-0.000 000 000 742 147 966 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 966(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
-0.000 000 000 742 147 966(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:

|-0.000 000 000 742 147 966| = 0.000 000 000 742 147 966


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: 0.000 000 000 742 147 966.

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) 0.000 000 000 742 147 966 × 2 = 0 + 0.000 000 001 484 295 932;
  • 2) 0.000 000 001 484 295 932 × 2 = 0 + 0.000 000 002 968 591 864;
  • 3) 0.000 000 002 968 591 864 × 2 = 0 + 0.000 000 005 937 183 728;
  • 4) 0.000 000 005 937 183 728 × 2 = 0 + 0.000 000 011 874 367 456;
  • 5) 0.000 000 011 874 367 456 × 2 = 0 + 0.000 000 023 748 734 912;
  • 6) 0.000 000 023 748 734 912 × 2 = 0 + 0.000 000 047 497 469 824;
  • 7) 0.000 000 047 497 469 824 × 2 = 0 + 0.000 000 094 994 939 648;
  • 8) 0.000 000 094 994 939 648 × 2 = 0 + 0.000 000 189 989 879 296;
  • 9) 0.000 000 189 989 879 296 × 2 = 0 + 0.000 000 379 979 758 592;
  • 10) 0.000 000 379 979 758 592 × 2 = 0 + 0.000 000 759 959 517 184;
  • 11) 0.000 000 759 959 517 184 × 2 = 0 + 0.000 001 519 919 034 368;
  • 12) 0.000 001 519 919 034 368 × 2 = 0 + 0.000 003 039 838 068 736;
  • 13) 0.000 003 039 838 068 736 × 2 = 0 + 0.000 006 079 676 137 472;
  • 14) 0.000 006 079 676 137 472 × 2 = 0 + 0.000 012 159 352 274 944;
  • 15) 0.000 012 159 352 274 944 × 2 = 0 + 0.000 024 318 704 549 888;
  • 16) 0.000 024 318 704 549 888 × 2 = 0 + 0.000 048 637 409 099 776;
  • 17) 0.000 048 637 409 099 776 × 2 = 0 + 0.000 097 274 818 199 552;
  • 18) 0.000 097 274 818 199 552 × 2 = 0 + 0.000 194 549 636 399 104;
  • 19) 0.000 194 549 636 399 104 × 2 = 0 + 0.000 389 099 272 798 208;
  • 20) 0.000 389 099 272 798 208 × 2 = 0 + 0.000 778 198 545 596 416;
  • 21) 0.000 778 198 545 596 416 × 2 = 0 + 0.001 556 397 091 192 832;
  • 22) 0.001 556 397 091 192 832 × 2 = 0 + 0.003 112 794 182 385 664;
  • 23) 0.003 112 794 182 385 664 × 2 = 0 + 0.006 225 588 364 771 328;
  • 24) 0.006 225 588 364 771 328 × 2 = 0 + 0.012 451 176 729 542 656;
  • 25) 0.012 451 176 729 542 656 × 2 = 0 + 0.024 902 353 459 085 312;
  • 26) 0.024 902 353 459 085 312 × 2 = 0 + 0.049 804 706 918 170 624;
  • 27) 0.049 804 706 918 170 624 × 2 = 0 + 0.099 609 413 836 341 248;
  • 28) 0.099 609 413 836 341 248 × 2 = 0 + 0.199 218 827 672 682 496;
  • 29) 0.199 218 827 672 682 496 × 2 = 0 + 0.398 437 655 345 364 992;
  • 30) 0.398 437 655 345 364 992 × 2 = 0 + 0.796 875 310 690 729 984;
  • 31) 0.796 875 310 690 729 984 × 2 = 1 + 0.593 750 621 381 459 968;
  • 32) 0.593 750 621 381 459 968 × 2 = 1 + 0.187 501 242 762 919 936;
  • 33) 0.187 501 242 762 919 936 × 2 = 0 + 0.375 002 485 525 839 872;
  • 34) 0.375 002 485 525 839 872 × 2 = 0 + 0.750 004 971 051 679 744;
  • 35) 0.750 004 971 051 679 744 × 2 = 1 + 0.500 009 942 103 359 488;
  • 36) 0.500 009 942 103 359 488 × 2 = 1 + 0.000 019 884 206 718 976;
  • 37) 0.000 019 884 206 718 976 × 2 = 0 + 0.000 039 768 413 437 952;
  • 38) 0.000 039 768 413 437 952 × 2 = 0 + 0.000 079 536 826 875 904;
  • 39) 0.000 079 536 826 875 904 × 2 = 0 + 0.000 159 073 653 751 808;
  • 40) 0.000 159 073 653 751 808 × 2 = 0 + 0.000 318 147 307 503 616;
  • 41) 0.000 318 147 307 503 616 × 2 = 0 + 0.000 636 294 615 007 232;
  • 42) 0.000 636 294 615 007 232 × 2 = 0 + 0.001 272 589 230 014 464;
  • 43) 0.001 272 589 230 014 464 × 2 = 0 + 0.002 545 178 460 028 928;
  • 44) 0.002 545 178 460 028 928 × 2 = 0 + 0.005 090 356 920 057 856;
  • 45) 0.005 090 356 920 057 856 × 2 = 0 + 0.010 180 713 840 115 712;
  • 46) 0.010 180 713 840 115 712 × 2 = 0 + 0.020 361 427 680 231 424;
  • 47) 0.020 361 427 680 231 424 × 2 = 0 + 0.040 722 855 360 462 848;
  • 48) 0.040 722 855 360 462 848 × 2 = 0 + 0.081 445 710 720 925 696;
  • 49) 0.081 445 710 720 925 696 × 2 = 0 + 0.162 891 421 441 851 392;
  • 50) 0.162 891 421 441 851 392 × 2 = 0 + 0.325 782 842 883 702 784;
  • 51) 0.325 782 842 883 702 784 × 2 = 0 + 0.651 565 685 767 405 568;
  • 52) 0.651 565 685 767 405 568 × 2 = 1 + 0.303 131 371 534 811 136;
  • 53) 0.303 131 371 534 811 136 × 2 = 0 + 0.606 262 743 069 622 272;
  • 54) 0.606 262 743 069 622 272 × 2 = 1 + 0.212 525 486 139 244 544;

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:


0.000 000 000 742 147 966(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 966(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:


0.000 000 000 742 147 966(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 -0.000 000 000 742 147 966 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