-0.000 000 000 742 167 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 167(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
-0.000 000 000 742 167(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:

|-0.000 000 000 742 167| = 0.000 000 000 742 167


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: 0.000 000 000 742 167.

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) 0.000 000 000 742 167 × 2 = 0 + 0.000 000 001 484 334;
  • 2) 0.000 000 001 484 334 × 2 = 0 + 0.000 000 002 968 668;
  • 3) 0.000 000 002 968 668 × 2 = 0 + 0.000 000 005 937 336;
  • 4) 0.000 000 005 937 336 × 2 = 0 + 0.000 000 011 874 672;
  • 5) 0.000 000 011 874 672 × 2 = 0 + 0.000 000 023 749 344;
  • 6) 0.000 000 023 749 344 × 2 = 0 + 0.000 000 047 498 688;
  • 7) 0.000 000 047 498 688 × 2 = 0 + 0.000 000 094 997 376;
  • 8) 0.000 000 094 997 376 × 2 = 0 + 0.000 000 189 994 752;
  • 9) 0.000 000 189 994 752 × 2 = 0 + 0.000 000 379 989 504;
  • 10) 0.000 000 379 989 504 × 2 = 0 + 0.000 000 759 979 008;
  • 11) 0.000 000 759 979 008 × 2 = 0 + 0.000 001 519 958 016;
  • 12) 0.000 001 519 958 016 × 2 = 0 + 0.000 003 039 916 032;
  • 13) 0.000 003 039 916 032 × 2 = 0 + 0.000 006 079 832 064;
  • 14) 0.000 006 079 832 064 × 2 = 0 + 0.000 012 159 664 128;
  • 15) 0.000 012 159 664 128 × 2 = 0 + 0.000 024 319 328 256;
  • 16) 0.000 024 319 328 256 × 2 = 0 + 0.000 048 638 656 512;
  • 17) 0.000 048 638 656 512 × 2 = 0 + 0.000 097 277 313 024;
  • 18) 0.000 097 277 313 024 × 2 = 0 + 0.000 194 554 626 048;
  • 19) 0.000 194 554 626 048 × 2 = 0 + 0.000 389 109 252 096;
  • 20) 0.000 389 109 252 096 × 2 = 0 + 0.000 778 218 504 192;
  • 21) 0.000 778 218 504 192 × 2 = 0 + 0.001 556 437 008 384;
  • 22) 0.001 556 437 008 384 × 2 = 0 + 0.003 112 874 016 768;
  • 23) 0.003 112 874 016 768 × 2 = 0 + 0.006 225 748 033 536;
  • 24) 0.006 225 748 033 536 × 2 = 0 + 0.012 451 496 067 072;
  • 25) 0.012 451 496 067 072 × 2 = 0 + 0.024 902 992 134 144;
  • 26) 0.024 902 992 134 144 × 2 = 0 + 0.049 805 984 268 288;
  • 27) 0.049 805 984 268 288 × 2 = 0 + 0.099 611 968 536 576;
  • 28) 0.099 611 968 536 576 × 2 = 0 + 0.199 223 937 073 152;
  • 29) 0.199 223 937 073 152 × 2 = 0 + 0.398 447 874 146 304;
  • 30) 0.398 447 874 146 304 × 2 = 0 + 0.796 895 748 292 608;
  • 31) 0.796 895 748 292 608 × 2 = 1 + 0.593 791 496 585 216;
  • 32) 0.593 791 496 585 216 × 2 = 1 + 0.187 582 993 170 432;
  • 33) 0.187 582 993 170 432 × 2 = 0 + 0.375 165 986 340 864;
  • 34) 0.375 165 986 340 864 × 2 = 0 + 0.750 331 972 681 728;
  • 35) 0.750 331 972 681 728 × 2 = 1 + 0.500 663 945 363 456;
  • 36) 0.500 663 945 363 456 × 2 = 1 + 0.001 327 890 726 912;
  • 37) 0.001 327 890 726 912 × 2 = 0 + 0.002 655 781 453 824;
  • 38) 0.002 655 781 453 824 × 2 = 0 + 0.005 311 562 907 648;
  • 39) 0.005 311 562 907 648 × 2 = 0 + 0.010 623 125 815 296;
  • 40) 0.010 623 125 815 296 × 2 = 0 + 0.021 246 251 630 592;
  • 41) 0.021 246 251 630 592 × 2 = 0 + 0.042 492 503 261 184;
  • 42) 0.042 492 503 261 184 × 2 = 0 + 0.084 985 006 522 368;
  • 43) 0.084 985 006 522 368 × 2 = 0 + 0.169 970 013 044 736;
  • 44) 0.169 970 013 044 736 × 2 = 0 + 0.339 940 026 089 472;
  • 45) 0.339 940 026 089 472 × 2 = 0 + 0.679 880 052 178 944;
  • 46) 0.679 880 052 178 944 × 2 = 1 + 0.359 760 104 357 888;
  • 47) 0.359 760 104 357 888 × 2 = 0 + 0.719 520 208 715 776;
  • 48) 0.719 520 208 715 776 × 2 = 1 + 0.439 040 417 431 552;
  • 49) 0.439 040 417 431 552 × 2 = 0 + 0.878 080 834 863 104;
  • 50) 0.878 080 834 863 104 × 2 = 1 + 0.756 161 669 726 208;
  • 51) 0.756 161 669 726 208 × 2 = 1 + 0.512 323 339 452 416;
  • 52) 0.512 323 339 452 416 × 2 = 1 + 0.024 646 678 904 832;
  • 53) 0.024 646 678 904 832 × 2 = 0 + 0.049 293 357 809 664;
  • 54) 0.049 293 357 809 664 × 2 = 0 + 0.098 586 715 619 328;

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:


0.000 000 000 742 167(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0101 0111 00(2)

6. Positive number before normalization:

0.000 000 000 742 167(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0101 0111 00(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:


0.000 000 000 742 167(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0101 0111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0101 0111 00(2) × 20 =

1.1001 1000 0000 0010 1011 100(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 1000 0000 0010 1011 100


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 1100 0000 0001 0101 1100 =

100 1100 0000 0001 0101 1100


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 1100 0000 0001 0101 1100


Decimal number -0.000 000 000 742 167 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0001 0101 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