-0.016 738 891 601 562 496 530 553 042 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 042(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
-0.016 738 891 601 562 496 530 553 042(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:

|-0.016 738 891 601 562 496 530 553 042| = 0.016 738 891 601 562 496 530 553 042


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: 0.016 738 891 601 562 496 530 553 042.

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) 0.016 738 891 601 562 496 530 553 042 × 2 = 0 + 0.033 477 783 203 124 993 061 106 084;
  • 2) 0.033 477 783 203 124 993 061 106 084 × 2 = 0 + 0.066 955 566 406 249 986 122 212 168;
  • 3) 0.066 955 566 406 249 986 122 212 168 × 2 = 0 + 0.133 911 132 812 499 972 244 424 336;
  • 4) 0.133 911 132 812 499 972 244 424 336 × 2 = 0 + 0.267 822 265 624 999 944 488 848 672;
  • 5) 0.267 822 265 624 999 944 488 848 672 × 2 = 0 + 0.535 644 531 249 999 888 977 697 344;
  • 6) 0.535 644 531 249 999 888 977 697 344 × 2 = 1 + 0.071 289 062 499 999 777 955 394 688;
  • 7) 0.071 289 062 499 999 777 955 394 688 × 2 = 0 + 0.142 578 124 999 999 555 910 789 376;
  • 8) 0.142 578 124 999 999 555 910 789 376 × 2 = 0 + 0.285 156 249 999 999 111 821 578 752;
  • 9) 0.285 156 249 999 999 111 821 578 752 × 2 = 0 + 0.570 312 499 999 998 223 643 157 504;
  • 10) 0.570 312 499 999 998 223 643 157 504 × 2 = 1 + 0.140 624 999 999 996 447 286 315 008;
  • 11) 0.140 624 999 999 996 447 286 315 008 × 2 = 0 + 0.281 249 999 999 992 894 572 630 016;
  • 12) 0.281 249 999 999 992 894 572 630 016 × 2 = 0 + 0.562 499 999 999 985 789 145 260 032;
  • 13) 0.562 499 999 999 985 789 145 260 032 × 2 = 1 + 0.124 999 999 999 971 578 290 520 064;
  • 14) 0.124 999 999 999 971 578 290 520 064 × 2 = 0 + 0.249 999 999 999 943 156 581 040 128;
  • 15) 0.249 999 999 999 943 156 581 040 128 × 2 = 0 + 0.499 999 999 999 886 313 162 080 256;
  • 16) 0.499 999 999 999 886 313 162 080 256 × 2 = 0 + 0.999 999 999 999 772 626 324 160 512;
  • 17) 0.999 999 999 999 772 626 324 160 512 × 2 = 1 + 0.999 999 999 999 545 252 648 321 024;
  • 18) 0.999 999 999 999 545 252 648 321 024 × 2 = 1 + 0.999 999 999 999 090 505 296 642 048;
  • 19) 0.999 999 999 999 090 505 296 642 048 × 2 = 1 + 0.999 999 999 998 181 010 593 284 096;
  • 20) 0.999 999 999 998 181 010 593 284 096 × 2 = 1 + 0.999 999 999 996 362 021 186 568 192;
  • 21) 0.999 999 999 996 362 021 186 568 192 × 2 = 1 + 0.999 999 999 992 724 042 373 136 384;
  • 22) 0.999 999 999 992 724 042 373 136 384 × 2 = 1 + 0.999 999 999 985 448 084 746 272 768;
  • 23) 0.999 999 999 985 448 084 746 272 768 × 2 = 1 + 0.999 999 999 970 896 169 492 545 536;
  • 24) 0.999 999 999 970 896 169 492 545 536 × 2 = 1 + 0.999 999 999 941 792 338 985 091 072;
  • 25) 0.999 999 999 941 792 338 985 091 072 × 2 = 1 + 0.999 999 999 883 584 677 970 182 144;
  • 26) 0.999 999 999 883 584 677 970 182 144 × 2 = 1 + 0.999 999 999 767 169 355 940 364 288;
  • 27) 0.999 999 999 767 169 355 940 364 288 × 2 = 1 + 0.999 999 999 534 338 711 880 728 576;
  • 28) 0.999 999 999 534 338 711 880 728 576 × 2 = 1 + 0.999 999 999 068 677 423 761 457 152;
  • 29) 0.999 999 999 068 677 423 761 457 152 × 2 = 1 + 0.999 999 998 137 354 847 522 914 304;
