-0.000 000 000 742 147 457 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 457(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
-0.000 000 000 742 147 457(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:

|-0.000 000 000 742 147 457| = 0.000 000 000 742 147 457


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: 0.000 000 000 742 147 457.

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) 0.000 000 000 742 147 457 × 2 = 0 + 0.000 000 001 484 294 914;
  • 2) 0.000 000 001 484 294 914 × 2 = 0 + 0.000 000 002 968 589 828;
  • 3) 0.000 000 002 968 589 828 × 2 = 0 + 0.000 000 005 937 179 656;
  • 4) 0.000 000 005 937 179 656 × 2 = 0 + 0.000 000 011 874 359 312;
  • 5) 0.000 000 011 874 359 312 × 2 = 0 + 0.000 000 023 748 718 624;
  • 6) 0.000 000 023 748 718 624 × 2 = 0 + 0.000 000 047 497 437 248;
  • 7) 0.000 000 047 497 437 248 × 2 = 0 + 0.000 000 094 994 874 496;
  • 8) 0.000 000 094 994 874 496 × 2 = 0 + 0.000 000 189 989 748 992;
  • 9) 0.000 000 189 989 748 992 × 2 = 0 + 0.000 000 379 979 497 984;
  • 10) 0.000 000 379 979 497 984 × 2 = 0 + 0.000 000 759 958 995 968;
  • 11) 0.000 000 759 958 995 968 × 2 = 0 + 0.000 001 519 917 991 936;
  • 12) 0.000 001 519 917 991 936 × 2 = 0 + 0.000 003 039 835 983 872;
  • 13) 0.000 003 039 835 983 872 × 2 = 0 + 0.000 006 079 671 967 744;
  • 14) 0.000 006 079 671 967 744 × 2 = 0 + 0.000 012 159 343 935 488;
  • 15) 0.000 012 159 343 935 488 × 2 = 0 + 0.000 024 318 687 870 976;
  • 16) 0.000 024 318 687 870 976 × 2 = 0 + 0.000 048 637 375 741 952;
  • 17) 0.000 048 637 375 741 952 × 2 = 0 + 0.000 097 274 751 483 904;
  • 18) 0.000 097 274 751 483 904 × 2 = 0 + 0.000 194 549 502 967 808;
  • 19) 0.000 194 549 502 967 808 × 2 = 0 + 0.000 389 099 005 935 616;
  • 20) 0.000 389 099 005 935 616 × 2 = 0 + 0.000 778 198 011 871 232;
  • 21) 0.000 778 198 011 871 232 × 2 = 0 + 0.001 556 396 023 742 464;
  • 22) 0.001 556 396 023 742 464 × 2 = 0 + 0.003 112 792 047 484 928;
  • 23) 0.003 112 792 047 484 928 × 2 = 0 + 0.006 225 584 094 969 856;
  • 24) 0.006 225 584 094 969 856 × 2 = 0 + 0.012 451 168 189 939 712;
  • 25) 0.012 451 168 189 939 712 × 2 = 0 + 0.024 902 336 379 879 424;
  • 26) 0.024 902 336 379 879 424 × 2 = 0 + 0.049 804 672 759 758 848;
  • 27) 0.049 804 672 759 758 848 × 2 = 0 + 0.099 609 345 519 517 696;
  • 28) 0.099 609 345 519 517 696 × 2 = 0 + 0.199 218 691 039 035 392;
  • 29) 0.199 218 691 039 035 392 × 2 = 0 + 0.398 437 382 078 070 784;
  • 30) 0.398 437 382 078 070 784 × 2 = 0 + 0.796 874 764 156 141 568;
  • 31) 0.796 874 764 156 141 568 × 2 = 1 + 0.593 749 528 312 283 136;
  • 32) 0.593 749 528 312 283 136 × 2 = 1 + 0.187 499 056 624 566 272;
  • 33) 0.187 499 056 624 566 272 × 2 = 0 + 0.374 998 113 249 132 544;
  • 34) 0.374 998 113 249 132 544 × 2 = 0 + 0.749 996 226 498 265 088;
  • 35) 0.749 996 226 498 265 088 × 2 = 1 + 0.499 992 452 996 530 176;
  • 36) 0.499 992 452 996 530 176 × 2 = 0 + 0.999 984 905 993 060 352;
  • 37) 0.999 984 905 993 060 352 × 2 = 1 + 0.999 969 811 986 120 704;
  • 38) 0.999 969 811 986 120 704 × 2 = 1 + 0.999 939 623 972 241 408;
  • 39) 0.999 939 623 972 241 408 × 2 = 1 + 0.999 879 247 944 482 816;
  • 40) 0.999 879 247 944 482 816 × 2 = 1 + 0.999 758 495 888 965 632;
  • 41) 0.999 758 495 888 965 632 × 2 = 1 + 0.999 516 991 777 931 264;
  • 42) 0.999 516 991 777 931 264 × 2 = 1 + 0.999 033 983 555 862 528;
  • 43) 0.999 033 983 555 862 528 × 2 = 1 + 0.998 067 967 111 725 056;
  • 44) 0.998 067 967 111 725 056 × 2 = 1 + 0.996 135 934 223 450 112;
  • 45) 0.996 135 934 223 450 112 × 2 = 1 + 0.992 271 868 446 900 224;
  • 46) 0.992 271 868 446 900 224 × 2 = 1 + 0.984 543 736 893 800 448;
  • 47) 0.984 543 736 893 800 448 × 2 = 1 + 0.969 087 473 787 600 896;
  • 48) 0.969 087 473 787 600 896 × 2 = 1 + 0.938 174 947 575 201 792;
  • 49) 0.938 174 947 575 201 792 × 2 = 1 + 0.876 349 895 150 403 584;
  • 50) 0.876 349 895 150 403 584 × 2 = 1 + 0.752 699 790 300 807 168;
  • 51) 0.752 699 790 300 807 168 × 2 = 1 + 0.505 399 580 601 614 336;
  • 52) 0.505 399 580 601 614 336 × 2 = 1 + 0.010 799 161 203 228 672;
  • 53) 0.010 799 161 203 228 672 × 2 = 0 + 0.021 598 322 406 457 344;
  • 54) 0.021 598 322 406 457 344 × 2 = 0 + 0.043 196 644 812 914 688;

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:


0.000 000 000 742 147 457(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 457(10) =

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


0.000 000 000 742 147 457(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 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 0111 1111 1111 1111 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 1011 1111 1111 1111 1100 =

100 1011 1111 1111 1111 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 1011 1111 1111 1111 1100


Decimal number -0.000 000 000 742 147 457 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 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