-0.000 000 000 742 147 395 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 395(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
-0.000 000 000 742 147 395(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:

|-0.000 000 000 742 147 395| = 0.000 000 000 742 147 395


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: 0.000 000 000 742 147 395.

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) 0.000 000 000 742 147 395 × 2 = 0 + 0.000 000 001 484 294 79;
  • 2) 0.000 000 001 484 294 79 × 2 = 0 + 0.000 000 002 968 589 58;
  • 3) 0.000 000 002 968 589 58 × 2 = 0 + 0.000 000 005 937 179 16;
  • 4) 0.000 000 005 937 179 16 × 2 = 0 + 0.000 000 011 874 358 32;
  • 5) 0.000 000 011 874 358 32 × 2 = 0 + 0.000 000 023 748 716 64;
  • 6) 0.000 000 023 748 716 64 × 2 = 0 + 0.000 000 047 497 433 28;
  • 7) 0.000 000 047 497 433 28 × 2 = 0 + 0.000 000 094 994 866 56;
  • 8) 0.000 000 094 994 866 56 × 2 = 0 + 0.000 000 189 989 733 12;
  • 9) 0.000 000 189 989 733 12 × 2 = 0 + 0.000 000 379 979 466 24;
  • 10) 0.000 000 379 979 466 24 × 2 = 0 + 0.000 000 759 958 932 48;
  • 11) 0.000 000 759 958 932 48 × 2 = 0 + 0.000 001 519 917 864 96;
  • 12) 0.000 001 519 917 864 96 × 2 = 0 + 0.000 003 039 835 729 92;
  • 13) 0.000 003 039 835 729 92 × 2 = 0 + 0.000 006 079 671 459 84;
  • 14) 0.000 006 079 671 459 84 × 2 = 0 + 0.000 012 159 342 919 68;
  • 15) 0.000 012 159 342 919 68 × 2 = 0 + 0.000 024 318 685 839 36;
  • 16) 0.000 024 318 685 839 36 × 2 = 0 + 0.000 048 637 371 678 72;
  • 17) 0.000 048 637 371 678 72 × 2 = 0 + 0.000 097 274 743 357 44;
  • 18) 0.000 097 274 743 357 44 × 2 = 0 + 0.000 194 549 486 714 88;
  • 19) 0.000 194 549 486 714 88 × 2 = 0 + 0.000 389 098 973 429 76;
  • 20) 0.000 389 098 973 429 76 × 2 = 0 + 0.000 778 197 946 859 52;
  • 21) 0.000 778 197 946 859 52 × 2 = 0 + 0.001 556 395 893 719 04;
  • 22) 0.001 556 395 893 719 04 × 2 = 0 + 0.003 112 791 787 438 08;
  • 23) 0.003 112 791 787 438 08 × 2 = 0 + 0.006 225 583 574 876 16;
  • 24) 0.006 225 583 574 876 16 × 2 = 0 + 0.012 451 167 149 752 32;
  • 25) 0.012 451 167 149 752 32 × 2 = 0 + 0.024 902 334 299 504 64;
  • 26) 0.024 902 334 299 504 64 × 2 = 0 + 0.049 804 668 599 009 28;
  • 27) 0.049 804 668 599 009 28 × 2 = 0 + 0.099 609 337 198 018 56;
  • 28) 0.099 609 337 198 018 56 × 2 = 0 + 0.199 218 674 396 037 12;
  • 29) 0.199 218 674 396 037 12 × 2 = 0 + 0.398 437 348 792 074 24;
  • 30) 0.398 437 348 792 074 24 × 2 = 0 + 0.796 874 697 584 148 48;
  • 31) 0.796 874 697 584 148 48 × 2 = 1 + 0.593 749 395 168 296 96;
  • 32) 0.593 749 395 168 296 96 × 2 = 1 + 0.187 498 790 336 593 92;
  • 33) 0.187 498 790 336 593 92 × 2 = 0 + 0.374 997 580 673 187 84;
  • 34) 0.374 997 580 673 187 84 × 2 = 0 + 0.749 995 161 346 375 68;
  • 35) 0.749 995 161 346 375 68 × 2 = 1 + 0.499 990 322 692 751 36;
  • 36) 0.499 990 322 692 751 36 × 2 = 0 + 0.999 980 645 385 502 72;
  • 37) 0.999 980 645 385 502 72 × 2 = 1 + 0.999 961 290 771 005 44;
  • 38) 0.999 961 290 771 005 44 × 2 = 1 + 0.999 922 581 542 010 88;
  • 39) 0.999 922 581 542 010 88 × 2 = 1 + 0.999 845 163 084 021 76;
  • 40) 0.999 845 163 084 021 76 × 2 = 1 + 0.999 690 326 168 043 52;
  • 41) 0.999 690 326 168 043 52 × 2 = 1 + 0.999 380 652 336 087 04;
  • 42) 0.999 380 652 336 087 04 × 2 = 1 + 0.998 761 304 672 174 08;
  • 43) 0.998 761 304 672 174 08 × 2 = 1 + 0.997 522 609 344 348 16;
  • 44) 0.997 522 609 344 348 16 × 2 = 1 + 0.995 045 218 688 696 32;
  • 45) 0.995 045 218 688 696 32 × 2 = 1 + 0.990 090 437 377 392 64;
  • 46) 0.990 090 437 377 392 64 × 2 = 1 + 0.980 180 874 754 785 28;
  • 47) 0.980 180 874 754 785 28 × 2 = 1 + 0.960 361 749 509 570 56;
  • 48) 0.960 361 749 509 570 56 × 2 = 1 + 0.920 723 499 019 141 12;
  • 49) 0.920 723 499 019 141 12 × 2 = 1 + 0.841 446 998 038 282 24;
  • 50) 0.841 446 998 038 282 24 × 2 = 1 + 0.682 893 996 076 564 48;
  • 51) 0.682 893 996 076 564 48 × 2 = 1 + 0.365 787 992 153 128 96;
  • 52) 0.365 787 992 153 128 96 × 2 = 0 + 0.731 575 984 306 257 92;
  • 53) 0.731 575 984 306 257 92 × 2 = 1 + 0.463 151 968 612 515 84;
  • 54) 0.463 151 968 612 515 84 × 2 = 0 + 0.926 303 937 225 031 68;

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:


0.000 000 000 742 147 395(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 395(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 395(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) × 20 =

1.1001 0111 1111 1111 1111 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 1111 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 1010 =

100 1011 1111 1111 1111 1010


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 1010


Decimal number -0.000 000 000 742 147 395 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 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