-0.000 000 000 742 147 518 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 518(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
-0.000 000 000 742 147 518(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:

|-0.000 000 000 742 147 518| = 0.000 000 000 742 147 518


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: 0.000 000 000 742 147 518.

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) 0.000 000 000 742 147 518 × 2 = 0 + 0.000 000 001 484 295 036;
  • 2) 0.000 000 001 484 295 036 × 2 = 0 + 0.000 000 002 968 590 072;
  • 3) 0.000 000 002 968 590 072 × 2 = 0 + 0.000 000 005 937 180 144;
  • 4) 0.000 000 005 937 180 144 × 2 = 0 + 0.000 000 011 874 360 288;
  • 5) 0.000 000 011 874 360 288 × 2 = 0 + 0.000 000 023 748 720 576;
  • 6) 0.000 000 023 748 720 576 × 2 = 0 + 0.000 000 047 497 441 152;
  • 7) 0.000 000 047 497 441 152 × 2 = 0 + 0.000 000 094 994 882 304;
  • 8) 0.000 000 094 994 882 304 × 2 = 0 + 0.000 000 189 989 764 608;
  • 9) 0.000 000 189 989 764 608 × 2 = 0 + 0.000 000 379 979 529 216;
  • 10) 0.000 000 379 979 529 216 × 2 = 0 + 0.000 000 759 959 058 432;
  • 11) 0.000 000 759 959 058 432 × 2 = 0 + 0.000 001 519 918 116 864;
  • 12) 0.000 001 519 918 116 864 × 2 = 0 + 0.000 003 039 836 233 728;
  • 13) 0.000 003 039 836 233 728 × 2 = 0 + 0.000 006 079 672 467 456;
  • 14) 0.000 006 079 672 467 456 × 2 = 0 + 0.000 012 159 344 934 912;
  • 15) 0.000 012 159 344 934 912 × 2 = 0 + 0.000 024 318 689 869 824;
  • 16) 0.000 024 318 689 869 824 × 2 = 0 + 0.000 048 637 379 739 648;
  • 17) 0.000 048 637 379 739 648 × 2 = 0 + 0.000 097 274 759 479 296;
  • 18) 0.000 097 274 759 479 296 × 2 = 0 + 0.000 194 549 518 958 592;
  • 19) 0.000 194 549 518 958 592 × 2 = 0 + 0.000 389 099 037 917 184;
  • 20) 0.000 389 099 037 917 184 × 2 = 0 + 0.000 778 198 075 834 368;
  • 21) 0.000 778 198 075 834 368 × 2 = 0 + 0.001 556 396 151 668 736;
  • 22) 0.001 556 396 151 668 736 × 2 = 0 + 0.003 112 792 303 337 472;
  • 23) 0.003 112 792 303 337 472 × 2 = 0 + 0.006 225 584 606 674 944;
  • 24) 0.006 225 584 606 674 944 × 2 = 0 + 0.012 451 169 213 349 888;
  • 25) 0.012 451 169 213 349 888 × 2 = 0 + 0.024 902 338 426 699 776;
  • 26) 0.024 902 338 426 699 776 × 2 = 0 + 0.049 804 676 853 399 552;
  • 27) 0.049 804 676 853 399 552 × 2 = 0 + 0.099 609 353 706 799 104;
  • 28) 0.099 609 353 706 799 104 × 2 = 0 + 0.199 218 707 413 598 208;
  • 29) 0.199 218 707 413 598 208 × 2 = 0 + 0.398 437 414 827 196 416;
  • 30) 0.398 437 414 827 196 416 × 2 = 0 + 0.796 874 829 654 392 832;
  • 31) 0.796 874 829 654 392 832 × 2 = 1 + 0.593 749 659 308 785 664;
  • 32) 0.593 749 659 308 785 664 × 2 = 1 + 0.187 499 318 617 571 328;
  • 33) 0.187 499 318 617 571 328 × 2 = 0 + 0.374 998 637 235 142 656;
  • 34) 0.374 998 637 235 142 656 × 2 = 0 + 0.749 997 274 470 285 312;
  • 35) 0.749 997 274 470 285 312 × 2 = 1 + 0.499 994 548 940 570 624;
  • 36) 0.499 994 548 940 570 624 × 2 = 0 + 0.999 989 097 881 141 248;
  • 37) 0.999 989 097 881 141 248 × 2 = 1 + 0.999 978 195 762 282 496;
  • 38) 0.999 978 195 762 282 496 × 2 = 1 + 0.999 956 391 524 564 992;
  • 39) 0.999 956 391 524 564 992 × 2 = 1 + 0.999 912 783 049 129 984;
  • 40) 0.999 912 783 049 129 984 × 2 = 1 + 0.999 825 566 098 259 968;
  • 41) 0.999 825 566 098 259 968 × 2 = 1 + 0.999 651 132 196 519 936;
  • 42) 0.999 651 132 196 519 936 × 2 = 1 + 0.999 302 264 393 039 872;
  • 43) 0.999 302 264 393 039 872 × 2 = 1 + 0.998 604 528 786 079 744;
  • 44) 0.998 604 528 786 079 744 × 2 = 1 + 0.997 209 057 572 159 488;
  • 45) 0.997 209 057 572 159 488 × 2 = 1 + 0.994 418 115 144 318 976;
  • 46) 0.994 418 115 144 318 976 × 2 = 1 + 0.988 836 230 288 637 952;
  • 47) 0.988 836 230 288 637 952 × 2 = 1 + 0.977 672 460 577 275 904;
  • 48) 0.977 672 460 577 275 904 × 2 = 1 + 0.955 344 921 154 551 808;
  • 49) 0.955 344 921 154 551 808 × 2 = 1 + 0.910 689 842 309 103 616;
  • 50) 0.910 689 842 309 103 616 × 2 = 1 + 0.821 379 684 618 207 232;
  • 51) 0.821 379 684 618 207 232 × 2 = 1 + 0.642 759 369 236 414 464;
  • 52) 0.642 759 369 236 414 464 × 2 = 1 + 0.285 518 738 472 828 928;
  • 53) 0.285 518 738 472 828 928 × 2 = 0 + 0.571 037 476 945 657 856;
  • 54) 0.571 037 476 945 657 856 × 2 = 1 + 0.142 074 953 891 315 712;

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:


0.000 000 000 742 147 518(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 518(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 518(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) × 20 =

1.1001 0111 1111 1111 1111 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 0111 1111 1111 1111 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 1011 1111 1111 1111 1101 =

100 1011 1111 1111 1111 1101


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 1011 1111 1111 1111 1101


Decimal number -0.000 000 000 742 147 518 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1101

Share

» Convert decimal -0.000 000 000 742 147 418 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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