-0.000 000 000 742 163 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 163(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
-0.000 000 000 742 163(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:

|-0.000 000 000 742 163| = 0.000 000 000 742 163


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: 0.000 000 000 742 163.

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) 0.000 000 000 742 163 × 2 = 0 + 0.000 000 001 484 326;
  • 2) 0.000 000 001 484 326 × 2 = 0 + 0.000 000 002 968 652;
  • 3) 0.000 000 002 968 652 × 2 = 0 + 0.000 000 005 937 304;
  • 4) 0.000 000 005 937 304 × 2 = 0 + 0.000 000 011 874 608;
  • 5) 0.000 000 011 874 608 × 2 = 0 + 0.000 000 023 749 216;
  • 6) 0.000 000 023 749 216 × 2 = 0 + 0.000 000 047 498 432;
  • 7) 0.000 000 047 498 432 × 2 = 0 + 0.000 000 094 996 864;
  • 8) 0.000 000 094 996 864 × 2 = 0 + 0.000 000 189 993 728;
  • 9) 0.000 000 189 993 728 × 2 = 0 + 0.000 000 379 987 456;
  • 10) 0.000 000 379 987 456 × 2 = 0 + 0.000 000 759 974 912;
  • 11) 0.000 000 759 974 912 × 2 = 0 + 0.000 001 519 949 824;
  • 12) 0.000 001 519 949 824 × 2 = 0 + 0.000 003 039 899 648;
  • 13) 0.000 003 039 899 648 × 2 = 0 + 0.000 006 079 799 296;
  • 14) 0.000 006 079 799 296 × 2 = 0 + 0.000 012 159 598 592;
  • 15) 0.000 012 159 598 592 × 2 = 0 + 0.000 024 319 197 184;
  • 16) 0.000 024 319 197 184 × 2 = 0 + 0.000 048 638 394 368;
  • 17) 0.000 048 638 394 368 × 2 = 0 + 0.000 097 276 788 736;
  • 18) 0.000 097 276 788 736 × 2 = 0 + 0.000 194 553 577 472;
  • 19) 0.000 194 553 577 472 × 2 = 0 + 0.000 389 107 154 944;
  • 20) 0.000 389 107 154 944 × 2 = 0 + 0.000 778 214 309 888;
  • 21) 0.000 778 214 309 888 × 2 = 0 + 0.001 556 428 619 776;
  • 22) 0.001 556 428 619 776 × 2 = 0 + 0.003 112 857 239 552;
  • 23) 0.003 112 857 239 552 × 2 = 0 + 0.006 225 714 479 104;
  • 24) 0.006 225 714 479 104 × 2 = 0 + 0.012 451 428 958 208;
  • 25) 0.012 451 428 958 208 × 2 = 0 + 0.024 902 857 916 416;
  • 26) 0.024 902 857 916 416 × 2 = 0 + 0.049 805 715 832 832;
  • 27) 0.049 805 715 832 832 × 2 = 0 + 0.099 611 431 665 664;
  • 28) 0.099 611 431 665 664 × 2 = 0 + 0.199 222 863 331 328;
  • 29) 0.199 222 863 331 328 × 2 = 0 + 0.398 445 726 662 656;
  • 30) 0.398 445 726 662 656 × 2 = 0 + 0.796 891 453 325 312;
  • 31) 0.796 891 453 325 312 × 2 = 1 + 0.593 782 906 650 624;
  • 32) 0.593 782 906 650 624 × 2 = 1 + 0.187 565 813 301 248;
  • 33) 0.187 565 813 301 248 × 2 = 0 + 0.375 131 626 602 496;
  • 34) 0.375 131 626 602 496 × 2 = 0 + 0.750 263 253 204 992;
  • 35) 0.750 263 253 204 992 × 2 = 1 + 0.500 526 506 409 984;
  • 36) 0.500 526 506 409 984 × 2 = 1 + 0.001 053 012 819 968;
  • 37) 0.001 053 012 819 968 × 2 = 0 + 0.002 106 025 639 936;
  • 38) 0.002 106 025 639 936 × 2 = 0 + 0.004 212 051 279 872;
  • 39) 0.004 212 051 279 872 × 2 = 0 + 0.008 424 102 559 744;
  • 40) 0.008 424 102 559 744 × 2 = 0 + 0.016 848 205 119 488;
  • 41) 0.016 848 205 119 488 × 2 = 0 + 0.033 696 410 238 976;
  • 42) 0.033 696 410 238 976 × 2 = 0 + 0.067 392 820 477 952;
  • 43) 0.067 392 820 477 952 × 2 = 0 + 0.134 785 640 955 904;
  • 44) 0.134 785 640 955 904 × 2 = 0 + 0.269 571 281 911 808;
  • 45) 0.269 571 281 911 808 × 2 = 0 + 0.539 142 563 823 616;
  • 46) 0.539 142 563 823 616 × 2 = 1 + 0.078 285 127 647 232;
  • 47) 0.078 285 127 647 232 × 2 = 0 + 0.156 570 255 294 464;
  • 48) 0.156 570 255 294 464 × 2 = 0 + 0.313 140 510 588 928;
  • 49) 0.313 140 510 588 928 × 2 = 0 + 0.626 281 021 177 856;
  • 50) 0.626 281 021 177 856 × 2 = 1 + 0.252 562 042 355 712;
  • 51) 0.252 562 042 355 712 × 2 = 0 + 0.505 124 084 711 424;
  • 52) 0.505 124 084 711 424 × 2 = 1 + 0.010 248 169 422 848;
  • 53) 0.010 248 169 422 848 × 2 = 0 + 0.020 496 338 845 696;
  • 54) 0.020 496 338 845 696 × 2 = 0 + 0.040 992 677 691 392;

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:


0.000 000 000 742 163(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0100 0101 00(2)

6. Positive number before normalization:

0.000 000 000 742 163(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0100 0101 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:


0.000 000 000 742 163(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0100 0101 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0100 0101 00(2) × 20 =

1.1001 1000 0000 0010 0010 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 0010 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 0001 0100 =

100 1100 0000 0001 0001 0100


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 0001 0100


Decimal number -0.000 000 000 742 163 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0001 0001 0100

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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