-0.000 000 000 654 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 654(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
-0.000 000 000 654(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:

|-0.000 000 000 654| = 0.000 000 000 654


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: 0.000 000 000 654.

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) 0.000 000 000 654 × 2 = 0 + 0.000 000 001 308;
  • 2) 0.000 000 001 308 × 2 = 0 + 0.000 000 002 616;
  • 3) 0.000 000 002 616 × 2 = 0 + 0.000 000 005 232;
  • 4) 0.000 000 005 232 × 2 = 0 + 0.000 000 010 464;
  • 5) 0.000 000 010 464 × 2 = 0 + 0.000 000 020 928;
  • 6) 0.000 000 020 928 × 2 = 0 + 0.000 000 041 856;
  • 7) 0.000 000 041 856 × 2 = 0 + 0.000 000 083 712;
  • 8) 0.000 000 083 712 × 2 = 0 + 0.000 000 167 424;
  • 9) 0.000 000 167 424 × 2 = 0 + 0.000 000 334 848;
  • 10) 0.000 000 334 848 × 2 = 0 + 0.000 000 669 696;
  • 11) 0.000 000 669 696 × 2 = 0 + 0.000 001 339 392;
  • 12) 0.000 001 339 392 × 2 = 0 + 0.000 002 678 784;
  • 13) 0.000 002 678 784 × 2 = 0 + 0.000 005 357 568;
  • 14) 0.000 005 357 568 × 2 = 0 + 0.000 010 715 136;
  • 15) 0.000 010 715 136 × 2 = 0 + 0.000 021 430 272;
  • 16) 0.000 021 430 272 × 2 = 0 + 0.000 042 860 544;
  • 17) 0.000 042 860 544 × 2 = 0 + 0.000 085 721 088;
  • 18) 0.000 085 721 088 × 2 = 0 + 0.000 171 442 176;
  • 19) 0.000 171 442 176 × 2 = 0 + 0.000 342 884 352;
  • 20) 0.000 342 884 352 × 2 = 0 + 0.000 685 768 704;
  • 21) 0.000 685 768 704 × 2 = 0 + 0.001 371 537 408;
  • 22) 0.001 371 537 408 × 2 = 0 + 0.002 743 074 816;
  • 23) 0.002 743 074 816 × 2 = 0 + 0.005 486 149 632;
  • 24) 0.005 486 149 632 × 2 = 0 + 0.010 972 299 264;
  • 25) 0.010 972 299 264 × 2 = 0 + 0.021 944 598 528;
  • 26) 0.021 944 598 528 × 2 = 0 + 0.043 889 197 056;
  • 27) 0.043 889 197 056 × 2 = 0 + 0.087 778 394 112;
  • 28) 0.087 778 394 112 × 2 = 0 + 0.175 556 788 224;
  • 29) 0.175 556 788 224 × 2 = 0 + 0.351 113 576 448;
  • 30) 0.351 113 576 448 × 2 = 0 + 0.702 227 152 896;
  • 31) 0.702 227 152 896 × 2 = 1 + 0.404 454 305 792;
  • 32) 0.404 454 305 792 × 2 = 0 + 0.808 908 611 584;
  • 33) 0.808 908 611 584 × 2 = 1 + 0.617 817 223 168;
  • 34) 0.617 817 223 168 × 2 = 1 + 0.235 634 446 336;
  • 35) 0.235 634 446 336 × 2 = 0 + 0.471 268 892 672;
  • 36) 0.471 268 892 672 × 2 = 0 + 0.942 537 785 344;
  • 37) 0.942 537 785 344 × 2 = 1 + 0.885 075 570 688;
  • 38) 0.885 075 570 688 × 2 = 1 + 0.770 151 141 376;
  • 39) 0.770 151 141 376 × 2 = 1 + 0.540 302 282 752;
  • 40) 0.540 302 282 752 × 2 = 1 + 0.080 604 565 504;
  • 41) 0.080 604 565 504 × 2 = 0 + 0.161 209 131 008;
  • 42) 0.161 209 131 008 × 2 = 0 + 0.322 418 262 016;
  • 43) 0.322 418 262 016 × 2 = 0 + 0.644 836 524 032;
  • 44) 0.644 836 524 032 × 2 = 1 + 0.289 673 048 064;
  • 45) 0.289 673 048 064 × 2 = 0 + 0.579 346 096 128;
  • 46) 0.579 346 096 128 × 2 = 1 + 0.158 692 192 256;
  • 47) 0.158 692 192 256 × 2 = 0 + 0.317 384 384 512;
  • 48) 0.317 384 384 512 × 2 = 0 + 0.634 768 769 024;
  • 49) 0.634 768 769 024 × 2 = 1 + 0.269 537 538 048;
  • 50) 0.269 537 538 048 × 2 = 0 + 0.539 075 076 096;
  • 51) 0.539 075 076 096 × 2 = 1 + 0.078 150 152 192;
  • 52) 0.078 150 152 192 × 2 = 0 + 0.156 300 304 384;
  • 53) 0.156 300 304 384 × 2 = 0 + 0.312 600 608 768;
  • 54) 0.312 600 608 768 × 2 = 0 + 0.625 201 217 536;

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:


0.000 000 000 654(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1100 1111 0001 0100 1010 00(2)

6. Positive number before normalization:

0.000 000 000 654(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1100 1111 0001 0100 1010 00(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:


0.000 000 000 654(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1100 1111 0001 0100 1010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1100 1111 0001 0100 1010 00(2) × 20 =

1.0110 0111 1000 1010 0101 000(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):
1.0110 0111 1000 1010 0101 000


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. 011 0011 1100 0101 0010 1000 =

011 0011 1100 0101 0010 1000


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) =
011 0011 1100 0101 0010 1000


Decimal number -0.000 000 000 654 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 011 0011 1100 0101 0010 1000

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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