-0.000 000 000 742 147 743 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 743(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
-0.000 000 000 742 147 743(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:

|-0.000 000 000 742 147 743| = 0.000 000 000 742 147 743


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: 0.000 000 000 742 147 743.

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) 0.000 000 000 742 147 743 × 2 = 0 + 0.000 000 001 484 295 486;
  • 2) 0.000 000 001 484 295 486 × 2 = 0 + 0.000 000 002 968 590 972;
  • 3) 0.000 000 002 968 590 972 × 2 = 0 + 0.000 000 005 937 181 944;
  • 4) 0.000 000 005 937 181 944 × 2 = 0 + 0.000 000 011 874 363 888;
  • 5) 0.000 000 011 874 363 888 × 2 = 0 + 0.000 000 023 748 727 776;
  • 6) 0.000 000 023 748 727 776 × 2 = 0 + 0.000 000 047 497 455 552;
  • 7) 0.000 000 047 497 455 552 × 2 = 0 + 0.000 000 094 994 911 104;
  • 8) 0.000 000 094 994 911 104 × 2 = 0 + 0.000 000 189 989 822 208;
  • 9) 0.000 000 189 989 822 208 × 2 = 0 + 0.000 000 379 979 644 416;
  • 10) 0.000 000 379 979 644 416 × 2 = 0 + 0.000 000 759 959 288 832;
  • 11) 0.000 000 759 959 288 832 × 2 = 0 + 0.000 001 519 918 577 664;
  • 12) 0.000 001 519 918 577 664 × 2 = 0 + 0.000 003 039 837 155 328;
  • 13) 0.000 003 039 837 155 328 × 2 = 0 + 0.000 006 079 674 310 656;
  • 14) 0.000 006 079 674 310 656 × 2 = 0 + 0.000 012 159 348 621 312;
  • 15) 0.000 012 159 348 621 312 × 2 = 0 + 0.000 024 318 697 242 624;
  • 16) 0.000 024 318 697 242 624 × 2 = 0 + 0.000 048 637 394 485 248;
  • 17) 0.000 048 637 394 485 248 × 2 = 0 + 0.000 097 274 788 970 496;
  • 18) 0.000 097 274 788 970 496 × 2 = 0 + 0.000 194 549 577 940 992;
  • 19) 0.000 194 549 577 940 992 × 2 = 0 + 0.000 389 099 155 881 984;
  • 20) 0.000 389 099 155 881 984 × 2 = 0 + 0.000 778 198 311 763 968;
  • 21) 0.000 778 198 311 763 968 × 2 = 0 + 0.001 556 396 623 527 936;
  • 22) 0.001 556 396 623 527 936 × 2 = 0 + 0.003 112 793 247 055 872;
  • 23) 0.003 112 793 247 055 872 × 2 = 0 + 0.006 225 586 494 111 744;
  • 24) 0.006 225 586 494 111 744 × 2 = 0 + 0.012 451 172 988 223 488;
  • 25) 0.012 451 172 988 223 488 × 2 = 0 + 0.024 902 345 976 446 976;
  • 26) 0.024 902 345 976 446 976 × 2 = 0 + 0.049 804 691 952 893 952;
  • 27) 0.049 804 691 952 893 952 × 2 = 0 + 0.099 609 383 905 787 904;
  • 28) 0.099 609 383 905 787 904 × 2 = 0 + 0.199 218 767 811 575 808;
  • 29) 0.199 218 767 811 575 808 × 2 = 0 + 0.398 437 535 623 151 616;
  • 30) 0.398 437 535 623 151 616 × 2 = 0 + 0.796 875 071 246 303 232;
  • 31) 0.796 875 071 246 303 232 × 2 = 1 + 0.593 750 142 492 606 464;
  • 32) 0.593 750 142 492 606 464 × 2 = 1 + 0.187 500 284 985 212 928;
  • 33) 0.187 500 284 985 212 928 × 2 = 0 + 0.375 000 569 970 425 856;
  • 34) 0.375 000 569 970 425 856 × 2 = 0 + 0.750 001 139 940 851 712;
  • 35) 0.750 001 139 940 851 712 × 2 = 1 + 0.500 002 279 881 703 424;
  • 36) 0.500 002 279 881 703 424 × 2 = 1 + 0.000 004 559 763 406 848;
  • 37) 0.000 004 559 763 406 848 × 2 = 0 + 0.000 009 119 526 813 696;
  • 38) 0.000 009 119 526 813 696 × 2 = 0 + 0.000 018 239 053 627 392;
  • 39) 0.000 018 239 053 627 392 × 2 = 0 + 0.000 036 478 107 254 784;
  • 40) 0.000 036 478 107 254 784 × 2 = 0 + 0.000 072 956 214 509 568;
  • 41) 0.000 072 956 214 509 568 × 2 = 0 + 0.000 145 912 429 019 136;
  • 42) 0.000 145 912 429 019 136 × 2 = 0 + 0.000 291 824 858 038 272;
  • 43) 0.000 291 824 858 038 272 × 2 = 0 + 0.000 583 649 716 076 544;
  • 44) 0.000 583 649 716 076 544 × 2 = 0 + 0.001 167 299 432 153 088;
  • 45) 0.001 167 299 432 153 088 × 2 = 0 + 0.002 334 598 864 306 176;
  • 46) 0.002 334 598 864 306 176 × 2 = 0 + 0.004 669 197 728 612 352;
  • 47) 0.004 669 197 728 612 352 × 2 = 0 + 0.009 338 395 457 224 704;
  • 48) 0.009 338 395 457 224 704 × 2 = 0 + 0.018 676 790 914 449 408;
  • 49) 0.018 676 790 914 449 408 × 2 = 0 + 0.037 353 581 828 898 816;
  • 50) 0.037 353 581 828 898 816 × 2 = 0 + 0.074 707 163 657 797 632;
  • 51) 0.074 707 163 657 797 632 × 2 = 0 + 0.149 414 327 315 595 264;
  • 52) 0.149 414 327 315 595 264 × 2 = 0 + 0.298 828 654 631 190 528;
  • 53) 0.298 828 654 631 190 528 × 2 = 0 + 0.597 657 309 262 381 056;
  • 54) 0.597 657 309 262 381 056 × 2 = 1 + 0.195 314 618 524 762 112;

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:


0.000 000 000 742 147 743(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 743(10) =

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


0.000 000 000 742 147 743(10) =

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

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

1.1001 1000 0000 0000 0000 001(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 001


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 0001 =

100 1100 0000 0000 0000 0001


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 0001


Decimal number -0.000 000 000 742 147 743 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0001

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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