0.000 000 000 000 000 000 176 181 893 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 176 181 893(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 000 000 000 000 176 181 893(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 000 176 181 893.

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) 0.000 000 000 000 000 000 176 181 893 × 2 = 0 + 0.000 000 000 000 000 000 352 363 786;
  • 2) 0.000 000 000 000 000 000 352 363 786 × 2 = 0 + 0.000 000 000 000 000 000 704 727 572;
  • 3) 0.000 000 000 000 000 000 704 727 572 × 2 = 0 + 0.000 000 000 000 000 001 409 455 144;
  • 4) 0.000 000 000 000 000 001 409 455 144 × 2 = 0 + 0.000 000 000 000 000 002 818 910 288;
  • 5) 0.000 000 000 000 000 002 818 910 288 × 2 = 0 + 0.000 000 000 000 000 005 637 820 576;
  • 6) 0.000 000 000 000 000 005 637 820 576 × 2 = 0 + 0.000 000 000 000 000 011 275 641 152;
  • 7) 0.000 000 000 000 000 011 275 641 152 × 2 = 0 + 0.000 000 000 000 000 022 551 282 304;
  • 8) 0.000 000 000 000 000 022 551 282 304 × 2 = 0 + 0.000 000 000 000 000 045 102 564 608;
  • 9) 0.000 000 000 000 000 045 102 564 608 × 2 = 0 + 0.000 000 000 000 000 090 205 129 216;
  • 10) 0.000 000 000 000 000 090 205 129 216 × 2 = 0 + 0.000 000 000 000 000 180 410 258 432;
  • 11) 0.000 000 000 000 000 180 410 258 432 × 2 = 0 + 0.000 000 000 000 000 360 820 516 864;
  • 12) 0.000 000 000 000 000 360 820 516 864 × 2 = 0 + 0.000 000 000 000 000 721 641 033 728;
  • 13) 0.000 000 000 000 000 721 641 033 728 × 2 = 0 + 0.000 000 000 000 001 443 282 067 456;
  • 14) 0.000 000 000 000 001 443 282 067 456 × 2 = 0 + 0.000 000 000 000 002 886 564 134 912;
  • 15) 0.000 000 000 000 002 886 564 134 912 × 2 = 0 + 0.000 000 000 000 005 773 128 269 824;
  • 16) 0.000 000 000 000 005 773 128 269 824 × 2 = 0 + 0.000 000 000 000 011 546 256 539 648;
  • 17) 0.000 000 000 000 011 546 256 539 648 × 2 = 0 + 0.000 000 000 000 023 092 513 079 296;
  • 18) 0.000 000 000 000 023 092 513 079 296 × 2 = 0 + 0.000 000 000 000 046 185 026 158 592;
  • 19) 0.000 000 000 000 046 185 026 158 592 × 2 = 0 + 0.000 000 000 000 092 370 052 317 184;
  • 20) 0.000 000 000 000 092 370 052 317 184 × 2 = 0 + 0.000 000 000 000 184 740 104 634 368;
  • 21) 0.000 000 000 000 184 740 104 634 368 × 2 = 0 + 0.000 000 000 000 369 480 209 268 736;
  • 22) 0.000 000 000 000 369 480 209 268 736 × 2 = 0 + 0.000 000 000 000 738 960 418 537 472;
  • 23) 0.000 000 000 000 738 960 418 537 472 × 2 = 0 + 0.000 000 000 001 477 920 837 074 944;
  • 24) 0.000 000 000 001 477 920 837 074 944 × 2 = 0 + 0.000 000 000 002 955 841 674 149 888;
  • 25) 0.000 000 000 002 955 841 674 149 888 × 2 = 0 + 0.000 000 000 005 911 683 348 299 776;
  • 26) 0.000 000 000 005 911 683 348 299 776 × 2 = 0 + 0.000 000 000 011 823 366 696 599 552;
  • 27) 0.000 000 000 011 823 366 696 599 552 × 2 = 0 + 0.000 000 000 023 646 733 393 199 104;
  • 28) 0.000 000 000 023 646 733 393 199 104 × 2 = 0 + 0.000 000 000 047 293 466 786 398 208;
  • 29) 0.000 000 000 047 293 466 786 398 208 × 2 = 0 + 0.000 000 000 094 586 933 572 796 416;
  • 30) 0.000 000 000 094 586 933 572 796 416 × 2 = 0 + 0.000 000 000 189 173 867 145 592 832;
  • 31) 0.000 000 000 189 173 867 145 592 832 × 2 = 0 + 0.000 000 000 378 347 734 291 185 664;
  • 32) 0.000 000 000 378 347 734 291 185 664 × 2 = 0 + 0.000 000 000 756 695 468 582 371 328;
