-0.000 000 000 000 000 000 019 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 019(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
-0.000 000 000 000 000 000 019(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:

|-0.000 000 000 000 000 000 019| = 0.000 000 000 000 000 000 019


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: 0.000 000 000 000 000 000 019.

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) 0.000 000 000 000 000 000 019 × 2 = 0 + 0.000 000 000 000 000 000 038;
  • 2) 0.000 000 000 000 000 000 038 × 2 = 0 + 0.000 000 000 000 000 000 076;
  • 3) 0.000 000 000 000 000 000 076 × 2 = 0 + 0.000 000 000 000 000 000 152;
  • 4) 0.000 000 000 000 000 000 152 × 2 = 0 + 0.000 000 000 000 000 000 304;
  • 5) 0.000 000 000 000 000 000 304 × 2 = 0 + 0.000 000 000 000 000 000 608;
  • 6) 0.000 000 000 000 000 000 608 × 2 = 0 + 0.000 000 000 000 000 001 216;
  • 7) 0.000 000 000 000 000 001 216 × 2 = 0 + 0.000 000 000 000 000 002 432;
  • 8) 0.000 000 000 000 000 002 432 × 2 = 0 + 0.000 000 000 000 000 004 864;
  • 9) 0.000 000 000 000 000 004 864 × 2 = 0 + 0.000 000 000 000 000 009 728;
  • 10) 0.000 000 000 000 000 009 728 × 2 = 0 + 0.000 000 000 000 000 019 456;
  • 11) 0.000 000 000 000 000 019 456 × 2 = 0 + 0.000 000 000 000 000 038 912;
  • 12) 0.000 000 000 000 000 038 912 × 2 = 0 + 0.000 000 000 000 000 077 824;
  • 13) 0.000 000 000 000 000 077 824 × 2 = 0 + 0.000 000 000 000 000 155 648;
  • 14) 0.000 000 000 000 000 155 648 × 2 = 0 + 0.000 000 000 000 000 311 296;
  • 15) 0.000 000 000 000 000 311 296 × 2 = 0 + 0.000 000 000 000 000 622 592;
  • 16) 0.000 000 000 000 000 622 592 × 2 = 0 + 0.000 000 000 000 001 245 184;
  • 17) 0.000 000 000 000 001 245 184 × 2 = 0 + 0.000 000 000 000 002 490 368;
  • 18) 0.000 000 000 000 002 490 368 × 2 = 0 + 0.000 000 000 000 004 980 736;
  • 19) 0.000 000 000 000 004 980 736 × 2 = 0 + 0.000 000 000 000 009 961 472;
  • 20) 0.000 000 000 000 009 961 472 × 2 = 0 + 0.000 000 000 000 019 922 944;
  • 21) 0.000 000 000 000 019 922 944 × 2 = 0 + 0.000 000 000 000 039 845 888;
  • 22) 0.000 000 000 000 039 845 888 × 2 = 0 + 0.000 000 000 000 079 691 776;
  • 23) 0.000 000 000 000 079 691 776 × 2 = 0 + 0.000 000 000 000 159 383 552;
  • 24) 0.000 000 000 000 159 383 552 × 2 = 0 + 0.000 000 000 000 318 767 104;
  • 25) 0.000 000 000 000 318 767 104 × 2 = 0 + 0.000 000 000 000 637 534 208;
  • 26) 0.000 000 000 000 637 534 208 × 2 = 0 + 0.000 000 000 001 275 068 416;
  • 27) 0.000 000 000 001 275 068 416 × 2 = 0 + 0.000 000 000 002 550 136 832;
  • 28) 0.000 000 000 002 550 136 832 × 2 = 0 + 0.000 000 000 005 100 273 664;
  • 29) 0.000 000 000 005 100 273 664 × 2 = 0 + 0.000 000 000 010 200 547 328;
  • 30) 0.000 000 000 010 200 547 328 × 2 = 0 + 0.000 000 000 020 401 094 656;
  • 31) 0.000 000 000 020 401 094 656 × 2 = 0 + 0.000 000 000 040 802 189 312;
  • 32) 0.000 000 000 040 802 189 312 × 2 = 0 + 0.000 000 000 081 604 378 624;
  • 33) 0.000 000 000 081 604 378 624 × 2 = 0 + 0.000 000 000 163 208 757 248;
  • 34) 0.000 000 000 163 208 757 248 × 2 = 0 + 0.000 000 000 326 417 514 496;
  • 35) 0.000 000 000 326 417 514 496 × 2 = 0 + 0.000 000 000 652 835 028 992;
  • 36) 0.000 000 000 652 835 028 992 × 2 = 0 + 0.000 000 001 305 670 057 984;
