0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6(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
0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6(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: 0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6.

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) 0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6 × 2 = 0 + 0.000 002 222 200 000 222 220 000 022 222 000 002 222 199 987 2;
  • 2) 0.000 002 222 200 000 222 220 000 022 222 000 002 222 199 987 2 × 2 = 0 + 0.000 004 444 400 000 444 440 000 044 444 000 004 444 399 974 4;
  • 3) 0.000 004 444 400 000 444 440 000 044 444 000 004 444 399 974 4 × 2 = 0 + 0.000 008 888 800 000 888 880 000 088 888 000 008 888 799 948 8;
  • 4) 0.000 008 888 800 000 888 880 000 088 888 000 008 888 799 948 8 × 2 = 0 + 0.000 017 777 600 001 777 760 000 177 776 000 017 777 599 897 6;
  • 5) 0.000 017 777 600 001 777 760 000 177 776 000 017 777 599 897 6 × 2 = 0 + 0.000 035 555 200 003 555 520 000 355 552 000 035 555 199 795 2;
  • 6) 0.000 035 555 200 003 555 520 000 355 552 000 035 555 199 795 2 × 2 = 0 + 0.000 071 110 400 007 111 040 000 711 104 000 071 110 399 590 4;
  • 7) 0.000 071 110 400 007 111 040 000 711 104 000 071 110 399 590 4 × 2 = 0 + 0.000 142 220 800 014 222 080 001 422 208 000 142 220 799 180 8;
  • 8) 0.000 142 220 800 014 222 080 001 422 208 000 142 220 799 180 8 × 2 = 0 + 0.000 284 441 600 028 444 160 002 844 416 000 284 441 598 361 6;
  • 9) 0.000 284 441 600 028 444 160 002 844 416 000 284 441 598 361 6 × 2 = 0 + 0.000 568 883 200 056 888 320 005 688 832 000 568 883 196 723 2;
  • 10) 0.000 568 883 200 056 888 320 005 688 832 000 568 883 196 723 2 × 2 = 0 + 0.001 137 766 400 113 776 640 011 377 664 001 137 766 393 446 4;
  • 11) 0.001 137 766 400 113 776 640 011 377 664 001 137 766 393 446 4 × 2 = 0 + 0.002 275 532 800 227 553 280 022 755 328 002 275 532 786 892 8;
  • 12) 0.002 275 532 800 227 553 280 022 755 328 002 275 532 786 892 8 × 2 = 0 + 0.004 551 065 600 455 106 560 045 510 656 004 551 065 573 785 6;
  • 13) 0.004 551 065 600 455 106 560 045 510 656 004 551 065 573 785 6 × 2 = 0 + 0.009 102 131 200 910 213 120 091 021 312 009 102 131 147 571 2;
  • 14) 0.009 102 131 200 910 213 120 091 021 312 009 102 131 147 571 2 × 2 = 0 + 0.018 204 262 401 820 426 240 182 042 624 018 204 262 295 142 4;
  • 15) 0.018 204 262 401 820 426 240 182 042 624 018 204 262 295 142 4 × 2 = 0 + 0.036 408 524 803 640 852 480 364 085 248 036 408 524 590 284 8;
  • 16) 0.036 408 524 803 640 852 480 364 085 248 036 408 524 590 284 8 × 2 = 0 + 0.072 817 049 607 281 704 960 728 170 496 072 817 049 180 569 6;
  • 17) 0.072 817 049 607 281 704 960 728 170 496 072 817 049 180 569 6 × 2 = 0 + 0.145 634 099 214 563 409 921 456 340 992 145 634 098 361 139 2;
  • 18) 0.145 634 099 214 563 409 921 456 340 992 145 634 098 361 139 2 × 2 = 0 + 0.291 268 198 429 126 819 842 912 681 984 291 268 196 722 278 4;
  • 19) 0.291 268 198 429 126 819 842 912 681 984 291 268 196 722 278 4 × 2 = 0 + 0.582 536 396 858 253 639 685 825 363 968 582 536 393 444 556 8;
  • 20) 0.582 536 396 858 253 639 685 825 363 968 582 536 393 444 556 8 × 2 = 1 + 0.165 072 793 716 507 279 371 650 727 937 165 072 786 889 113 6;
  • 21) 0.165 072 793 716 507 279 371 650 727 937 165 072 786 889 113 6 × 2 = 0 + 0.330 145 587 433 014 558 743 301 455 874 330 145 573 778 227 2;
  • 22) 0.330 145 587 433 014 558 743 301 455 874 330 145 573 778 227 2 × 2 = 0 + 0.660 291 174 866 029 117 486 602 911 748 660 291 147 556 454 4;
  • 23) 0.660 291 174 866 029 117 486 602 911 748 660 291 147 556 454 4 × 2 = 1 + 0.320 582 349 732 058 234 973 205 823 497 320 582 295 112 908 8;
  • 24) 0.320 582 349 732 058 234 973 205 823 497 320 582 295 112 908 8 × 2 = 0 + 0.641 164 699 464 116 469 946 411 646 994 641 164 590 225 817 6;
  • 25) 0.641 164 699 464 116 469 946 411 646 994 641 164 590 225 817 6 × 2 = 1 + 0.282 329 398 928 232 939 892 823 293 989 282 329 180 451 635 2;
