-0.000 000 000 742 151 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 151(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
-0.000 000 000 742 151(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:

|-0.000 000 000 742 151| = 0.000 000 000 742 151


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: 0.000 000 000 742 151.

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) 0.000 000 000 742 151 × 2 = 0 + 0.000 000 001 484 302;
  • 2) 0.000 000 001 484 302 × 2 = 0 + 0.000 000 002 968 604;
  • 3) 0.000 000 002 968 604 × 2 = 0 + 0.000 000 005 937 208;
  • 4) 0.000 000 005 937 208 × 2 = 0 + 0.000 000 011 874 416;
  • 5) 0.000 000 011 874 416 × 2 = 0 + 0.000 000 023 748 832;
  • 6) 0.000 000 023 748 832 × 2 = 0 + 0.000 000 047 497 664;
  • 7) 0.000 000 047 497 664 × 2 = 0 + 0.000 000 094 995 328;
  • 8) 0.000 000 094 995 328 × 2 = 0 + 0.000 000 189 990 656;
  • 9) 0.000 000 189 990 656 × 2 = 0 + 0.000 000 379 981 312;
  • 10) 0.000 000 379 981 312 × 2 = 0 + 0.000 000 759 962 624;
  • 11) 0.000 000 759 962 624 × 2 = 0 + 0.000 001 519 925 248;
  • 12) 0.000 001 519 925 248 × 2 = 0 + 0.000 003 039 850 496;
  • 13) 0.000 003 039 850 496 × 2 = 0 + 0.000 006 079 700 992;
  • 14) 0.000 006 079 700 992 × 2 = 0 + 0.000 012 159 401 984;
  • 15) 0.000 012 159 401 984 × 2 = 0 + 0.000 024 318 803 968;
  • 16) 0.000 024 318 803 968 × 2 = 0 + 0.000 048 637 607 936;
  • 17) 0.000 048 637 607 936 × 2 = 0 + 0.000 097 275 215 872;
  • 18) 0.000 097 275 215 872 × 2 = 0 + 0.000 194 550 431 744;
  • 19) 0.000 194 550 431 744 × 2 = 0 + 0.000 389 100 863 488;
  • 20) 0.000 389 100 863 488 × 2 = 0 + 0.000 778 201 726 976;
  • 21) 0.000 778 201 726 976 × 2 = 0 + 0.001 556 403 453 952;
  • 22) 0.001 556 403 453 952 × 2 = 0 + 0.003 112 806 907 904;
  • 23) 0.003 112 806 907 904 × 2 = 0 + 0.006 225 613 815 808;
  • 24) 0.006 225 613 815 808 × 2 = 0 + 0.012 451 227 631 616;
  • 25) 0.012 451 227 631 616 × 2 = 0 + 0.024 902 455 263 232;
  • 26) 0.024 902 455 263 232 × 2 = 0 + 0.049 804 910 526 464;
  • 27) 0.049 804 910 526 464 × 2 = 0 + 0.099 609 821 052 928;
  • 28) 0.099 609 821 052 928 × 2 = 0 + 0.199 219 642 105 856;
  • 29) 0.199 219 642 105 856 × 2 = 0 + 0.398 439 284 211 712;
  • 30) 0.398 439 284 211 712 × 2 = 0 + 0.796 878 568 423 424;
  • 31) 0.796 878 568 423 424 × 2 = 1 + 0.593 757 136 846 848;
  • 32) 0.593 757 136 846 848 × 2 = 1 + 0.187 514 273 693 696;
  • 33) 0.187 514 273 693 696 × 2 = 0 + 0.375 028 547 387 392;
  • 34) 0.375 028 547 387 392 × 2 = 0 + 0.750 057 094 774 784;
  • 35) 0.750 057 094 774 784 × 2 = 1 + 0.500 114 189 549 568;
  • 36) 0.500 114 189 549 568 × 2 = 1 + 0.000 228 379 099 136;
  • 37) 0.000 228 379 099 136 × 2 = 0 + 0.000 456 758 198 272;
  • 38) 0.000 456 758 198 272 × 2 = 0 + 0.000 913 516 396 544;
  • 39) 0.000 913 516 396 544 × 2 = 0 + 0.001 827 032 793 088;
  • 40) 0.001 827 032 793 088 × 2 = 0 + 0.003 654 065 586 176;
  • 41) 0.003 654 065 586 176 × 2 = 0 + 0.007 308 131 172 352;
  • 42) 0.007 308 131 172 352 × 2 = 0 + 0.014 616 262 344 704;
  • 43) 0.014 616 262 344 704 × 2 = 0 + 0.029 232 524 689 408;
  • 44) 0.029 232 524 689 408 × 2 = 0 + 0.058 465 049 378 816;
  • 45) 0.058 465 049 378 816 × 2 = 0 + 0.116 930 098 757 632;
  • 46) 0.116 930 098 757 632 × 2 = 0 + 0.233 860 197 515 264;
  • 47) 0.233 860 197 515 264 × 2 = 0 + 0.467 720 395 030 528;
  • 48) 0.467 720 395 030 528 × 2 = 0 + 0.935 440 790 061 056;
  • 49) 0.935 440 790 061 056 × 2 = 1 + 0.870 881 580 122 112;
  • 50) 0.870 881 580 122 112 × 2 = 1 + 0.741 763 160 244 224;
  • 51) 0.741 763 160 244 224 × 2 = 1 + 0.483 526 320 488 448;
  • 52) 0.483 526 320 488 448 × 2 = 0 + 0.967 052 640 976 896;
  • 53) 0.967 052 640 976 896 × 2 = 1 + 0.934 105 281 953 792;
  • 54) 0.934 105 281 953 792 × 2 = 1 + 0.868 210 563 907 584;

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:


0.000 000 000 742 151(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 151(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1110 11(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:


0.000 000 000 742 151(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1110 11(2) × 20 =

1.1001 1000 0000 0000 0111 011(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 0111 011


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 0011 1011 =

100 1100 0000 0000 0011 1011


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 0011 1011


Decimal number -0.000 000 000 742 151 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0011 1011

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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