-0.000 000 000 742 147 817 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 817(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
-0.000 000 000 742 147 817(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:

|-0.000 000 000 742 147 817| = 0.000 000 000 742 147 817


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: 0.000 000 000 742 147 817.

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) 0.000 000 000 742 147 817 × 2 = 0 + 0.000 000 001 484 295 634;
  • 2) 0.000 000 001 484 295 634 × 2 = 0 + 0.000 000 002 968 591 268;
  • 3) 0.000 000 002 968 591 268 × 2 = 0 + 0.000 000 005 937 182 536;
  • 4) 0.000 000 005 937 182 536 × 2 = 0 + 0.000 000 011 874 365 072;
  • 5) 0.000 000 011 874 365 072 × 2 = 0 + 0.000 000 023 748 730 144;
  • 6) 0.000 000 023 748 730 144 × 2 = 0 + 0.000 000 047 497 460 288;
  • 7) 0.000 000 047 497 460 288 × 2 = 0 + 0.000 000 094 994 920 576;
  • 8) 0.000 000 094 994 920 576 × 2 = 0 + 0.000 000 189 989 841 152;
  • 9) 0.000 000 189 989 841 152 × 2 = 0 + 0.000 000 379 979 682 304;
  • 10) 0.000 000 379 979 682 304 × 2 = 0 + 0.000 000 759 959 364 608;
  • 11) 0.000 000 759 959 364 608 × 2 = 0 + 0.000 001 519 918 729 216;
  • 12) 0.000 001 519 918 729 216 × 2 = 0 + 0.000 003 039 837 458 432;
  • 13) 0.000 003 039 837 458 432 × 2 = 0 + 0.000 006 079 674 916 864;
  • 14) 0.000 006 079 674 916 864 × 2 = 0 + 0.000 012 159 349 833 728;
  • 15) 0.000 012 159 349 833 728 × 2 = 0 + 0.000 024 318 699 667 456;
  • 16) 0.000 024 318 699 667 456 × 2 = 0 + 0.000 048 637 399 334 912;
  • 17) 0.000 048 637 399 334 912 × 2 = 0 + 0.000 097 274 798 669 824;
  • 18) 0.000 097 274 798 669 824 × 2 = 0 + 0.000 194 549 597 339 648;
  • 19) 0.000 194 549 597 339 648 × 2 = 0 + 0.000 389 099 194 679 296;
  • 20) 0.000 389 099 194 679 296 × 2 = 0 + 0.000 778 198 389 358 592;
  • 21) 0.000 778 198 389 358 592 × 2 = 0 + 0.001 556 396 778 717 184;
  • 22) 0.001 556 396 778 717 184 × 2 = 0 + 0.003 112 793 557 434 368;
  • 23) 0.003 112 793 557 434 368 × 2 = 0 + 0.006 225 587 114 868 736;
  • 24) 0.006 225 587 114 868 736 × 2 = 0 + 0.012 451 174 229 737 472;
  • 25) 0.012 451 174 229 737 472 × 2 = 0 + 0.024 902 348 459 474 944;
  • 26) 0.024 902 348 459 474 944 × 2 = 0 + 0.049 804 696 918 949 888;
  • 27) 0.049 804 696 918 949 888 × 2 = 0 + 0.099 609 393 837 899 776;
  • 28) 0.099 609 393 837 899 776 × 2 = 0 + 0.199 218 787 675 799 552;
  • 29) 0.199 218 787 675 799 552 × 2 = 0 + 0.398 437 575 351 599 104;
  • 30) 0.398 437 575 351 599 104 × 2 = 0 + 0.796 875 150 703 198 208;
  • 31) 0.796 875 150 703 198 208 × 2 = 1 + 0.593 750 301 406 396 416;
  • 32) 0.593 750 301 406 396 416 × 2 = 1 + 0.187 500 602 812 792 832;
  • 33) 0.187 500 602 812 792 832 × 2 = 0 + 0.375 001 205 625 585 664;
  • 34) 0.375 001 205 625 585 664 × 2 = 0 + 0.750 002 411 251 171 328;
  • 35) 0.750 002 411 251 171 328 × 2 = 1 + 0.500 004 822 502 342 656;
  • 36) 0.500 004 822 502 342 656 × 2 = 1 + 0.000 009 645 004 685 312;
  • 37) 0.000 009 645 004 685 312 × 2 = 0 + 0.000 019 290 009 370 624;
  • 38) 0.000 019 290 009 370 624 × 2 = 0 + 0.000 038 580 018 741 248;
  • 39) 0.000 038 580 018 741 248 × 2 = 0 + 0.000 077 160 037 482 496;
  • 40) 0.000 077 160 037 482 496 × 2 = 0 + 0.000 154 320 074 964 992;
  • 41) 0.000 154 320 074 964 992 × 2 = 0 + 0.000 308 640 149 929 984;
  • 42) 0.000 308 640 149 929 984 × 2 = 0 + 0.000 617 280 299 859 968;
  • 43) 0.000 617 280 299 859 968 × 2 = 0 + 0.001 234 560 599 719 936;
  • 44) 0.001 234 560 599 719 936 × 2 = 0 + 0.002 469 121 199 439 872;
  • 45) 0.002 469 121 199 439 872 × 2 = 0 + 0.004 938 242 398 879 744;
  • 46) 0.004 938 242 398 879 744 × 2 = 0 + 0.009 876 484 797 759 488;
  • 47) 0.009 876 484 797 759 488 × 2 = 0 + 0.019 752 969 595 518 976;
  • 48) 0.019 752 969 595 518 976 × 2 = 0 + 0.039 505 939 191 037 952;
  • 49) 0.039 505 939 191 037 952 × 2 = 0 + 0.079 011 878 382 075 904;
  • 50) 0.079 011 878 382 075 904 × 2 = 0 + 0.158 023 756 764 151 808;
  • 51) 0.158 023 756 764 151 808 × 2 = 0 + 0.316 047 513 528 303 616;
  • 52) 0.316 047 513 528 303 616 × 2 = 0 + 0.632 095 027 056 607 232;
  • 53) 0.632 095 027 056 607 232 × 2 = 1 + 0.264 190 054 113 214 464;
  • 54) 0.264 190 054 113 214 464 × 2 = 0 + 0.528 380 108 226 428 928;

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:


0.000 000 000 742 147 817(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 817(10) =

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


0.000 000 000 742 147 817(10) =

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

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

1.1001 1000 0000 0000 0000 010(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 010


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 0010 =

100 1100 0000 0000 0000 0010


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 0010


Decimal number -0.000 000 000 742 147 817 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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