0.000 001 111 100 000 111 110 000 011 111 000 001 111 87 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 87(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 87(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 87.

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 87 × 2 = 0 + 0.000 002 222 200 000 222 220 000 022 222 000 002 223 74;
  • 2) 0.000 002 222 200 000 222 220 000 022 222 000 002 223 74 × 2 = 0 + 0.000 004 444 400 000 444 440 000 044 444 000 004 447 48;
  • 3) 0.000 004 444 400 000 444 440 000 044 444 000 004 447 48 × 2 = 0 + 0.000 008 888 800 000 888 880 000 088 888 000 008 894 96;
  • 4) 0.000 008 888 800 000 888 880 000 088 888 000 008 894 96 × 2 = 0 + 0.000 017 777 600 001 777 760 000 177 776 000 017 789 92;
  • 5) 0.000 017 777 600 001 777 760 000 177 776 000 017 789 92 × 2 = 0 + 0.000 035 555 200 003 555 520 000 355 552 000 035 579 84;
  • 6) 0.000 035 555 200 003 555 520 000 355 552 000 035 579 84 × 2 = 0 + 0.000 071 110 400 007 111 040 000 711 104 000 071 159 68;
  • 7) 0.000 071 110 400 007 111 040 000 711 104 000 071 159 68 × 2 = 0 + 0.000 142 220 800 014 222 080 001 422 208 000 142 319 36;
  • 8) 0.000 142 220 800 014 222 080 001 422 208 000 142 319 36 × 2 = 0 + 0.000 284 441 600 028 444 160 002 844 416 000 284 638 72;
  • 9) 0.000 284 441 600 028 444 160 002 844 416 000 284 638 72 × 2 = 0 + 0.000 568 883 200 056 888 320 005 688 832 000 569 277 44;
  • 10) 0.000 568 883 200 056 888 320 005 688 832 000 569 277 44 × 2 = 0 + 0.001 137 766 400 113 776 640 011 377 664 001 138 554 88;
  • 11) 0.001 137 766 400 113 776 640 011 377 664 001 138 554 88 × 2 = 0 + 0.002 275 532 800 227 553 280 022 755 328 002 277 109 76;
  • 12) 0.002 275 532 800 227 553 280 022 755 328 002 277 109 76 × 2 = 0 + 0.004 551 065 600 455 106 560 045 510 656 004 554 219 52;
  • 13) 0.004 551 065 600 455 106 560 045 510 656 004 554 219 52 × 2 = 0 + 0.009 102 131 200 910 213 120 091 021 312 009 108 439 04;
  • 14) 0.009 102 131 200 910 213 120 091 021 312 009 108 439 04 × 2 = 0 + 0.018 204 262 401 820 426 240 182 042 624 018 216 878 08;
  • 15) 0.018 204 262 401 820 426 240 182 042 624 018 216 878 08 × 2 = 0 + 0.036 408 524 803 640 852 480 364 085 248 036 433 756 16;
  • 16) 0.036 408 524 803 640 852 480 364 085 248 036 433 756 16 × 2 = 0 + 0.072 817 049 607 281 704 960 728 170 496 072 867 512 32;
  • 17) 0.072 817 049 607 281 704 960 728 170 496 072 867 512 32 × 2 = 0 + 0.145 634 099 214 563 409 921 456 340 992 145 735 024 64;
  • 18) 0.145 634 099 214 563 409 921 456 340 992 145 735 024 64 × 2 = 0 + 0.291 268 198 429 126 819 842 912 681 984 291 470 049 28;
  • 19) 0.291 268 198 429 126 819 842 912 681 984 291 470 049 28 × 2 = 0 + 0.582 536 396 858 253 639 685 825 363 968 582 940 098 56;
  • 20) 0.582 536 396 858 253 639 685 825 363 968 582 940 098 56 × 2 = 1 + 0.165 072 793 716 507 279 371 650 727 937 165 880 197 12;
  • 21) 0.165 072 793 716 507 279 371 650 727 937 165 880 197 12 × 2 = 0 + 0.330 145 587 433 014 558 743 301 455 874 331 760 394 24;
  • 22) 0.330 145 587 433 014 558 743 301 455 874 331 760 394 24 × 2 = 0 + 0.660 291 174 866 029 117 486 602 911 748 663 520 788 48;
