-0.000 000 000 742 127 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 127(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
-0.000 000 000 742 127(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:

|-0.000 000 000 742 127| = 0.000 000 000 742 127


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: 0.000 000 000 742 127.

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) 0.000 000 000 742 127 × 2 = 0 + 0.000 000 001 484 254;
  • 2) 0.000 000 001 484 254 × 2 = 0 + 0.000 000 002 968 508;
  • 3) 0.000 000 002 968 508 × 2 = 0 + 0.000 000 005 937 016;
  • 4) 0.000 000 005 937 016 × 2 = 0 + 0.000 000 011 874 032;
  • 5) 0.000 000 011 874 032 × 2 = 0 + 0.000 000 023 748 064;
  • 6) 0.000 000 023 748 064 × 2 = 0 + 0.000 000 047 496 128;
  • 7) 0.000 000 047 496 128 × 2 = 0 + 0.000 000 094 992 256;
  • 8) 0.000 000 094 992 256 × 2 = 0 + 0.000 000 189 984 512;
  • 9) 0.000 000 189 984 512 × 2 = 0 + 0.000 000 379 969 024;
  • 10) 0.000 000 379 969 024 × 2 = 0 + 0.000 000 759 938 048;
  • 11) 0.000 000 759 938 048 × 2 = 0 + 0.000 001 519 876 096;
  • 12) 0.000 001 519 876 096 × 2 = 0 + 0.000 003 039 752 192;
  • 13) 0.000 003 039 752 192 × 2 = 0 + 0.000 006 079 504 384;
  • 14) 0.000 006 079 504 384 × 2 = 0 + 0.000 012 159 008 768;
  • 15) 0.000 012 159 008 768 × 2 = 0 + 0.000 024 318 017 536;
  • 16) 0.000 024 318 017 536 × 2 = 0 + 0.000 048 636 035 072;
  • 17) 0.000 048 636 035 072 × 2 = 0 + 0.000 097 272 070 144;
  • 18) 0.000 097 272 070 144 × 2 = 0 + 0.000 194 544 140 288;
  • 19) 0.000 194 544 140 288 × 2 = 0 + 0.000 389 088 280 576;
  • 20) 0.000 389 088 280 576 × 2 = 0 + 0.000 778 176 561 152;
  • 21) 0.000 778 176 561 152 × 2 = 0 + 0.001 556 353 122 304;
  • 22) 0.001 556 353 122 304 × 2 = 0 + 0.003 112 706 244 608;
  • 23) 0.003 112 706 244 608 × 2 = 0 + 0.006 225 412 489 216;
  • 24) 0.006 225 412 489 216 × 2 = 0 + 0.012 450 824 978 432;
  • 25) 0.012 450 824 978 432 × 2 = 0 + 0.024 901 649 956 864;
  • 26) 0.024 901 649 956 864 × 2 = 0 + 0.049 803 299 913 728;
  • 27) 0.049 803 299 913 728 × 2 = 0 + 0.099 606 599 827 456;
  • 28) 0.099 606 599 827 456 × 2 = 0 + 0.199 213 199 654 912;
  • 29) 0.199 213 199 654 912 × 2 = 0 + 0.398 426 399 309 824;
  • 30) 0.398 426 399 309 824 × 2 = 0 + 0.796 852 798 619 648;
  • 31) 0.796 852 798 619 648 × 2 = 1 + 0.593 705 597 239 296;
  • 32) 0.593 705 597 239 296 × 2 = 1 + 0.187 411 194 478 592;
  • 33) 0.187 411 194 478 592 × 2 = 0 + 0.374 822 388 957 184;
  • 34) 0.374 822 388 957 184 × 2 = 0 + 0.749 644 777 914 368;
  • 35) 0.749 644 777 914 368 × 2 = 1 + 0.499 289 555 828 736;
  • 36) 0.499 289 555 828 736 × 2 = 0 + 0.998 579 111 657 472;
  • 37) 0.998 579 111 657 472 × 2 = 1 + 0.997 158 223 314 944;
  • 38) 0.997 158 223 314 944 × 2 = 1 + 0.994 316 446 629 888;
  • 39) 0.994 316 446 629 888 × 2 = 1 + 0.988 632 893 259 776;
  • 40) 0.988 632 893 259 776 × 2 = 1 + 0.977 265 786 519 552;
  • 41) 0.977 265 786 519 552 × 2 = 1 + 0.954 531 573 039 104;
  • 42) 0.954 531 573 039 104 × 2 = 1 + 0.909 063 146 078 208;
  • 43) 0.909 063 146 078 208 × 2 = 1 + 0.818 126 292 156 416;
  • 44) 0.818 126 292 156 416 × 2 = 1 + 0.636 252 584 312 832;
  • 45) 0.636 252 584 312 832 × 2 = 1 + 0.272 505 168 625 664;
  • 46) 0.272 505 168 625 664 × 2 = 0 + 0.545 010 337 251 328;
  • 47) 0.545 010 337 251 328 × 2 = 1 + 0.090 020 674 502 656;
  • 48) 0.090 020 674 502 656 × 2 = 0 + 0.180 041 349 005 312;
  • 49) 0.180 041 349 005 312 × 2 = 0 + 0.360 082 698 010 624;
  • 50) 0.360 082 698 010 624 × 2 = 0 + 0.720 165 396 021 248;
  • 51) 0.720 165 396 021 248 × 2 = 1 + 0.440 330 792 042 496;
  • 52) 0.440 330 792 042 496 × 2 = 0 + 0.880 661 584 084 992;
  • 53) 0.880 661 584 084 992 × 2 = 1 + 0.761 323 168 169 984;
  • 54) 0.761 323 168 169 984 × 2 = 1 + 0.522 646 336 339 968;

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:


0.000 000 000 742 127(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 0010 11(2)

6. Positive number before normalization:

0.000 000 000 742 127(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 0010 11(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:


0.000 000 000 742 127(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 0010 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 0010 11(2) × 20 =

1.1001 0111 1111 1101 0001 011(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 1101 0001 011


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 1110 1000 1011 =

100 1011 1111 1110 1000 1011


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 1110 1000 1011


Decimal number -0.000 000 000 742 127 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1000 1011

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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