-0.000 000 000 742 146 92 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 92(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
-0.000 000 000 742 146 92(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. Start with the positive version of the number:

|-0.000 000 000 742 146 92| = 0.000 000 000 742 146 92


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: 0.000 000 000 742 146 92.

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) 0.000 000 000 742 146 92 × 2 = 0 + 0.000 000 001 484 293 84;
  • 2) 0.000 000 001 484 293 84 × 2 = 0 + 0.000 000 002 968 587 68;
  • 3) 0.000 000 002 968 587 68 × 2 = 0 + 0.000 000 005 937 175 36;
  • 4) 0.000 000 005 937 175 36 × 2 = 0 + 0.000 000 011 874 350 72;
  • 5) 0.000 000 011 874 350 72 × 2 = 0 + 0.000 000 023 748 701 44;
  • 6) 0.000 000 023 748 701 44 × 2 = 0 + 0.000 000 047 497 402 88;
  • 7) 0.000 000 047 497 402 88 × 2 = 0 + 0.000 000 094 994 805 76;
  • 8) 0.000 000 094 994 805 76 × 2 = 0 + 0.000 000 189 989 611 52;
  • 9) 0.000 000 189 989 611 52 × 2 = 0 + 0.000 000 379 979 223 04;
  • 10) 0.000 000 379 979 223 04 × 2 = 0 + 0.000 000 759 958 446 08;
  • 11) 0.000 000 759 958 446 08 × 2 = 0 + 0.000 001 519 916 892 16;
  • 12) 0.000 001 519 916 892 16 × 2 = 0 + 0.000 003 039 833 784 32;
  • 13) 0.000 003 039 833 784 32 × 2 = 0 + 0.000 006 079 667 568 64;
  • 14) 0.000 006 079 667 568 64 × 2 = 0 + 0.000 012 159 335 137 28;
  • 15) 0.000 012 159 335 137 28 × 2 = 0 + 0.000 024 318 670 274 56;
  • 16) 0.000 024 318 670 274 56 × 2 = 0 + 0.000 048 637 340 549 12;
  • 17) 0.000 048 637 340 549 12 × 2 = 0 + 0.000 097 274 681 098 24;
  • 18) 0.000 097 274 681 098 24 × 2 = 0 + 0.000 194 549 362 196 48;
  • 19) 0.000 194 549 362 196 48 × 2 = 0 + 0.000 389 098 724 392 96;
  • 20) 0.000 389 098 724 392 96 × 2 = 0 + 0.000 778 197 448 785 92;
  • 21) 0.000 778 197 448 785 92 × 2 = 0 + 0.001 556 394 897 571 84;
  • 22) 0.001 556 394 897 571 84 × 2 = 0 + 0.003 112 789 795 143 68;
  • 23) 0.003 112 789 795 143 68 × 2 = 0 + 0.006 225 579 590 287 36;
  • 24) 0.006 225 579 590 287 36 × 2 = 0 + 0.012 451 159 180 574 72;
  • 25) 0.012 451 159 180 574 72 × 2 = 0 + 0.024 902 318 361 149 44;
  • 26) 0.024 902 318 361 149 44 × 2 = 0 + 0.049 804 636 722 298 88;
  • 27) 0.049 804 636 722 298 88 × 2 = 0 + 0.099 609 273 444 597 76;
  • 28) 0.099 609 273 444 597 76 × 2 = 0 + 0.199 218 546 889 195 52;
  • 29) 0.199 218 546 889 195 52 × 2 = 0 + 0.398 437 093 778 391 04;
  • 30) 0.398 437 093 778 391 04 × 2 = 0 + 0.796 874 187 556 782 08;
  • 31) 0.796 874 187 556 782 08 × 2 = 1 + 0.593 748 375 113 564 16;
  • 32) 0.593 748 375 113 564 16 × 2 = 1 + 0.187 496 750 227 128 32;
  • 33) 0.187 496 750 227 128 32 × 2 = 0 + 0.374 993 500 454 256 64;
  • 34) 0.374 993 500 454 256 64 × 2 = 0 + 0.749 987 000 908 513 28;
  • 35) 0.749 987 000 908 513 28 × 2 = 1 + 0.499 974 001 817 026 56;
  • 36) 0.499 974 001 817 026 56 × 2 = 0 + 0.999 948 003 634 053 12;
  • 37) 0.999 948 003 634 053 12 × 2 = 1 + 0.999 896 007 268 106 24;
  • 38) 0.999 896 007 268 106 24 × 2 = 1 + 0.999 792 014 536 212 48;
  • 39) 0.999 792 014 536 212 48 × 2 = 1 + 0.999 584 029 072 424 96;
  • 40) 0.999 584 029 072 424 96 × 2 = 1 + 0.999 168 058 144 849 92;
  • 41) 0.999 168 058 144 849 92 × 2 = 1 + 0.998 336 116 289 699 84;
  • 42) 0.998 336 116 289 699 84 × 2 = 1 + 0.996 672 232 579 399 68;
  • 43) 0.996 672 232 579 399 68 × 2 = 1 + 0.993 344 465 158 799 36;
  • 44) 0.993 344 465 158 799 36 × 2 = 1 + 0.986 688 930 317 598 72;
  • 45) 0.986 688 930 317 598 72 × 2 = 1 + 0.973 377 860 635 197 44;
  • 46) 0.973 377 860 635 197 44 × 2 = 1 + 0.946 755 721 270 394 88;
  • 47) 0.946 755 721 270 394 88 × 2 = 1 + 0.893 511 442 540 789 76;
  • 48) 0.893 511 442 540 789 76 × 2 = 1 + 0.787 022 885 081 579 52;
  • 49) 0.787 022 885 081 579 52 × 2 = 1 + 0.574 045 770 163 159 04;
  • 50) 0.574 045 770 163 159 04 × 2 = 1 + 0.148 091 540 326 318 08;
  • 51) 0.148 091 540 326 318 08 × 2 = 0 + 0.296 183 080 652 636 16;
  • 52) 0.296 183 080 652 636 16 × 2 = 0 + 0.592 366 161 305 272 32;
  • 53) 0.592 366 161 305 272 32 × 2 = 1 + 0.184 732 322 610 544 64;
  • 54) 0.184 732 322 610 544 64 × 2 = 0 + 0.369 464 645 221 089 28;

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:


0.000 000 000 742 146 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 10(2)

6. Positive number before normalization:

0.000 000 000 742 146 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 146 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 10(2) × 20 =

1.1001 0111 1111 1111 1110 010(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1110 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

96(10) =

0110 0000(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 0010 =

100 1011 1111 1111 1111 0010


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

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

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 0010


Decimal number -0.000 000 000 742 146 92 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 0010

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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