.562 499 999 999 985 789 144 973 312;
  • 13) 0.562 499 999 999 985 789 144 973 312 × 2 = 1 + 0.124 999 999 999 971 578 289 946 624;
  • 14) 0.124 999 999 999 971 578 289 946 624 × 2 = 0 + 0.249 999 999 999 943 156 579 893 248;
  • 15) 0.249 999 999 999 943 156 579 893 248 × 2 = 0 + 0.499 999 999 999 886 313 159 786 496;
  • 16) 0.499 999 999 999 886 313 159 786 496 × 2 = 0 + 0.999 999 999 999 772 626 319 572 992;
  • 17) 0.999 999 999 999 772 626 319 572 992 × 2 = 1 + 0.999 999 999 999 545 252 639 145 984;
  • 18) 0.999 999 999 999 545 252 639 145 984 × 2 = 1 + 0.999 999 999 999 090 505 278 291 968;
  • 19) 0.999 999 999 999 090 505 278 291 968 × 2 = 1 + 0.999 999 999 998 181 010 556 583 936;
  • 20) 0.999 999 999 998 181 010 556 583 936 × 2 = 1 + 0.999 999 999 996 362 021 113 167 872;
  • 21) 0.999 999 999 996 362 021 113 167 872 × 2 = 1 + 0.999 999 999 992 724 042 226 335 744;
  • 22) 0.999 999 999 992 724 042 226 335 744 × 2 = 1 + 0.999 999 999 985 448 084 452 671 488;
  • 23) 0.999 999 999 985 448 084 452 671 488 × 2 = 1 + 0.999 999 999 970 896 168 905 342 976;
  • 24) 0.999 999 999 970 896 168 905 342 976 × 2 = 1 + 0.999 999 999 941 792 337 810 685 952;
  • 25) 0.999 999 999 941 792 337 810 685 952 × 2 = 1 + 0.999 999 999 883 584 675 621 371 904;
  • 26) 0.999 999 999 883 584 675 621 371 904 × 2 = 1 + 0.999 999 999 767 169 351 242 743 808;
  • 27) 0.999 999 999 767 169 351 242 743 808 × 2 = 1 + 0.999 999 999 534 338 702 485 487 616;
  • 28) 0.999 999 999 534 338 702 485 487 616 × 2 = 1 + 0.999 999 999 068 677 404 970 975 232;
  • 29) 0.999 999 999 068 677 404 970 975 232 × 2 = 1 + 0.999 999 998 137 354 809 941 950 464;
  • 30) 0.999 999 998 137 354 809 941 950 464 × 2 = 1 + 0.999 999 996 274 709 619 883 900 928;
  • 31) 0.999 999 996 274 709 619 883 900 928 × 2 = 1 + 0.999 999 992 549 419 239 767 801 856;
  • 32) 0.999 999 992 549 419 239 767 801 856 × 2 = 1 + 0.999 999 985 098 838 479 535 603 712;
  • 33) 0.999 999 985 098 838 479 535 603 712 × 2 = 1 + 0.999 999 970 197 676 959 071 207 424;
  • 34) 0.999 999 970 197 676 959 071 207 424 × 2 = 1 + 0.999 999 940 395 353 918 142 414 848;
  • 35) 0.999 999 940 395 353 918 142 414 848 × 2 = 1 + 0.999 999 880 790 707 836 284 829 696;
  • 36) 0.999 999 880 790 707 836 284 829 696 × 2 = 1 + 0.999 999 761 581 415 672 569 659 392;
  • 37) 0.999 999 761 581 415 672 569 659 392 × 2 = 1 + 0.999 999 523 162 831 345 139 318 784;
  • 38) 0.999 999 523 162 831 345 139 318 784 × 2 = 1 + 0.999 999 046 325 662 690 278 637 568;
  • 39) 0.999 999 046 325 662 690 278 637 568 × 2 = 1 + 0.999 998 092 651 325 380 557 275 136;
  • 40) 0.999 998 092 651 325 380 557 275 136 × 2 = 1 + 0.999 996 185 302 650 761 114 550 272;
  • 41) 0.999 996 185 302 650 761 114 550 272 × 2 = 1 + 0.999 992 370 605 301 522 229 100 544;
  • 42) 0.999 992 370 605 301 522 229 100 544 × 2 = 1 + 0.999 984 741 210 603 044 458 201 088;
  • 43) 0.999 984 741 210 603 044 458 201 088 × 2 = 1 + 0.999 969 482 421 206 088 916 402 176;
  • 44) 0.999 969 482 421 206 088 916 402 176 × 2 = 1 + 0.999 938 964 842 412 177 832 804 352;
  • 45) 0.999 938 964 842 412 177 832 804 352 × 2 = 1 + 0.999 877 929 684 824 355 665 608 704;
  • 46) 0.999 877 929 684 824 355 665 608 704 × 2 = 1 + 0.999 755 859 369 648 711 331 217 408;
  • 47) 0.999 755 859 369 648 711 331 217 408 × 2 = 1 + 0.999 511 718 739 297 422 662 434 816;
  • 48) 0.999 511 718 739 297 422 662 434 816 × 2 = 1 + 0.999 023 437 478 594 845 324 869 632;
  • 49) 0.999 023 437 478 594 845 324 869 632 × 2 = 1 + 0.998 046 874 957 189 690 649 739 264;
  • 50) 0.998 046 874 957 189 690 649 739 264 × 2 = 1 + 0.996 093 749 914 379 381 299 478 528;
  • 51) 0.996 093 749 914 379 381 299 478 528 × 2 = 1 + 0.992 187 499 828 758 762 598 957 056;
  • 52) 0.992 187 499 828 758 762 598 957 056 × 2 = 1 + 0.984 374 999 657 517 525 197 914 112;
  • 53) 0.984 374 999 657 517 525 197 914 112 × 2 = 1 + 0.968 749 999 315 035 050 395 828 224;
  • 54) 0.968 749 999 315 035 050 395 828 224 × 2 = 1 + 0.937 499 998 630 070 100 791 656 448;
  • 55) 0.937 499 998 630 070 100 791 656 448 × 2 = 1 + 0.874 999 997 260 140 201 583 312 896;
  • 56) 0.874 999 997 260 140 201 583 312 896 × 2 = 1 + 0.749 999 994 520 280 403 166 625 792;
  • 57) 0.749 999 994 520 280 403 166 625 792 × 2 = 1 + 0.499 999 989 040 560 806 333 251 584;
  • 58) 0.499 999 989 040 560 806 333 251 584 × 2 = 0 + 0.999 999 978 081 121 612 666 503 168;

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:


0.016 738 891 601 562 496 530 552 972(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 552 972(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 530 552 972(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1017(10) =

011 1111 1001(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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 530 552 972 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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