-0.000 000 000 742 148 052 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 052(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
-0.000 000 000 742 148 052(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:

|-0.000 000 000 742 148 052| = 0.000 000 000 742 148 052


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: 0.000 000 000 742 148 052.

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) 0.000 000 000 742 148 052 × 2 = 0 + 0.000 000 001 484 296 104;
  • 2) 0.000 000 001 484 296 104 × 2 = 0 + 0.000 000 002 968 592 208;
  • 3) 0.000 000 002 968 592 208 × 2 = 0 + 0.000 000 005 937 184 416;
  • 4) 0.000 000 005 937 184 416 × 2 = 0 + 0.000 000 011 874 368 832;
  • 5) 0.000 000 011 874 368 832 × 2 = 0 + 0.000 000 023 748 737 664;
  • 6) 0.000 000 023 748 737 664 × 2 = 0 + 0.000 000 047 497 475 328;
  • 7) 0.000 000 047 497 475 328 × 2 = 0 + 0.000 000 094 994 950 656;
  • 8) 0.000 000 094 994 950 656 × 2 = 0 + 0.000 000 189 989 901 312;
  • 9) 0.000 000 189 989 901 312 × 2 = 0 + 0.000 000 379 979 802 624;
  • 10) 0.000 000 379 979 802 624 × 2 = 0 + 0.000 000 759 959 605 248;
  • 11) 0.000 000 759 959 605 248 × 2 = 0 + 0.000 001 519 919 210 496;
  • 12) 0.000 001 519 919 210 496 × 2 = 0 + 0.000 003 039 838 420 992;
  • 13) 0.000 003 039 838 420 992 × 2 = 0 + 0.000 006 079 676 841 984;
  • 14) 0.000 006 079 676 841 984 × 2 = 0 + 0.000 012 159 353 683 968;
  • 15) 0.000 012 159 353 683 968 × 2 = 0 + 0.000 024 318 707 367 936;
  • 16) 0.000 024 318 707 367 936 × 2 = 0 + 0.000 048 637 414 735 872;
  • 17) 0.000 048 637 414 735 872 × 2 = 0 + 0.000 097 274 829 471 744;
  • 18) 0.000 097 274 829 471 744 × 2 = 0 + 0.000 194 549 658 943 488;
  • 19) 0.000 194 549 658 943 488 × 2 = 0 + 0.000 389 099 317 886 976;
  • 20) 0.000 389 099 317 886 976 × 2 = 0 + 0.000 778 198 635 773 952;
  • 21) 0.000 778 198 635 773 952 × 2 = 0 + 0.001 556 397 271 547 904;
  • 22) 0.001 556 397 271 547 904 × 2 = 0 + 0.003 112 794 543 095 808;
  • 23) 0.003 112 794 543 095 808 × 2 = 0 + 0.006 225 589 086 191 616;
  • 24) 0.006 225 589 086 191 616 × 2 = 0 + 0.012 451 178 172 383 232;
  • 25) 0.012 451 178 172 383 232 × 2 = 0 + 0.024 902 356 344 766 464;
  • 26) 0.024 902 356 344 766 464 × 2 = 0 + 0.049 804 712 689 532 928;
  • 27) 0.049 804 712 689 532 928 × 2 = 0 + 0.099 609 425 379 065 856;
  • 28) 0.099 609 425 379 065 856 × 2 = 0 + 0.199 218 850 758 131 712;
  • 29) 0.199 218 850 758 131 712 × 2 = 0 + 0.398 437 701 516 263 424;
  • 30) 0.398 437 701 516 263 424 × 2 = 0 + 0.796 875 403 032 526 848;
  • 31) 0.796 875 403 032 526 848 × 2 = 1 + 0.593 750 806 065 053 696;
  • 32) 0.593 750 806 065 053 696 × 2 = 1 + 0.187 501 612 130 107 392;
  • 33) 0.187 501 612 130 107 392 × 2 = 0 + 0.375 003 224 260 214 784;
  • 34) 0.375 003 224 260 214 784 × 2 = 0 + 0.750 006 448 520 429 568;
  • 35) 0.750 006 448 520 429 568 × 2 = 1 + 0.500 012 897 040 859 136;
  • 36) 0.500 012 897 040 859 136 × 2 = 1 + 0.000 025 794 081 718 272;
  • 37) 0.000 025 794 081 718 272 × 2 = 0 + 0.000 051 588 163 436 544;
  • 38) 0.000 051 588 163 436 544 × 2 = 0 + 0.000 103 176 326 873 088;
  • 39) 0.000 103 176 326 873 088 × 2 = 0 + 0.000 206 352 653 746 176;
  • 40) 0.000 206 352 653 746 176 × 2 = 0 + 0.000 412 705 307 492 352;
  • 41) 0.000 412 705 307 492 352 × 2 = 0 + 0.000 825 410 614 984 704;
  • 42) 0.000 825 410 614 984 704 × 2 = 0 + 0.001 650 821 229 969 408;
  • 43) 0.001 650 821 229 969 408 × 2 = 0 + 0.003 301 642 459 938 816;
  • 44) 0.003 301 642 459 938 816 × 2 = 0 + 0.006 603 284 919 877 632;
  • 45) 0.006 603 284 919 877 632 × 2 = 0 + 0.013 206 569 839 755 264;
  • 46) 0.013 206 569 839 755 264 × 2 = 0 + 0.026 413 139 679 510 528;
  • 47) 0.026 413 139 679 510 528 × 2 = 0 + 0.052 826 279 359 021 056;
  • 48) 0.052 826 279 359 021 056 × 2 = 0 + 0.105 652 558 718 042 112;
  • 49) 0.105 652 558 718 042 112 × 2 = 0 + 0.211 305 117 436 084 224;
  • 50) 0.211 305 117 436 084 224 × 2 = 0 + 0.422 610 234 872 168 448;
  • 51) 0.422 610 234 872 168 448 × 2 = 0 + 0.845 220 469 744 336 896;
  • 52) 0.845 220 469 744 336 896 × 2 = 1 + 0.690 440 939 488 673 792;
  • 53) 0.690 440 939 488 673 792 × 2 = 1 + 0.380 881 878 977 347 584;
  • 54) 0.380 881 878 977 347 584 × 2 = 0 + 0.761 763 757 954 695 168;

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:


0.000 000 000 742 148 052(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2)

6. Positive number before normalization:

0.000 000 000 742 148 052(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(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:


0.000 000 000 742 148 052(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2) × 20 =

1.1001 1000 0000 0000 0000 110(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 110


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 0110 =

100 1100 0000 0000 0000 0110


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 0110


Decimal number -0.000 000 000 742 148 052 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0110

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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