  • 33) 0.000 000 000 756 695 468 582 371 328 × 2 = 0 + 0.000 000 001 513 390 937 164 742 656;
  • 34) 0.000 000 001 513 390 937 164 742 656 × 2 = 0 + 0.000 000 003 026 781 874 329 485 312;
  • 35) 0.000 000 003 026 781 874 329 485 312 × 2 = 0 + 0.000 000 006 053 563 748 658 970 624;
  • 36) 0.000 000 006 053 563 748 658 970 624 × 2 = 0 + 0.000 000 012 107 127 497 317 941 248;
  • 37) 0.000 000 012 107 127 497 317 941 248 × 2 = 0 + 0.000 000 024 214 254 994 635 882 496;
  • 38) 0.000 000 024 214 254 994 635 882 496 × 2 = 0 + 0.000 000 048 428 509 989 271 764 992;
  • 39) 0.000 000 048 428 509 989 271 764 992 × 2 = 0 + 0.000 000 096 857 019 978 543 529 984;
  • 40) 0.000 000 096 857 019 978 543 529 984 × 2 = 0 + 0.000 000 193 714 039 957 087 059 968;
  • 41) 0.000 000 193 714 039 957 087 059 968 × 2 = 0 + 0.000 000 387 428 079 914 174 119 936;
  • 42) 0.000 000 387 428 079 914 174 119 936 × 2 = 0 + 0.000 000 774 856 159 828 348 239 872;
  • 43) 0.000 000 774 856 159 828 348 239 872 × 2 = 0 + 0.000 001 549 712 319 656 696 479 744;
  • 44) 0.000 001 549 712 319 656 696 479 744 × 2 = 0 + 0.000 003 099 424 639 313 392 959 488;
  • 45) 0.000 003 099 424 639 313 392 959 488 × 2 = 0 + 0.000 006 198 849 278 626 785 918 976;
  • 46) 0.000 006 198 849 278 626 785 918 976 × 2 = 0 + 0.000 012 397 698 557 253 571 837 952;
  • 47) 0.000 012 397 698 557 253 571 837 952 × 2 = 0 + 0.000 024 795 397 114 507 143 675 904;
  • 48) 0.000 024 795 397 114 507 143 675 904 × 2 = 0 + 0.000 049 590 794 229 014 287 351 808;
  • 49) 0.000 049 590 794 229 014 287 351 808 × 2 = 0 + 0.000 099 181 588 458 028 574 703 616;
  • 50) 0.000 099 181 588 458 028 574 703 616 × 2 = 0 + 0.000 198 363 176 916 057 149 407 232;
  • 51) 0.000 198 363 176 916 057 149 407 232 × 2 = 0 + 0.000 396 726 353 832 114 298 814 464;
  • 52) 0.000 396 726 353 832 114 298 814 464 × 2 = 0 + 0.000 793 452 707 664 228 597 628 928;
  • 53) 0.000 793 452 707 664 228 597 628 928 × 2 = 0 + 0.001 586 905 415 328 457 195 257 856;
  • 54) 0.001 586 905 415 328 457 195 257 856 × 2 = 0 + 0.003 173 810 830 656 914 390 515 712;
  • 55) 0.003 173 810 830 656 914 390 515 712 × 2 = 0 + 0.006 347 621 661 313 828 781 031 424;
  • 56) 0.006 347 621 661 313 828 781 031 424 × 2 = 0 + 0.012 695 243 322 627 657 562 062 848;
  • 57) 0.012 695 243 322 627 657 562 062 848 × 2 = 0 + 0.025 390 486 645 255 315 124 125 696;
  • 58) 0.025 390 486 645 255 315 124 125 696 × 2 = 0 + 0.050 780 973 290 510 630 248 251 392;
  • 59) 0.050 780 973 290 510 630 248 251 392 × 2 = 0 + 0.101 561 946 581 021 260 496 502 784;
  • 60) 0.101 561 946 581 021 260 496 502 784 × 2 = 0 + 0.203 123 893 162 042 520 993 005 568;
  • 61) 0.203 123 893 162 042 520 993 005 568 × 2 = 0 + 0.406 247 786 324 085 041 986 011 136;
  • 62) 0.406 247 786 324 085 041 986 011 136 × 2 = 0 + 0.812 495 572 648 170 083 972 022 272;
  • 63) 0.812 495 572 648 170 083 972 022 272 × 2 = 1 + 0.624 991 145 296 340 167 944 044 544;
  • 64) 0.624 991 145 296 340 167 944 044 544 × 2 = 1 + 0.249 982 290 592 680 335 888 089 088;
  • 65) 0.249 982 290 592 680 335 888 089 088 × 2 = 0 + 0.499 964 581 185 360 671 776 178 176;
  • 66) 0.499 964 581 185 360 671 776 178 176 × 2 = 0 + 0.999 929 162 370 721 343 552 356 352;
  • 67) 0.999 929 162 370 721 343 552 356 352 × 2 = 1 + 0.999 858 324 741 442 687 104 712 704;