  • 37) 0.000 000 001 305 670 057 984 × 2 = 0 + 0.000 000 002 611 340 115 968;
  • 38) 0.000 000 002 611 340 115 968 × 2 = 0 + 0.000 000 005 222 680 231 936;
  • 39) 0.000 000 005 222 680 231 936 × 2 = 0 + 0.000 000 010 445 360 463 872;
  • 40) 0.000 000 010 445 360 463 872 × 2 = 0 + 0.000 000 020 890 720 927 744;
  • 41) 0.000 000 020 890 720 927 744 × 2 = 0 + 0.000 000 041 781 441 855 488;
  • 42) 0.000 000 041 781 441 855 488 × 2 = 0 + 0.000 000 083 562 883 710 976;
  • 43) 0.000 000 083 562 883 710 976 × 2 = 0 + 0.000 000 167 125 767 421 952;
  • 44) 0.000 000 167 125 767 421 952 × 2 = 0 + 0.000 000 334 251 534 843 904;
  • 45) 0.000 000 334 251 534 843 904 × 2 = 0 + 0.000 000 668 503 069 687 808;
  • 46) 0.000 000 668 503 069 687 808 × 2 = 0 + 0.000 001 337 006 139 375 616;
  • 47) 0.000 001 337 006 139 375 616 × 2 = 0 + 0.000 002 674 012 278 751 232;
  • 48) 0.000 002 674 012 278 751 232 × 2 = 0 + 0.000 005 348 024 557 502 464;
  • 49) 0.000 005 348 024 557 502 464 × 2 = 0 + 0.000 010 696 049 115 004 928;
  • 50) 0.000 010 696 049 115 004 928 × 2 = 0 + 0.000 021 392 098 230 009 856;
  • 51) 0.000 021 392 098 230 009 856 × 2 = 0 + 0.000 042 784 196 460 019 712;
  • 52) 0.000 042 784 196 460 019 712 × 2 = 0 + 0.000 085 568 392 920 039 424;
  • 53) 0.000 085 568 392 920 039 424 × 2 = 0 + 0.000 171 136 785 840 078 848;
  • 54) 0.000 171 136 785 840 078 848 × 2 = 0 + 0.000 342 273 571 680 157 696;
  • 55) 0.000 342 273 571 680 157 696 × 2 = 0 + 0.000 684 547 143 360 315 392;
  • 56) 0.000 684 547 143 360 315 392 × 2 = 0 + 0.001 369 094 286 720 630 784;
  • 57) 0.001 369 094 286 720 630 784 × 2 = 0 + 0.002 738 188 573 441 261 568;
  • 58) 0.002 738 188 573 441 261 568 × 2 = 0 + 0.005 476 377 146 882 523 136;
  • 59) 0.005 476 377 146 882 523 136 × 2 = 0 + 0.010 952 754 293 765 046 272;
  • 60) 0.010 952 754 293 765 046 272 × 2 = 0 + 0.021 905 508 587 530 092 544;
  • 61) 0.021 905 508 587 530 092 544 × 2 = 0 + 0.043 811 017 175 060 185 088;
  • 62) 0.043 811 017 175 060 185 088 × 2 = 0 + 0.087 622 034 350 120 370 176;
  • 63) 0.087 622 034 350 120 370 176 × 2 = 0 + 0.175 244 068 700 240 740 352;
  • 64) 0.175 244 068 700 240 740 352 × 2 = 0 + 0.350 488 137 400 481 480 704;
  • 65) 0.350 488 137 400 481 480 704 × 2 = 0 + 0.700 976 274 800 962 961 408;
  • 66) 0.700 976 274 800 962 961 408 × 2 = 1 + 0.401 952 549 601 925 922 816;
  • 67) 0.401 952 549 601 925 922 816 × 2 = 0 + 0.803 905 099 203 851 845 632;
  • 68) 0.803 905 099 203 851 845 632 × 2 = 1 + 0.607 810 198 407 703 691 264;
  • 69) 0.607 810 198 407 703 691 264 × 2 = 1 + 0.215 620 396 815 407 382 528;
  • 70) 0.215 620 396 815 407 382 528 × 2 = 0 + 0.431 240 793 630 814 765 056;
  • 71) 0.431 240 793 630 814 765 056 × 2 = 0 + 0.862 481 587 261 629 530 112;
  • 72) 0.862 481 587 261 629 530 112 × 2 = 1 + 0.724 963 174 523 259 060 224;
  • 73) 0.724 963 174 523 259 060 224 × 2 = 1 + 0.449 926 349 046 518 120 448;
  • 74) 0.449 926 349 046 518 120 448 × 2 = 0 + 0.899 852 698 093 036 240 896;
  • 75) 0.899 852 698 093 036 240 896 × 2 = 1 + 0.799 705 396 186 072 481 792;
  • 76) 0.799 705 396 186 072 481 792 × 2 = 1 + 0.599 410 792 372 144 963 584;
  • 77) 0.599 410 792 372 144 963 584 × 2 = 1 + 0.198 821 584 744 289 927 168;