  • 26) 0.282 329 398 928 232 939 892 823 293 989 282 329 180 451 635 2 × 2 = 0 + 0.564 658 797 856 465 879 785 646 587 978 564 658 360 903 270 4;
  • 27) 0.564 658 797 856 465 879 785 646 587 978 564 658 360 903 270 4 × 2 = 1 + 0.129 317 595 712 931 759 571 293 175 957 129 316 721 806 540 8;
  • 28) 0.129 317 595 712 931 759 571 293 175 957 129 316 721 806 540 8 × 2 = 0 + 0.258 635 191 425 863 519 142 586 351 914 258 633 443 613 081 6;
  • 29) 0.258 635 191 425 863 519 142 586 351 914 258 633 443 613 081 6 × 2 = 0 + 0.517 270 382 851 727 038 285 172 703 828 517 266 887 226 163 2;
  • 30) 0.517 270 382 851 727 038 285 172 703 828 517 266 887 226 163 2 × 2 = 1 + 0.034 540 765 703 454 076 570 345 407 657 034 533 774 452 326 4;
  • 31) 0.034 540 765 703 454 076 570 345 407 657 034 533 774 452 326 4 × 2 = 0 + 0.069 081 531 406 908 153 140 690 815 314 069 067 548 904 652 8;
  • 32) 0.069 081 531 406 908 153 140 690 815 314 069 067 548 904 652 8 × 2 = 0 + 0.138 163 062 813 816 306 281 381 630 628 138 135 097 809 305 6;
  • 33) 0.138 163 062 813 816 306 281 381 630 628 138 135 097 809 305 6 × 2 = 0 + 0.276 326 125 627 632 612 562 763 261 256 276 270 195 618 611 2;
  • 34) 0.276 326 125 627 632 612 562 763 261 256 276 270 195 618 611 2 × 2 = 0 + 0.552 652 251 255 265 225 125 526 522 512 552 540 391 237 222 4;
  • 35) 0.552 652 251 255 265 225 125 526 522 512 552 540 391 237 222 4 × 2 = 1 + 0.105 304 502 510 530 450 251 053 045 025 105 080 782 474 444 8;
  • 36) 0.105 304 502 510 530 450 251 053 045 025 105 080 782 474 444 8 × 2 = 0 + 0.210 609 005 021 060 900 502 106 090 050 210 161 564 948 889 6;
  • 37) 0.210 609 005 021 060 900 502 106 090 050 210 161 564 948 889 6 × 2 = 0 + 0.421 218 010 042 121 801 004 212 180 100 420 323 129 897 779 2;
  • 38) 0.421 218 010 042 121 801 004 212 180 100 420 323 129 897 779 2 × 2 = 0 + 0.842 436 020 084 243 602 008 424 360 200 840 646 259 795 558 4;
  • 39) 0.842 436 020 084 243 602 008 424 360 200 840 646 259 795 558 4 × 2 = 1 + 0.684 872 040 168 487 204 016 848 720 401 681 292 519 591 116 8;
  • 40) 0.684 872 040 168 487 204 016 848 720 401 681 292 519 591 116 8 × 2 = 1 + 0.369 744 080 336 974 408 033 697 440 803 362 585 039 182 233 6;
  • 41) 0.369 744 080 336 974 408 033 697 440 803 362 585 039 182 233 6 × 2 = 0 + 0.739 488 160 673 948 816 067 394 881 606 725 170 078 364 467 2;
  • 42) 0.739 488 160 673 948 816 067 394 881 606 725 170 078 364 467 2 × 2 = 1 + 0.478 976 321 347 897 632 134 789 763 213 450 340 156 728 934 4;
  • 43) 0.478 976 321 347 897 632 134 789 763 213 450 340 156 728 934 4 × 2 = 0 + 0.957 952 642 695 795 264 269 579 526 426 900 680 313 457 868 8;

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:


0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6(10) =

0.0000 0000 0000 0000 0001 0010 1010 0100 0010 0011 010(2)

5. Positive number before normalization:

0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6(10) =

0.0000 0000 0000 0000 0001 0010 1010 0100 0010 0011 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6(10) =

0.0000 0000 0000 0000 0001 0010 1010 0100 0010 0011 010(2) =

0.0000 0000 0000 0000 0001 0010 1010 0100 0010 0011 010(2) × 20 =

1.0010 1010 0100 0010 0011 010(2) × 2-20


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): -20

Mantissa (not normalized):
1.0010 1010 0100 0010 0011 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

107(10) =

0110 1011(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. 001 0101 0010 0001 0001 1010 =

001 0101 0010 0001 0001 1010


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) =
0110 1011

Mantissa (23 bits) =
001 0101 0010 0001 0001 1010


Decimal number 0.000 001 111 100 000 111 110 000 011 111 000 001 111 099 993 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1011 - 001 0101 0010 0001 0001 1010

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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