  • 23) 0.660 291 174 866 029 117 486 602 911 748 663 520 788 48 × 2 = 1 + 0.320 582 349 732 058 234 973 205 823 497 327 041 576 96;
  • 24) 0.320 582 349 732 058 234 973 205 823 497 327 041 576 96 × 2 = 0 + 0.641 164 699 464 116 469 946 411 646 994 654 083 153 92;
  • 25) 0.641 164 699 464 116 469 946 411 646 994 654 083 153 92 × 2 = 1 + 0.282 329 398 928 232 939 892 823 293 989 308 166 307 84;
  • 26) 0.282 329 398 928 232 939 892 823 293 989 308 166 307 84 × 2 = 0 + 0.564 658 797 856 465 879 785 646 587 978 616 332 615 68;
  • 27) 0.564 658 797 856 465 879 785 646 587 978 616 332 615 68 × 2 = 1 + 0.129 317 595 712 931 759 571 293 175 957 232 665 231 36;
  • 28) 0.129 317 595 712 931 759 571 293 175 957 232 665 231 36 × 2 = 0 + 0.258 635 191 425 863 519 142 586 351 914 465 330 462 72;
  • 29) 0.258 635 191 425 863 519 142 586 351 914 465 330 462 72 × 2 = 0 + 0.517 270 382 851 727 038 285 172 703 828 930 660 925 44;
  • 30) 0.517 270 382 851 727 038 285 172 703 828 930 660 925 44 × 2 = 1 + 0.034 540 765 703 454 076 570 345 407 657 861 321 850 88;
  • 31) 0.034 540 765 703 454 076 570 345 407 657 861 321 850 88 × 2 = 0 + 0.069 081 531 406 908 153 140 690 815 315 722 643 701 76;
  • 32) 0.069 081 531 406 908 153 140 690 815 315 722 643 701 76 × 2 = 0 + 0.138 163 062 813 816 306 281 381 630 631 445 287 403 52;
  • 33) 0.138 163 062 813 816 306 281 381 630 631 445 287 403 52 × 2 = 0 + 0.276 326 125 627 632 612 562 763 261 262 890 574 807 04;
  • 34) 0.276 326 125 627 632 612 562 763 261 262 890 574 807 04 × 2 = 0 + 0.552 652 251 255 265 225 125 526 522 525 781 149 614 08;
  • 35) 0.552 652 251 255 265 225 125 526 522 525 781 149 614 08 × 2 = 1 + 0.105 304 502 510 530 450 251 053 045 051 562 299 228 16;
  • 36) 0.105 304 502 510 530 450 251 053 045 051 562 299 228 16 × 2 = 0 + 0.210 609 005 021 060 900 502 106 090 103 124 598 456 32;
  • 37) 0.210 609 005 021 060 900 502 106 090 103 124 598 456 32 × 2 = 0 + 0.421 218 010 042 121 801 004 212 180 206 249 196 912 64;
  • 38) 0.421 218 010 042 121 801 004 212 180 206 249 196 912 64 × 2 = 0 + 0.842 436 020 084 243 602 008 424 360 412 498 393 825 28;
  • 39) 0.842 436 020 084 243 602 008 424 360 412 498 393 825 28 × 2 = 1 + 0.684 872 040 168 487 204 016 848 720 824 996 787 650 56;
  • 40) 0.684 872 040 168 487 204 016 848 720 824 996 787 650 56 × 2 = 1 + 0.369 744 080 336 974 408 033 697 441 649 993 575 301 12;
  • 41) 0.369 744 080 336 974 408 033 697 441 649 993 575 301 12 × 2 = 0 + 0.739 488 160 673 948 816 067 394 883 299 987 150 602 24;
  • 42) 0.739 488 160 673 948 816 067 394 883 299 987 150 602 24 × 2 = 1 + 0.478 976 321 347 897 632 134 789 766 599 974 301 204 48;
  • 43) 0.478 976 321 347 897 632 134 789 766 599 974 301 204 48 × 2 = 0 + 0.957 952 642 695 795 264 269 579 533 199 948 602 408 96;

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 87(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 87(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 87(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 87 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1011 - 001 0101 0010 0001 0001 1010

» Convert decimal 0.000 001 111 100 000 111 110 000 011 111 000 001 112 53 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