  • 78) 0.198 821 584 744 289 927 168 × 2 = 0 + 0.397 643 169 488 579 854 336;
  • 79) 0.397 643 169 488 579 854 336 × 2 = 0 + 0.795 286 338 977 159 708 672;
  • 80) 0.795 286 338 977 159 708 672 × 2 = 1 + 0.590 572 677 954 319 417 344;
  • 81) 0.590 572 677 954 319 417 344 × 2 = 1 + 0.181 145 355 908 638 834 688;
  • 82) 0.181 145 355 908 638 834 688 × 2 = 0 + 0.362 290 711 817 277 669 376;
  • 83) 0.362 290 711 817 277 669 376 × 2 = 0 + 0.724 581 423 634 555 338 752;
  • 84) 0.724 581 423 634 555 338 752 × 2 = 1 + 0.449 162 847 269 110 677 504;
  • 85) 0.449 162 847 269 110 677 504 × 2 = 0 + 0.898 325 694 538 221 355 008;
  • 86) 0.898 325 694 538 221 355 008 × 2 = 1 + 0.796 651 389 076 442 710 016;
  • 87) 0.796 651 389 076 442 710 016 × 2 = 1 + 0.593 302 778 152 885 420 032;
  • 88) 0.593 302 778 152 885 420 032 × 2 = 1 + 0.186 605 556 305 770 840 064;
  • 89) 0.186 605 556 305 770 840 064 × 2 = 0 + 0.373 211 112 611 541 680 128;
  • 90) 0.373 211 112 611 541 680 128 × 2 = 0 + 0.746 422 225 223 083 360 256;
  • 91) 0.746 422 225 223 083 360 256 × 2 = 1 + 0.492 844 450 446 166 720 512;
  • 92) 0.492 844 450 446 166 720 512 × 2 = 0 + 0.985 688 900 892 333 441 024;
  • 93) 0.985 688 900 892 333 441 024 × 2 = 1 + 0.971 377 801 784 666 882 048;
  • 94) 0.971 377 801 784 666 882 048 × 2 = 1 + 0.942 755 603 569 333 764 096;
  • 95) 0.942 755 603 569 333 764 096 × 2 = 1 + 0.885 511 207 138 667 528 192;
  • 96) 0.885 511 207 138 667 528 192 × 2 = 1 + 0.771 022 414 277 335 056 384;
  • 97) 0.771 022 414 277 335 056 384 × 2 = 1 + 0.542 044 828 554 670 112 768;
  • 98) 0.542 044 828 554 670 112 768 × 2 = 1 + 0.084 089 657 109 340 225 536;
  • 99) 0.084 089 657 109 340 225 536 × 2 = 0 + 0.168 179 314 218 680 451 072;
  • 100) 0.168 179 314 218 680 451 072 × 2 = 0 + 0.336 358 628 437 360 902 144;
  • 101) 0.336 358 628 437 360 902 144 × 2 = 0 + 0.672 717 256 874 721 804 288;
  • 102) 0.672 717 256 874 721 804 288 × 2 = 1 + 0.345 434 513 749 443 608 576;
  • 103) 0.345 434 513 749 443 608 576 × 2 = 0 + 0.690 869 027 498 887 217 152;
  • 104) 0.690 869 027 498 887 217 152 × 2 = 1 + 0.381 738 054 997 774 434 304;
  • 105) 0.381 738 054 997 774 434 304 × 2 = 0 + 0.763 476 109 995 548 868 608;
  • 106) 0.763 476 109 995 548 868 608 × 2 = 1 + 0.526 952 219 991 097 737 216;
  • 107) 0.526 952 219 991 097 737 216 × 2 = 1 + 0.053 904 439 982 195 474 432;
  • 108) 0.053 904 439 982 195 474 432 × 2 = 0 + 0.107 808 879 964 390 948 864;
  • 109) 0.107 808 879 964 390 948 864 × 2 = 0 + 0.215 617 759 928 781 897 728;
  • 110) 0.215 617 759 928 781 897 728 × 2 = 0 + 0.431 235 519 857 563 795 456;
  • 111) 0.431 235 519 857 563 795 456 × 2 = 0 + 0.862 471 039 715 127 590 912;
  • 112) 0.862 471 039 715 127 590 912 × 2 = 1 + 0.724 942 079 430 255 181 824;
  • 113) 0.724 942 079 430 255 181 824 × 2 = 1 + 0.449 884 158 860 510 363 648;
  • 114) 0.449 884 158 860 510 363 648 × 2 = 0 + 0.899 768 317 721 020 727 296;
  • 115) 0.899 768 317 721 020 727 296 × 2 = 1 + 0.799 536 635 442 041 454 592;
  • 116) 0.799 536 635 442 041 454 592 × 2 = 1 + 0.599 073 270 884 082 909 184;
  • 117) 0.599 073 270 884 082 909 184 × 2 = 1 + 0.198 146 541 768 165 818 368;
  • 118) 0.198 146 541 768 165 818 368 × 2 = 0 + 0.396 293 083 536 331 636 736;