  • 68) 0.999 858 324 741 442 687 104 712 704 × 2 = 1 + 0.999 716 649 482 885 374 209 425 408;
  • 69) 0.999 716 649 482 885 374 209 425 408 × 2 = 1 + 0.999 433 298 965 770 748 418 850 816;
  • 70) 0.999 433 298 965 770 748 418 850 816 × 2 = 1 + 0.998 866 597 931 541 496 837 701 632;
  • 71) 0.998 866 597 931 541 496 837 701 632 × 2 = 1 + 0.997 733 195 863 082 993 675 403 264;
  • 72) 0.997 733 195 863 082 993 675 403 264 × 2 = 1 + 0.995 466 391 726 165 987 350 806 528;
  • 73) 0.995 466 391 726 165 987 350 806 528 × 2 = 1 + 0.990 932 783 452 331 974 701 613 056;
  • 74) 0.990 932 783 452 331 974 701 613 056 × 2 = 1 + 0.981 865 566 904 663 949 403 226 112;
  • 75) 0.981 865 566 904 663 949 403 226 112 × 2 = 1 + 0.963 731 133 809 327 898 806 452 224;
  • 76) 0.963 731 133 809 327 898 806 452 224 × 2 = 1 + 0.927 462 267 618 655 797 612 904 448;
  • 77) 0.927 462 267 618 655 797 612 904 448 × 2 = 1 + 0.854 924 535 237 311 595 225 808 896;
  • 78) 0.854 924 535 237 311 595 225 808 896 × 2 = 1 + 0.709 849 070 474 623 190 451 617 792;
  • 79) 0.709 849 070 474 623 190 451 617 792 × 2 = 1 + 0.419 698 140 949 246 380 903 235 584;
  • 80) 0.419 698 140 949 246 380 903 235 584 × 2 = 0 + 0.839 396 281 898 492 761 806 471 168;
  • 81) 0.839 396 281 898 492 761 806 471 168 × 2 = 1 + 0.678 792 563 796 985 523 612 942 336;
  • 82) 0.678 792 563 796 985 523 612 942 336 × 2 = 1 + 0.357 585 127 593 971 047 225 884 672;
  • 83) 0.357 585 127 593 971 047 225 884 672 × 2 = 0 + 0.715 170 255 187 942 094 451 769 344;
  • 84) 0.715 170 255 187 942 094 451 769 344 × 2 = 1 + 0.430 340 510 375 884 188 903 538 688;
  • 85) 0.430 340 510 375 884 188 903 538 688 × 2 = 0 + 0.860 681 020 751 768 377 807 077 376;
  • 86) 0.860 681 020 751 768 377 807 077 376 × 2 = 1 + 0.721 362 041 503 536 755 614 154 752;

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).


4. 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:


0.000 000 000 000 000 000 176 181 893(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1110 1101 01(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 176 181 893(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1110 1101 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 63 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 176 181 893(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1110 1101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1110 1101 01(2) × 20 =

1.1001 1111 1111 1111 0110 101(2) × 2-63


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -63

Mantissa (not normalized):
1.1001 1111 1111 1111 0110 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-63 + 2(8-1) - 1 =

(-63 + 127)(10) =

64(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

64(10) =

0100 0000(2)


11. 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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1111 1111 1111 1011 0101 =

100 1111 1111 1111 1011 0101


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0100 0000

Mantissa (23 bits) =
100 1111 1111 1111 1011 0101


Decimal number 0.000 000 000 000 000 000 176 181 893 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0100 0000 - 100 1111 1111 1111 1011 0101

» Convert decimal 0.000 000 000 000 000 000 176 181 961 to 32 bit single precision IEEE 754 binary floating point representation standard

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111