0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9(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
0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9(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: 0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9.

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) 0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9 × 2 = 0 + 0.400 000 000 000 000 022 204 460 492 503 130 808 472 633 361 816 385 8;
  • 2) 0.400 000 000 000 000 022 204 460 492 503 130 808 472 633 361 816 385 8 × 2 = 0 + 0.800 000 000 000 000 044 408 920 985 006 261 616 945 266 723 632 771 6;
  • 3) 0.800 000 000 000 000 044 408 920 985 006 261 616 945 266 723 632 771 6 × 2 = 1 + 0.600 000 000 000 000 088 817 841 970 012 523 233 890 533 447 265 543 2;
  • 4) 0.600 000 000 000 000 088 817 841 970 012 523 233 890 533 447 265 543 2 × 2 = 1 + 0.200 000 000 000 000 177 635 683 940 025 046 467 781 066 894 531 086 4;
  • 5) 0.200 000 000 000 000 177 635 683 940 025 046 467 781 066 894 531 086 4 × 2 = 0 + 0.400 000 000 000 000 355 271 367 880 050 092 935 562 133 789 062 172 8;
  • 6) 0.400 000 000 000 000 355 271 367 880 050 092 935 562 133 789 062 172 8 × 2 = 0 + 0.800 000 000 000 000 710 542 735 760 100 185 871 124 267 578 124 345 6;
  • 7) 0.800 000 000 000 000 710 542 735 760 100 185 871 124 267 578 124 345 6 × 2 = 1 + 0.600 000 000 000 001 421 085 471 520 200 371 742 248 535 156 248 691 2;
  • 8) 0.600 000 000 000 001 421 085 471 520 200 371 742 248 535 156 248 691 2 × 2 = 1 + 0.200 000 000 000 002 842 170 943 040 400 743 484 497 070 312 497 382 4;
  • 9) 0.200 000 000 000 002 842 170 943 040 400 743 484 497 070 312 497 382 4 × 2 = 0 + 0.400 000 000 000 005 684 341 886 080 801 486 968 994 140 624 994 764 8;
  • 10) 0.400 000 000 000 005 684 341 886 080 801 486 968 994 140 624 994 764 8 × 2 = 0 + 0.800 000 000 000 011 368 683 772 161 602 973 937 988 281 249 989 529 6;
  • 11) 0.800 000 000 000 011 368 683 772 161 602 973 937 988 281 249 989 529 6 × 2 = 1 + 0.600 000 000 000 022 737 367 544 323 205 947 875 976 562 499 979 059 2;
  • 12) 0.600 000 000 000 022 737 367 544 323 205 947 875 976 562 499 979 059 2 × 2 = 1 + 0.200 000 000 000 045 474 735 088 646 411 895 751 953 124 999 958 118 4;
  • 13) 0.200 000 000 000 045 474 735 088 646 411 895 751 953 124 999 958 118 4 × 2 = 0 + 0.400 000 000 000 090 949 470 177 292 823 791 503 906 249 999 916 236 8;
  • 14) 0.400 000 000 000 090 949 470 177 292 823 791 503 906 249 999 916 236 8 × 2 = 0 + 0.800 000 000 000 181 898 940 354 585 647 583 007 812 499 999 832 473 6;
  • 15) 0.800 000 000 000 181 898 940 354 585 647 583 007 812 499 999 832 473 6 × 2 = 1 + 0.600 000 000 000 363 797 880 709 171 295 166 015 624 999 999 664 947 2;
  • 16) 0.600 000 000 000 363 797 880 709 171 295 166 015 624 999 999 664 947 2 × 2 = 1 + 0.200 000 000 000 727 595 761 418 342 590 332 031 249 999 999 329 894 4;
  • 17) 0.200 000 000 000 727 595 761 418 342 590 332 031 249 999 999 329 894 4 × 2 = 0 + 0.400 000 000 001 455 191 522 836 685 180 664 062 499 999 998 659 788 8;
  • 18) 0.400 000 000 001 455 191 522 836 685 180 664 062 499 999 998 659 788 8 × 2 = 0 + 0.800 000 000 002 910 383 045 673 370 361 328 124 999 999 997 319 577 6;
  • 19) 0.800 000 000 002 910 383 045 673 370 361 328 124 999 999 997 319 577 6 × 2 = 1 + 0.600 000 000 005 820 766 091 346 740 722 656 249 999 999 994 639 155 2;
  • 20) 0.600 000 000 005 820 766 091 346 740 722 656 249 999 999 994 639 155 2 × 2 = 1 + 0.200 000 000 011 641 532 182 693 481 445 312 499 999 999 989 278 310 4;
  • 21) 0.200 000 000 011 641 532 182 693 481 445 312 499 999 999 989 278 310 4 × 2 = 0 + 0.400 000 000 023 283 064 365 386 962 890 624 999 999 999 978 556 620 8;
  • 22) 0.400 000 000 023 283 064 365 386 962 890 624 999 999 999 978 556 620 8 × 2 = 0 + 0.800 000 000 046 566 128 730 773 925 781 249 999 999 999 957 113 241 6;
  • 23) 0.800 000 000 046 566 128 730 773 925 781 249 999 999 999 957 113 241 6 × 2 = 1 + 0.600 000 000 093 132 257 461 547 851 562 499 999 999 999 914 226 483 2;
  • 24) 0.600 000 000 093 132 257 461 547 851 562 499 999 999 999 914 226 483 2 × 2 = 1 + 0.200 000 000 186 264 514 923 095 703 124 999 999 999 999 828 452 966 4;
  • 25) 0.200 000 000 186 264 514 923 095 703 124 999 999 999 999 828 452 966 4 × 2 = 0 + 0.400 000 000 372 529 029 846 191 406 249 999 999 999 999 656 905 932 8;
  • 26) 0.400 000 000 372 529 029 846 191 406 249 999 999 999 999 656 905 932 8 × 2 = 0 + 0.800 000 000 745 058 059 692 382 812 499 999 999 999 999 313 811 865 6;

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:


0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9(10) =

0.0011 0011 0011 0011 0011 0011 00(2)

5. Positive number before normalization:

0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9(10) =

0.0011 0011 0011 0011 0011 0011 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9(10) =

0.0011 0011 0011 0011 0011 0011 00(2) =

0.0011 0011 0011 0011 0011 0011 00(2) × 20 =

1.1001 1001 1001 1001 1001 100(2) × 2-3


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): -3

Mantissa (not normalized):
1.1001 1001 1001 1001 1001 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(8-1) - 1 =

(-3 + 127)(10) =

124(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;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

124(10) =

0111 1100(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 1100 1100 1100 1100 1100 =

100 1100 1100 1100 1100 1100


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) =
0111 1100

Mantissa (23 bits) =
100 1100 1100 1100 1100 1100


Decimal number 0.200 000 000 000 000 011 102 230 246 251 565 404 236 316 680 908 192 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1100 - 100 1100 1100 1100 1100 1100

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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