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:


0.000 000 000 000 000 000 019(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1001 1011 1001 1001 0111 0010 1111 1100 0101 0110 0001 1011 10(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 019(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1001 1011 1001 1001 0111 0010 1111 1100 0101 0110 0001 1011 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 66 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 019(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1001 1011 1001 1001 0111 0010 1111 1100 0101 0110 0001 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1001 1011 1001 1001 0111 0010 1111 1100 0101 0110 0001 1011 10(2) × 20 =

1.0110 0110 1110 0110 0101 1100 1011 1111 0001 0101 1000 0110 1110(2) × 2-66


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): -66

Mantissa (not normalized):
1.0110 0110 1110 0110 0101 1100 1011 1111 0001 0101 1000 0110 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-66 + 2(11-1) - 1 =

(-66 + 1 023)(10) =

957(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;
  • 957 ÷ 2 = 478 + 1;
  • 478 ÷ 2 = 239 + 0;
  • 239 ÷ 2 = 119 + 1;
  • 119 ÷ 2 = 59 + 1;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 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) =

957(10) =

011 1011 1101(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. 0110 0110 1110 0110 0101 1100 1011 1111 0001 0101 1000 0110 1110 =

0110 0110 1110 0110 0101 1100 1011 1111 0001 0101 1000 0110 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 1011 1101

Mantissa (52 bits) =
0110 0110 1110 0110 0101 1100 1011 1111 0001 0101 1000 0110 1110


Decimal number -0.000 000 000 000 000 000 019 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1011 1101 - 0110 0110 1110 0110 0101 1100 1011 1111 0001 0101 1000 0110 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