  • 30) 0.999 999 998 137 354 847 522 914 304 × 2 = 1 + 0.999 999 996 274 709 695 045 828 608;
  • 31) 0.999 999 996 274 709 695 045 828 608 × 2 = 1 + 0.999 999 992 549 419 390 091 657 216;
  • 32) 0.999 999 992 549 419 390 091 657 216 × 2 = 1 + 0.999 999 985 098 838 780 183 314 432;
  • 33) 0.999 999 985 098 838 780 183 314 432 × 2 = 1 + 0.999 999 970 197 677 560 366 628 864;
  • 34) 0.999 999 970 197 677 560 366 628 864 × 2 = 1 + 0.999 999 940 395 355 120 733 257 728;
  • 35) 0.999 999 940 395 355 120 733 257 728 × 2 = 1 + 0.999 999 880 790 710 241 466 515 456;
  • 36) 0.999 999 880 790 710 241 466 515 456 × 2 = 1 + 0.999 999 761 581 420 482 933 030 912;
  • 37) 0.999 999 761 581 420 482 933 030 912 × 2 = 1 + 0.999 999 523 162 840 965 866 061 824;
  • 38) 0.999 999 523 162 840 965 866 061 824 × 2 = 1 + 0.999 999 046 325 681 931 732 123 648;
  • 39) 0.999 999 046 325 681 931 732 123 648 × 2 = 1 + 0.999 998 092 651 363 863 464 247 296;
  • 40) 0.999 998 092 651 363 863 464 247 296 × 2 = 1 + 0.999 996 185 302 727 726 928 494 592;
  • 41) 0.999 996 185 302 727 726 928 494 592 × 2 = 1 + 0.999 992 370 605 455 453 856 989 184;
  • 42) 0.999 992 370 605 455 453 856 989 184 × 2 = 1 + 0.999 984 741 210 910 907 713 978 368;
  • 43) 0.999 984 741 210 910 907 713 978 368 × 2 = 1 + 0.999 969 482 421 821 815 427 956 736;
  • 44) 0.999 969 482 421 821 815 427 956 736 × 2 = 1 + 0.999 938 964 843 643 630 855 913 472;
  • 45) 0.999 938 964 843 643 630 855 913 472 × 2 = 1 + 0.999 877 929 687 287 261 711 826 944;
  • 46) 0.999 877 929 687 287 261 711 826 944 × 2 = 1 + 0.999 755 859 374 574 523 423 653 888;
  • 47) 0.999 755 859 374 574 523 423 653 888 × 2 = 1 + 0.999 511 718 749 149 046 847 307 776;
  • 48) 0.999 511 718 749 149 046 847 307 776 × 2 = 1 + 0.999 023 437 498 298 093 694 615 552;
  • 49) 0.999 023 437 498 298 093 694 615 552 × 2 = 1 + 0.998 046 874 996 596 187 389 231 104;
  • 50) 0.998 046 874 996 596 187 389 231 104 × 2 = 1 + 0.996 093 749 993 192 374 778 462 208;
  • 51) 0.996 093 749 993 192 374 778 462 208 × 2 = 1 + 0.992 187 499 986 384 749 556 924 416;
  • 52) 0.992 187 499 986 384 749 556 924 416 × 2 = 1 + 0.984 374 999 972 769 499 113 848 832;
  • 53) 0.984 374 999 972 769 499 113 848 832 × 2 = 1 + 0.968 749 999 945 538 998 227 697 664;
  • 54) 0.968 749 999 945 538 998 227 697 664 × 2 = 1 + 0.937 499 999 891 077 996 455 395 328;
  • 55) 0.937 499 999 891 077 996 455 395 328 × 2 = 1 + 0.874 999 999 782 155 992 910 790 656;
  • 56) 0.874 999 999 782 155 992 910 790 656 × 2 = 1 + 0.749 999 999 564 311 985 821 581 312;
  • 57) 0.749 999 999 564 311 985 821 581 312 × 2 = 1 + 0.499 999 999 128 623 971 643 162 624;
  • 58) 0.499 999 999 128 623 971 643 162 624 × 2 = 0 + 0.999 999 998 257 247 943 286 325 248;

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:


0.016 738 891 601 562 496 530 553 042(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 042(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:


0.016 738 891 601 562 496 530 553 042(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 -0.016 738 891 601 562 496 530 553 042 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