-0.000 000 000 742 147 956 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 956(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
-0.000 000 000 742 147 956(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:

|-0.000 000 000 742 147 956| = 0.000 000 000 742 147 956


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: 0.000 000 000 742 147 956.

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) 0.000 000 000 742 147 956 × 2 = 0 + 0.000 000 001 484 295 912;
  • 2) 0.000 000 001 484 295 912 × 2 = 0 + 0.000 000 002 968 591 824;
  • 3) 0.000 000 002 968 591 824 × 2 = 0 + 0.000 000 005 937 183 648;
  • 4) 0.000 000 005 937 183 648 × 2 = 0 + 0.000 000 011 874 367 296;
  • 5) 0.000 000 011 874 367 296 × 2 = 0 + 0.000 000 023 748 734 592;
  • 6) 0.000 000 023 748 734 592 × 2 = 0 + 0.000 000 047 497 469 184;
  • 7) 0.000 000 047 497 469 184 × 2 = 0 + 0.000 000 094 994 938 368;
  • 8) 0.000 000 094 994 938 368 × 2 = 0 + 0.000 000 189 989 876 736;
  • 9) 0.000 000 189 989 876 736 × 2 = 0 + 0.000 000 379 979 753 472;
  • 10) 0.000 000 379 979 753 472 × 2 = 0 + 0.000 000 759 959 506 944;
  • 11) 0.000 000 759 959 506 944 × 2 = 0 + 0.000 001 519 919 013 888;
  • 12) 0.000 001 519 919 013 888 × 2 = 0 + 0.000 003 039 838 027 776;
  • 13) 0.000 003 039 838 027 776 × 2 = 0 + 0.000 006 079 676 055 552;
  • 14) 0.000 006 079 676 055 552 × 2 = 0 + 0.000 012 159 352 111 104;
  • 15) 0.000 012 159 352 111 104 × 2 = 0 + 0.000 024 318 704 222 208;
  • 16) 0.000 024 318 704 222 208 × 2 = 0 + 0.000 048 637 408 444 416;
  • 17) 0.000 048 637 408 444 416 × 2 = 0 + 0.000 097 274 816 888 832;
  • 18) 0.000 097 274 816 888 832 × 2 = 0 + 0.000 194 549 633 777 664;
  • 19) 0.000 194 549 633 777 664 × 2 = 0 + 0.000 389 099 267 555 328;
  • 20) 0.000 389 099 267 555 328 × 2 = 0 + 0.000 778 198 535 110 656;
  • 21) 0.000 778 198 535 110 656 × 2 = 0 + 0.001 556 397 070 221 312;
  • 22) 0.001 556 397 070 221 312 × 2 = 0 + 0.003 112 794 140 442 624;
  • 23) 0.003 112 794 140 442 624 × 2 = 0 + 0.006 225 588 280 885 248;
  • 24) 0.006 225 588 280 885 248 × 2 = 0 + 0.012 451 176 561 770 496;
  • 25) 0.012 451 176 561 770 496 × 2 = 0 + 0.024 902 353 123 540 992;
  • 26) 0.024 902 353 123 540 992 × 2 = 0 + 0.049 804 706 247 081 984;
  • 27) 0.049 804 706 247 081 984 × 2 = 0 + 0.099 609 412 494 163 968;
  • 28) 0.099 609 412 494 163 968 × 2 = 0 + 0.199 218 824 988 327 936;
  • 29) 0.199 218 824 988 327 936 × 2 = 0 + 0.398 437 649 976 655 872;
  • 30) 0.398 437 649 976 655 872 × 2 = 0 + 0.796 875 299 953 311 744;
  • 31) 0.796 875 299 953 311 744 × 2 = 1 + 0.593 750 599 906 623 488;
  • 32) 0.593 750 599 906 623 488 × 2 = 1 + 0.187 501 199 813 246 976;
  • 33) 0.187 501 199 813 246 976 × 2 = 0 + 0.375 002 399 626 493 952;
  • 34) 0.375 002 399 626 493 952 × 2 = 0 + 0.750 004 799 252 987 904;
  • 35) 0.750 004 799 252 987 904 × 2 = 1 + 0.500 009 598 505 975 808;
  • 36) 0.500 009 598 505 975 808 × 2 = 1 + 0.000 019 197 011 951 616;
  • 37) 0.000 019 197 011 951 616 × 2 = 0 + 0.000 038 394 023 903 232;
  • 38) 0.000 038 394 023 903 232 × 2 = 0 + 0.000 076 788 047 806 464;
  • 39) 0.000 076 788 047 806 464 × 2 = 0 + 0.000 153 576 095 612 928;
  • 40) 0.000 153 576 095 612 928 × 2 = 0 + 0.000 307 152 191 225 856;
  • 41) 0.000 307 152 191 225 856 × 2 = 0 + 0.000 614 304 382 451 712;
  • 42) 0.000 614 304 382 451 712 × 2 = 0 + 0.001 228 608 764 903 424;
  • 43) 0.001 228 608 764 903 424 × 2 = 0 + 0.002 457 217 529 806 848;
  • 44) 0.002 457 217 529 806 848 × 2 = 0 + 0.004 914 435 059 613 696;
  • 45) 0.004 914 435 059 613 696 × 2 = 0 + 0.009 828 870 119 227 392;
  • 46) 0.009 828 870 119 227 392 × 2 = 0 + 0.019 657 740 238 454 784;
  • 47) 0.019 657 740 238 454 784 × 2 = 0 + 0.039 315 480 476 909 568;
  • 48) 0.039 315 480 476 909 568 × 2 = 0 + 0.078 630 960 953 819 136;
  • 49) 0.078 630 960 953 819 136 × 2 = 0 + 0.157 261 921 907 638 272;
  • 50) 0.157 261 921 907 638 272 × 2 = 0 + 0.314 523 843 815 276 544;
  • 51) 0.314 523 843 815 276 544 × 2 = 0 + 0.629 047 687 630 553 088;
  • 52) 0.629 047 687 630 553 088 × 2 = 1 + 0.258 095 375 261 106 176;
  • 53) 0.258 095 375 261 106 176 × 2 = 0 + 0.516 190 750 522 212 352;
  • 54) 0.516 190 750 522 212 352 × 2 = 1 + 0.032 381 501 044 424 704;

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:


0.000 000 000 742 147 956(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 956(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 01(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:


0.000 000 000 742 147 956(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 01(2) × 20 =

1.1001 1000 0000 0000 0000 101(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 101


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 0101 =

100 1100 0000 0000 0000 0101


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 0101


Decimal number -0.000 000 000 742 147 956 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0101

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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