-0.000 000 000 742 147 718 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 718(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
-0.000 000 000 742 147 718(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:

|-0.000 000 000 742 147 718| = 0.000 000 000 742 147 718


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: 0.000 000 000 742 147 718.

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) 0.000 000 000 742 147 718 × 2 = 0 + 0.000 000 001 484 295 436;
  • 2) 0.000 000 001 484 295 436 × 2 = 0 + 0.000 000 002 968 590 872;
  • 3) 0.000 000 002 968 590 872 × 2 = 0 + 0.000 000 005 937 181 744;
  • 4) 0.000 000 005 937 181 744 × 2 = 0 + 0.000 000 011 874 363 488;
  • 5) 0.000 000 011 874 363 488 × 2 = 0 + 0.000 000 023 748 726 976;
  • 6) 0.000 000 023 748 726 976 × 2 = 0 + 0.000 000 047 497 453 952;
  • 7) 0.000 000 047 497 453 952 × 2 = 0 + 0.000 000 094 994 907 904;
  • 8) 0.000 000 094 994 907 904 × 2 = 0 + 0.000 000 189 989 815 808;
  • 9) 0.000 000 189 989 815 808 × 2 = 0 + 0.000 000 379 979 631 616;
  • 10) 0.000 000 379 979 631 616 × 2 = 0 + 0.000 000 759 959 263 232;
  • 11) 0.000 000 759 959 263 232 × 2 = 0 + 0.000 001 519 918 526 464;
  • 12) 0.000 001 519 918 526 464 × 2 = 0 + 0.000 003 039 837 052 928;
  • 13) 0.000 003 039 837 052 928 × 2 = 0 + 0.000 006 079 674 105 856;
  • 14) 0.000 006 079 674 105 856 × 2 = 0 + 0.000 012 159 348 211 712;
  • 15) 0.000 012 159 348 211 712 × 2 = 0 + 0.000 024 318 696 423 424;
  • 16) 0.000 024 318 696 423 424 × 2 = 0 + 0.000 048 637 392 846 848;
  • 17) 0.000 048 637 392 846 848 × 2 = 0 + 0.000 097 274 785 693 696;
  • 18) 0.000 097 274 785 693 696 × 2 = 0 + 0.000 194 549 571 387 392;
  • 19) 0.000 194 549 571 387 392 × 2 = 0 + 0.000 389 099 142 774 784;
  • 20) 0.000 389 099 142 774 784 × 2 = 0 + 0.000 778 198 285 549 568;
  • 21) 0.000 778 198 285 549 568 × 2 = 0 + 0.001 556 396 571 099 136;
  • 22) 0.001 556 396 571 099 136 × 2 = 0 + 0.003 112 793 142 198 272;
  • 23) 0.003 112 793 142 198 272 × 2 = 0 + 0.006 225 586 284 396 544;
  • 24) 0.006 225 586 284 396 544 × 2 = 0 + 0.012 451 172 568 793 088;
  • 25) 0.012 451 172 568 793 088 × 2 = 0 + 0.024 902 345 137 586 176;
  • 26) 0.024 902 345 137 586 176 × 2 = 0 + 0.049 804 690 275 172 352;
  • 27) 0.049 804 690 275 172 352 × 2 = 0 + 0.099 609 380 550 344 704;
  • 28) 0.099 609 380 550 344 704 × 2 = 0 + 0.199 218 761 100 689 408;
  • 29) 0.199 218 761 100 689 408 × 2 = 0 + 0.398 437 522 201 378 816;
  • 30) 0.398 437 522 201 378 816 × 2 = 0 + 0.796 875 044 402 757 632;
  • 31) 0.796 875 044 402 757 632 × 2 = 1 + 0.593 750 088 805 515 264;
  • 32) 0.593 750 088 805 515 264 × 2 = 1 + 0.187 500 177 611 030 528;
  • 33) 0.187 500 177 611 030 528 × 2 = 0 + 0.375 000 355 222 061 056;
  • 34) 0.375 000 355 222 061 056 × 2 = 0 + 0.750 000 710 444 122 112;
  • 35) 0.750 000 710 444 122 112 × 2 = 1 + 0.500 001 420 888 244 224;
  • 36) 0.500 001 420 888 244 224 × 2 = 1 + 0.000 002 841 776 488 448;
  • 37) 0.000 002 841 776 488 448 × 2 = 0 + 0.000 005 683 552 976 896;
  • 38) 0.000 005 683 552 976 896 × 2 = 0 + 0.000 011 367 105 953 792;
  • 39) 0.000 011 367 105 953 792 × 2 = 0 + 0.000 022 734 211 907 584;
  • 40) 0.000 022 734 211 907 584 × 2 = 0 + 0.000 045 468 423 815 168;
  • 41) 0.000 045 468 423 815 168 × 2 = 0 + 0.000 090 936 847 630 336;
  • 42) 0.000 090 936 847 630 336 × 2 = 0 + 0.000 181 873 695 260 672;
  • 43) 0.000 181 873 695 260 672 × 2 = 0 + 0.000 363 747 390 521 344;
  • 44) 0.000 363 747 390 521 344 × 2 = 0 + 0.000 727 494 781 042 688;
  • 45) 0.000 727 494 781 042 688 × 2 = 0 + 0.001 454 989 562 085 376;
  • 46) 0.001 454 989 562 085 376 × 2 = 0 + 0.002 909 979 124 170 752;
  • 47) 0.002 909 979 124 170 752 × 2 = 0 + 0.005 819 958 248 341 504;
  • 48) 0.005 819 958 248 341 504 × 2 = 0 + 0.011 639 916 496 683 008;
  • 49) 0.011 639 916 496 683 008 × 2 = 0 + 0.023 279 832 993 366 016;
  • 50) 0.023 279 832 993 366 016 × 2 = 0 + 0.046 559 665 986 732 032;
  • 51) 0.046 559 665 986 732 032 × 2 = 0 + 0.093 119 331 973 464 064;
  • 52) 0.093 119 331 973 464 064 × 2 = 0 + 0.186 238 663 946 928 128;
  • 53) 0.186 238 663 946 928 128 × 2 = 0 + 0.372 477 327 893 856 256;
  • 54) 0.372 477 327 893 856 256 × 2 = 0 + 0.744 954 655 787 712 512;

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:


0.000 000 000 742 147 718(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 718(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:


0.000 000 000 742 147 718(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 -0.000 000 000 742 147 718 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