-0.000 000 000 742 147 694 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 694 8(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
-0.000 000 000 742 147 694 8(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:

|-0.000 000 000 742 147 694 8| = 0.000 000 000 742 147 694 8


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: 0.000 000 000 742 147 694 8.

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) 0.000 000 000 742 147 694 8 × 2 = 0 + 0.000 000 001 484 295 389 6;
  • 2) 0.000 000 001 484 295 389 6 × 2 = 0 + 0.000 000 002 968 590 779 2;
  • 3) 0.000 000 002 968 590 779 2 × 2 = 0 + 0.000 000 005 937 181 558 4;
  • 4) 0.000 000 005 937 181 558 4 × 2 = 0 + 0.000 000 011 874 363 116 8;
  • 5) 0.000 000 011 874 363 116 8 × 2 = 0 + 0.000 000 023 748 726 233 6;
  • 6) 0.000 000 023 748 726 233 6 × 2 = 0 + 0.000 000 047 497 452 467 2;
  • 7) 0.000 000 047 497 452 467 2 × 2 = 0 + 0.000 000 094 994 904 934 4;
  • 8) 0.000 000 094 994 904 934 4 × 2 = 0 + 0.000 000 189 989 809 868 8;
  • 9) 0.000 000 189 989 809 868 8 × 2 = 0 + 0.000 000 379 979 619 737 6;
  • 10) 0.000 000 379 979 619 737 6 × 2 = 0 + 0.000 000 759 959 239 475 2;
  • 11) 0.000 000 759 959 239 475 2 × 2 = 0 + 0.000 001 519 918 478 950 4;
  • 12) 0.000 001 519 918 478 950 4 × 2 = 0 + 0.000 003 039 836 957 900 8;
  • 13) 0.000 003 039 836 957 900 8 × 2 = 0 + 0.000 006 079 673 915 801 6;
  • 14) 0.000 006 079 673 915 801 6 × 2 = 0 + 0.000 012 159 347 831 603 2;
  • 15) 0.000 012 159 347 831 603 2 × 2 = 0 + 0.000 024 318 695 663 206 4;
  • 16) 0.000 024 318 695 663 206 4 × 2 = 0 + 0.000 048 637 391 326 412 8;
  • 17) 0.000 048 637 391 326 412 8 × 2 = 0 + 0.000 097 274 782 652 825 6;
  • 18) 0.000 097 274 782 652 825 6 × 2 = 0 + 0.000 194 549 565 305 651 2;
  • 19) 0.000 194 549 565 305 651 2 × 2 = 0 + 0.000 389 099 130 611 302 4;
  • 20) 0.000 389 099 130 611 302 4 × 2 = 0 + 0.000 778 198 261 222 604 8;
  • 21) 0.000 778 198 261 222 604 8 × 2 = 0 + 0.001 556 396 522 445 209 6;
  • 22) 0.001 556 396 522 445 209 6 × 2 = 0 + 0.003 112 793 044 890 419 2;
  • 23) 0.003 112 793 044 890 419 2 × 2 = 0 + 0.006 225 586 089 780 838 4;
  • 24) 0.006 225 586 089 780 838 4 × 2 = 0 + 0.012 451 172 179 561 676 8;
  • 25) 0.012 451 172 179 561 676 8 × 2 = 0 + 0.024 902 344 359 123 353 6;
  • 26) 0.024 902 344 359 123 353 6 × 2 = 0 + 0.049 804 688 718 246 707 2;
  • 27) 0.049 804 688 718 246 707 2 × 2 = 0 + 0.099 609 377 436 493 414 4;
  • 28) 0.099 609 377 436 493 414 4 × 2 = 0 + 0.199 218 754 872 986 828 8;
  • 29) 0.199 218 754 872 986 828 8 × 2 = 0 + 0.398 437 509 745 973 657 6;
  • 30) 0.398 437 509 745 973 657 6 × 2 = 0 + 0.796 875 019 491 947 315 2;
  • 31) 0.796 875 019 491 947 315 2 × 2 = 1 + 0.593 750 038 983 894 630 4;
  • 32) 0.593 750 038 983 894 630 4 × 2 = 1 + 0.187 500 077 967 789 260 8;
  • 33) 0.187 500 077 967 789 260 8 × 2 = 0 + 0.375 000 155 935 578 521 6;
  • 34) 0.375 000 155 935 578 521 6 × 2 = 0 + 0.750 000 311 871 157 043 2;
  • 35) 0.750 000 311 871 157 043 2 × 2 = 1 + 0.500 000 623 742 314 086 4;
  • 36) 0.500 000 623 742 314 086 4 × 2 = 1 + 0.000 001 247 484 628 172 8;
  • 37) 0.000 001 247 484 628 172 8 × 2 = 0 + 0.000 002 494 969 256 345 6;
  • 38) 0.000 002 494 969 256 345 6 × 2 = 0 + 0.000 004 989 938 512 691 2;
  • 39) 0.000 004 989 938 512 691 2 × 2 = 0 + 0.000 009 979 877 025 382 4;
  • 40) 0.000 009 979 877 025 382 4 × 2 = 0 + 0.000 019 959 754 050 764 8;
  • 41) 0.000 019 959 754 050 764 8 × 2 = 0 + 0.000 039 919 508 101 529 6;
  • 42) 0.000 039 919 508 101 529 6 × 2 = 0 + 0.000 079 839 016 203 059 2;
  • 43) 0.000 079 839 016 203 059 2 × 2 = 0 + 0.000 159 678 032 406 118 4;
  • 44) 0.000 159 678 032 406 118 4 × 2 = 0 + 0.000 319 356 064 812 236 8;
  • 45) 0.000 319 356 064 812 236 8 × 2 = 0 + 0.000 638 712 129 624 473 6;
  • 46) 0.000 638 712 129 624 473 6 × 2 = 0 + 0.001 277 424 259 248 947 2;
  • 47) 0.001 277 424 259 248 947 2 × 2 = 0 + 0.002 554 848 518 497 894 4;
  • 48) 0.002 554 848 518 497 894 4 × 2 = 0 + 0.005 109 697 036 995 788 8;
  • 49) 0.005 109 697 036 995 788 8 × 2 = 0 + 0.010 219 394 073 991 577 6;
  • 50) 0.010 219 394 073 991 577 6 × 2 = 0 + 0.020 438 788 147 983 155 2;
  • 51) 0.020 438 788 147 983 155 2 × 2 = 0 + 0.040 877 576 295 966 310 4;
  • 52) 0.040 877 576 295 966 310 4 × 2 = 0 + 0.081 755 152 591 932 620 8;
  • 53) 0.081 755 152 591 932 620 8 × 2 = 0 + 0.163 510 305 183 865 241 6;
  • 54) 0.163 510 305 183 865 241 6 × 2 = 0 + 0.327 020 610 367 730 483 2;

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:


0.000 000 000 742 147 694 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 694 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 694 8(10) =

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

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

1.1001 1000 0000 0000 0000 000(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 0000 0000 000


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 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 0000 0000 0000


Decimal number -0.000 000 000 742 147 694 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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